7. ARITHMETIC & NUMBER THEORETIC RECREATIONS
7.A. FIBONACCI NUMBERS
We use the standard form: F_{0} = 0, F_{1} = 1, F_{n+1} = F_{n} + F_{n1}, with the auxiliary Lucas numbers being given by: L_{0} = 2, L_{1} = 1, L_{n+1} = L_{n} + L_{n1}.
Parmanand Singh. The socalled Fibonacci numbers in ancient and medieval India. HM 12 (1985) 229244. In early Indian poetry, letters had weights of 1 or 2 and meters were classified both by the number of letters and by the weight. Classifying by weight gives the number of sequences of 1s and 2s which add to the weight n and this is F_{n+1}.
Pińgala [NOTE: ń denotes n with an overdot.] (c450) studied prosody and gives cryptic rules which have been interpreted as methods for generating the next set of sequences, either classified by number of letters or by weight and several later writers have given similar rules. The generation implies F_{n+1} = F_{n} + F_{n1}. Virahāńka [NOTE: ń denotes n with an overdot.] (c7C) is slightly more explicit. Gopāla (c1134) gives a commentary on Virahāńka [NOTE: ń denotes n with an overdot.] which explicitly gives the numbers as 3, 5, 8, 13, 21. Hemacandra (c1150) states "Sum of the last and last but one numbers ... is ... next." This is repeated by later authors.
The Prākŗta [NOTE: ŗ denotes r with an underdot] Paińgala [NOTE: ń denotes n with an overdot.] (c1315) gives rules for finding the kth sequence of weight n and for finding the position of a particular sequence in the list of sequences of weight n and the positions of those sequences having a given number of 2s (and hence a given number of letters). It also gives the relation F_{n+1} = Σ_{i} BC(ni,i).
Narayana Pandita (= Nārāyaņa Paņdita’s [NOTE: ņ denotes n with an overdot and the d should have an underdot.]) Gaņita[NOTE: ņ denotes n with an underdot.] Kaumudī (1356) studies additive sequences in chap. 13, where each term is the sum of the last q terms. He gives rules which are equivalent to finding the coefficients of (1 + x + ... + x^{q1})^{p} and relates to ordered partitions using 1, 2, ..., q.
Narayana Pandita (= Nārāyaņa Paņdita [NOTE: ņ denotes n with an overdot and the d should have an underdot.]). Gaņita[NOTE: ņ denotes n with an underdot.] Kaumudī (1356). Part I, (p. 126 of the Sanskrit ed. by P. Dvivedi, Indian Press, Benares, 1942), ??NYS  quoted by Kripa Shankar Shukla in the Introduction to his edition of: Narayana Pandita (= Nārāyaņa Paņdita); Bījagaņitāvatamsa [NOTE: the ņ denotes an n with an under dot and there should be a dot over the m.]; Part I; Akhila Bharatiya Sanskrit Parishad, Lucknow, 1970, p. iv. "A cow gives birth to a calf every year. The calves become young and themselves begin giving birth to calves when they are three years old. Tell me, O learned man, the number of progeny produced during twenty years by one cow."
WESTERN HISTORIES
H. S. M. Coxeter. The golden section, phyllotaxis, and Wythoff's game. SM 19 (1953) 135 143. Sketches history and interconnections.
H. S. M. Coxeter. Introduction to Geometry. Wiley, 1961. Chap. 11: The golden section and phyllotaxis, pp. 160172. Extends his 1953 material.
Maxey Brooke. Fibonacci numbers: Their history through 1900. Fibonacci Quarterly 2:2 (1964) 149 153. Brief sketch, with lots of typographical errors. Doesn't know of Bernoulli's work.
Leonard Curchin & Roger HerzFischler. De quand date le premier rapprochement entre la suite de Fibonacci et la division en extrême et moyenne raison? Centaurus 28 (1985) 129138. Discusses the history of the result that the ratio F_{n+1}/F_{n} approaches φ. Pacioli and Kepler, described below, seem to be the first to find this.
Roger Herz Fischler. Letter to the Editor. Fibonacci Quarterly 24:4 (1986) 382.
Roger HerzFischler. A Mathematical History of Division in Extreme and Mean Ratio. Wilfrid Laurier University Press, Waterloo, Ontario, 1987. Retitled: A Mathematical History of the Golden Number, with new preface and corrections and additions, Dover, 1998. Pp. 157162 discuss early work relating the Fibonacci sequence to division in extreme and mean ratio. 15 pages of references.
Georg Markovsky. Misconceptions about the Golden Ratio. CMJ 23 (1992) 219. This surveys many of the common misconceptions  e.g. that  appears in the Great Pyramid, the Parthenon, Renaissance paintings and/or the human body and that the Golden Rectangle is the most pleasing  with 59 references. He also discusses the origin of the term 'golden section', sketching the results given in HerzFischler's book.
Thomas Koshy. Fibonacci and Lucas Numbers with Applications. WileyInterscience, Wiley, 2001. Claims to be 'the first attempt to compile a definitive history and authoritative analysis' of the Fibonacci numbers, but the history is generally secondhand and marred with a substantial number of errors, The mathematical work is extensive, covering many topics not organised before, and is better done, but there are more errors than one would like.
Ron Knott has a huge website on Fibonacci numbers and their applications, with material on many related topics, e.g. continued fractions, π, etc. with some history. www.ee.surrey.ac.uk/personal/r.knott/fibonacci/fibnat.html .
Fibonacci. 1202. Pp. 283 284 (S: 404405): Quot paria coniculorum in uno anno ex uno pario germinentur [How many pairs of rabbits are created by one pair in one year]. Rabbit problem  the pair propagate in the first month so there are F_{n+2} pairs at the end of the nth month. (English translation in: Struik, Source Book, pp. 2 3.) I have colour slides of this from L.IV.20 & 21 and Conventi Soppresi, C. I. 2616. This is on ff. 130r130v of L.IV.20, f. 225v of L.IV.21, f. 124r of CS.C.I.2616.
Unknown early 16C annotator. Marginal note to II.11 in Luca Pacioli's copy of his 1509 edition of Euclid. Reproduced and discussed in Curchin & HerzFischler and discussed in HerzFischler's book, pp. 157158. II.11 involves division in mean and extreme ratio. Uses 89, 144, 233 and that 144^{2} = 89 * 233 + 1. Also refers to 5, 8, 13.
Gori. Libro di arimetricha. 1571. F. 73r (p.81). Rabbit problem as in Fibonacci.
J. Kepler. Letter of Oct 1597 to Mästlin. ??NYS  described in HerzFischler's book, p. 158. This gives a construction for division in extreme and mean ration. On the original, Mästlin has added his numerical calculations, getting 1/φ = .6180340, which HerzFischler believes to be the first time anyone actually calculated this number.
J. Kepler. Letter of 12 May 1608 to Joachim Tanckius. ??NYS  described in HerzFischler (1986), Curchin & HerzFischler and HerzFishler's book, pp. 160161. Shows that he knows that the ratio F_{n+1}/F_{n} approaches φ and that F_{n}^{2} + (1)^{n} = F_{n1}F_{n+1}.
J. Kepler. The Six Cornered Snowflake. Op. cit. in 6.AT.3. 1611. P. 12 (20 21). Mentions golden section in polyhedra and that the ratio F_{n+1}/F_{n} approaches φ. See HerzFischler's book, p. 161.
Albert Girard, ed. Les Œuvres Mathematiques de Simon Stevin de Bruges. Elsevier, Leiden, 1634. Pp. 169 170, at the end of Stevin's edition of Diophantos (but I have seen other page references). Notes the recurrence property of the Fibonacci numbers, starting with 0, and asserts that the ratio F_{n+1}/F_{n} approaches the ratio of segments of a line cut in mean and extreme ratio, i.e. φ, though he doesn't even give its value  but he says 13, 13, 21 'rather precisely constitutes an isosceles triangle having the angle of a pentagon'. HerzFischler's book, p. 162, notes that Girard describes it as a new result and includes 0 as the starting point of the sequence.
Abraham de Moivre. The Doctrine of Chances: or, A Method of Calculating the Probability of Events in Play. W. Pearson for the Author, London, 1718. Lemmas II & III, pp. 128 134. Describes how to find the generating function of a recurrence. One of his illustrations is the Lucas numbers for which he gets:
x + 3x^{2} + 4x^{3} + 7x^{4} + 11x^{5} + ... = (2x + x^{2})/(1  x  x^{2}). However, he does not have the Fibonacci numbers and he does not use the generating function to determine the individual coefficients of the sequence. In Lemma III, he describes how to find the recurrence of p(n) a^{n} where p(n) is a polynomial. Koshy [p. 215] says De Moivre invented generating functions to solve the Fibonacci recurrence, which seems to be reading much more into De Moivre than De Moivre wrote. The second edition is considerably revised, cf below.
Daniel Bernoulli. Observationes de seriebus quae formantur ex additione vel substractione quacunque terminorum se mutuo consequentium, ubi praesertim earundem insignis usus pro inveniendis radicum omnium aequationum algebraicarum ostenditur. Comm. Acad. Sci. Petropolitanae 3 (1728(1732)) 85 100, ??NYS. = Die Werke von Daniel Bernoulli, ed. by L. P. Bouckaert & B. L. van der Waerden, Birkhäuser, 1982, vol. 2, pp. 49+??. Section 6, p. 52, gives the general solution of a linear recurrence when the roots of the auxiliary equation are distinct. Section 7, pp. 52 53, gives the 'Binet' formula for F_{n}. [Binet's presentation is so much less clear that I suggest the formula should be called the Bernoulli formula.]
Abraham de Moivre. The Doctrine of Chances: or, A Method of Calculating the Probability of Events in Play. 2nd ed, H. Woodfall for the Author, London, 1738. Of the Summation of recurring Series, pp. 193206. This is a much revised and extended version of the material, but he says it is just a summary, without demonstrations, as he has given the demonstrations in his Miscellanea Analytica of 1730 (??NYS). Gives the generating functions for various recurrences and even for a finite number of terms. Prop. VI is: In a recurring series, any term may be obtained whose place is assigned. He assumes the roots of the auxiliary equation are real and distinct. E.g., for a second order recurrence with distinct roots m, p, he says the general term is Am^{n} + Bp^{n} where he has given A and B in terms of the first two values of the recurrence. He even gives the general solution for a fourth order recurrence and expresses A, B, C, D in terms of the first four values of the recurrence. Describes how to take the even terms and the odd terms of a recurrence separately and how to deal with sum and product of recurrences.
R. Simson. An explication of an obscure passage in Albert Girard's commentary upon Simon Stevin's works. Phil. Trans. Roy. Soc. 48 (1753) 368 377. Proves that F_{n}^{2} + ( 1)^{n} = F_{n1}F_{n+1}. This says that the triple F_{n1}, F_{n}, F_{n+1} "nearly express the segments of a line cut in extreme and mean proportion, and the whole line;" from which he concludes that the ratio F_{n+1}/F_{n} does converge to φ. HerzFischler's book, p. 162, notes that his proof is essentially an induction. (He also spells the author Simpson, but it is definitely Simson on the paper.) [Koshy, p. 74, says F_{n}^{2} + ( 1)^{n} = F_{n1}F_{n+1} was discovered in 1680 by Giovanni Domenico Cassini, but he gives no reference and neither Poggendorff nor BDM help to determine what paper this might be.]
Ch. Bonnet. Recherches sur l'usage des feuilles dans les plantes. 1754, pp. 164 188. Supposed to be about phyllotaxis but only shows some spirals without any numbers. Refers to Calandrini. Nice plates.
Master J. Paty (at the Mathematical Academy, Bristol), proposer; W. Spicer, solver. Ladies' Diary, 176869 = T. Leybourn, II: 293, quest. 584. Cow calves at age two and every year thereafter. How many offspring in 40 years? Answer is: 0 + 1 + 1 + 2 + 3 + ... + F_{39}. We would give this as: F_{41}  1, but he gets it as: 2F_{39} + F_{38}  1.
Eadon. Repository. 1794. P. 389, no. 47. Same as the previous problem.
Jackson. Rational Amusement. 1821. Curious Arithmetical Questions, no. 32, pp. 22 & 82. Similar to Fibonacci, with a cow and going for 20 generations.
Martin Ohm. Die reine ElementarMathematik. 2nd ed., Jonas VerlagsBuchhandlung, Berlin, 1835. P. 194, footnote to Prop. 5. ??NYS  extensively discussed in HerzFischler's book, p. 168. This is the oldest known usage of 'goldene Schnitt'. It does not appear in the 1st ed. of 1826 and here occurs as: "... nennt man wohl auch den goldenen Schnitt" (... one also appropriately calls [this] the golden section). The word 'wohl' has many, rather vague, meanings, giving different senses to Ohm's phrase. HerzFischler interprets it as 'habitually', which would tend to imply that Ohm and/or his colleagues had been using the term for some time. I don't really see this meaning and interpreting 'wohl' as 'appropriately' would give no necessity for anyone else to know of the phrase before Ohm. However the term is used in several other German books by 1847. [Incidentally, this is not the Ohm of Ohm's Law, but his brother.]
A. F. W. Schimper & A. Braun. Flora. 1835. Pp. 145 & 737. ??NYS
J. Binet. Mémoire sur l'integration des équations linéaires aux différences finies, d'un ordre quelconque, à coefficients variables. (Extrait par l'auteur). CR Acad. Sci. Paris 17 (1843) 559 567. States the Binet formula as an example of a general technique for solving recurrences of the form: v(n+2) = v(n+1) + r(n)v(n), but the general technique is not clearly described, nor is the linear case.
B. Peirce. Mathematical investigation of the fractions which occur in phyllotaxis. Proc. Amer. Assoc. Adv. Sci. 2 (1849) 444 447. Not very interesting.
Gustav Theodor Fechner. Vorschule der Ästhetik. Breitkopf & Härtel, Leipzig, 1876. ??NYS. Origin of the aesthetic experiments on golden rectangles.
Koshy [p. 5] says Lucas originally called F_{n} the 'série de Lamé', but introduced the name Fibonacci numbers in May 1876. However, he doesn't give a reference. There are several papers by Lucas which might be the desired paper.
Note sur le triangle arithmétique de Pascal et sur la série de Lamé. Nouvelle Correspondence Mathématique 2 (1876) 7075; which might be the desired paper.
L'arithmétique, la série de Lamé, le problème de BehaEddin, etc. Nouvelles Annales de Mathématiques 15 (1876) 20pp.
Édouard Lucas. Théorie des fonctions numériques simplement périodiques. AJM 1 (1878) 184240 (Sections 123) & 289321 (Sections 2430). [There is a translation by Sidney Kravitz of the first part as: The Theory of Simply Periodic Numerical Functions, edited by Douglas Lind, The Fibonacci Association, 1969. Dickson I 400, says this consists of 7 previous papers in Nouv. Corresp. Math. in 18771878 with some corrections and additions. Robert D. Carmichael; Annals of Math. (2) 15 (1913) 3070, ??NX, gives corrections.] The classic work which begins the modern study of recurrences.
Koshy, p. 273, says Adolf Zeising's Der goldene Schnitt of 1884 put forth the theory that "the golden ratio is the most artistically pleasing of all proportions ...." But cf Fechner, 1876.
Pearson. 1907. Part II, no. 63: A prolific cow, pp. 126 & 203. Same as Fibonacci's rabbits, but wants the total after 16 generations.
Koshy, p. 242, asserts that Mark Barr, an American mathematician, introduced the symbol φ (from Phidias) for the Golden Ratio, (1 + 5)/2, about 1900, but he gives no reference.
Coxeter, 1953, takes τ from the initial letter of τoμη, the Greek word for section, but I have no idea if this was used before him.
There is a magic trick where you ask someone to pick two numbers and extend them to a sequence of ten by adding the last two numbers each time. You then ask him to add up the ten numbers and you tell him the answer, which is 11 times the seventh number. In general, if the two starting numbers are A and B, the nth term is F_{n2}A + F_{n1}B and the sum of the first 2n terms is F_{2n}A + (F_{2n+1}1)B = L_{n} (F_{n}A + F_{n+1}B), but only the case n = 5 is interesting! I saw Johnny Ball do this in 1989 and I have found it in: Shari Lewis; Abracadabra! Magic and Other Tricks; (World Almanac Publications, NY, 1984); Puffin, 1985; Sum trick!, p. 14, but it seems likely to be much older.
7.B. JOSEPHUS OR SURVIVOR PROBLEM
See Tropfke 652.
This is the problem of counting out every k th from a circle of n. Early versions counted out half the group; later authors and the Japanese are interested in the last man  the survivor. Euler (1775) seems to be the first to ask for the last man in general which we denote as L(n, k). Cardan, 1539, is the first to associate this process with Josephus. Some later authors derive this from the Roman practice of decimation.
For last man versions, see the general entries and: Michinori?, Kenkō, Cardan, Coburg, Bachet, van Etten, Yoshida, Muramatsu, Schnippel, Ozanam (1696 & 1725), Les Amusemens, Fujita, Euler, Miyake, Matuoka, Boy's Own Book, Nuts to Crack, The Sociable, Indoor & Outdoor, Secret Out (UK), Leske, Le Vallois, Hanky Panky, Kemp, Mittenzwey, Gaidoz, Ducret, Lemoine, Akar et al, Lucas, Schubert, Busche, Tait, Ahrens, Rudin, MacFhraing, Mendelsohn, Barnard, Zabell, Richards, Dean, Richards,
2 to last, counted by 9s: Boy's Own Book,
3 to last, counted by 9s: Boy's Own Book,
4 to last, counted by 9s: Boy's Own Book,
5 to last, counted by 9s: Boy's Own Book,
6 to last, counted by 9s: Boy's Own Book,
7 to last, counted by 9s: Boy's Own Book,
9 to last, counted by 9s: Boy's Own Book,
10 to last, counted by 9s: Boy's Own Book,
11 to last, counted by 9s: Boy's Own Book,
12 to last, counted by 9s: Boy's Own Book, Secret Out (UK),
12 to last, counting number unspecified: Coburg,
13 to last, counted by 2s: Ducret, Leeming,
13 to last, counted by 9s: Boy's Own Book, Secret Out (UK), Leske, Rudin,
14 to all!, counted by 6s: Secret Out,
14 to last, counted by 10s: Mittenzwey,
17 to last, counted by 3s: Barnard,
21 to last, counted by 5s: Hyde,
21 to last, counted by 7s: Nuts to Crack, The Sociable, Indoor & Outdoor, Hanky Panky, H. D. Northrop,
21 to last, counted by 8s: Mittenzwey,
21 to last, counted by 10s: Hyde,
24 to last, counted by 9s: Kemp,
28 to last, counted by 9s: Kemp,
30 to last, counted by 9s: Schnippel,
30 to last, counted by 10s: see entries in next table for 15 & 15 counted by 10s
40 to last man, counted by 3s: van Etten (erroneous),
41 to last man, counted by 3s: van Etten, Ozanam (1725), Vinot, Ducret, Lucas (1895),
General case: Euler, Lemoine, Akar et al., Schubert, Busche, Tait, Ahrens, MacFhraing, Mendelsohn, Robinson, Jakóbczyk, Herstein & Kaplansky, Zabell, Richards,
There are a few examples where one counts down to the last two persons  see references to Josephus and: Pacioli, Muramatsu, Mittenzwey, Ducret, Les Bourgeois Punis.
Almost all the authors cited consider 15 & 15 counted by 9s, so I will only index other versions.
2 & 2 counted by 3s: Ball (1911),
2 & 2 counted by 4s: Ball (1911),
3 & 3 counted by 7s: Ball (1911),
3 & 3 counted by 8s: Ball (1911),
4 & 4 counted by 2s: Leeming,
4 & 4 counted by 5s: Ball (1911),
4 & 4 counted by 9s: Ball (1911),
5 & 5 counted by ??: Dudeney (1905), Pearson, Ball (1911), Ball (1920), Shaw,
6 & 6 counted by ??: Dudeney (1900),
8 & 2 counted by ??: Les Bourgeois Punis,
8 & 8 counted by 8s: Kanchusen,
8 & 8 counted by ??: Dudeney (1899),
12 & 12 counted by 6s: Harrison,
15 & 15 counted by 3s: Tartaglia, Alberti,
15 & 15 counted by 4s: Tartaglia,
15 & 15 counted by 5s: Tartaglia,
15 & 15 counted by 6s: AR, Codex lat. Monacensis 14908, Tartaglia,
15 & 15 counted by 7s: Tartaglia, Schnippel,
15 & 15 counted by 8s: Codex lat. Monacensis 14836, AR, Codex lat. Monacensis 14908, Tartaglia, Alberti,
15 & 15 counted by 10s: Michinori?, Reimar von Zweiter, AR, Codex lat. Monacensis 14908, Chuquet, Tartaglia, Buteo, Hunt, Yoshida, Muramatsu, Wingate/Kersey, Schnippel, Alberti, Shinpen Kinkoki, Fujita, Miyake, Matuoka, Sanpo Chie Bukuro, Hoffmann, Brandreth, Benson, Williams, Collins, Dean. (Almost all of these actually continue to the last person.)
15 & 15 counted by 11s: Tartaglia, Schnippel,
15 & 15 counted by 12s: AR, Codex lat. Monacensis 14908, Tartaglia,
15 & 15 counted by other values, not specified  ??check: Codex lat. Monacensis 14836, Meermanische Codex, at Tilimsâni, Bartoli, Murray 643, Chuquet, Keasby,
17 & 15 counted by 10s: Schnippel,
17 & 15 counted by 12s: Mittenzwey,
18 & 2 counted by 12s: Pacioli, Rudin,
18 & 6 counted by 8s: Manuel des Sorciers,
18 & 18 counted by 9s: Chuquet,
24 & 24 counted by 9s: Chuquet,
30 & 2 counted by 7s: Pacioli,
30 & 2 counted by 9s: Pacioli,
30 & 6 counted by 10s: Ducret, 30 & 10 counted by 12s: Endless Amusement II, Magician's Own Book, The Sociable, Boy's Own Conjuring Book, Lucas (1895),
30 & 30 counted by 12s: Sarma,
36 & 4 counted by 10s: Jackson,
n^{2}n+1 & n1 counted by n: Lucas (1894), Cesarò, Franel, Akar,
Many authors provide a mnemonic for the case of 15 and 15 counted by 9s. In this case, the longest group of the same type is five, so a common device is to encode the numbers 1, 2, 3, 4, 5 by the vowels a, e, i, o, u and then produce a phrase with the vowels in the correct order. I will call this a vowel mnemonic. The most popular form is: Populeam virgam Mater Regina ferebat, giving the numerical sequence: 4, 5, 2, 1, 3, 1, 1, 2, 2, 3, 1, 2, 2, 1. The first group of 4 are good guys, followed by 5 bad guys, etc. Below I list the mnemonics and where they occur, but I did not always record them in my notes below, so I must check a number of the sources again ??  the classification was inspired by seeing that Franci (op. cit. in 3.A) describes a vowel mnemonic in Benedetto da Firenze which I had overlooked. Ahrens gives many more verse and vowel mnemonics  to be added below. Hyde gives an Arabic mnemonic due to alSafadi using the first letters of the Arabic alphabet: a, b, gj, d, h.
Dahbagja Ababgja Baba [= Da h b a gj a A ba b gj a Ba b a]: Hyde from alSafadi.
Unspecified(?) verse mnemonic: ibn Ezra
Populea irgam mater regina reserra: Pacioli;
Populea virga pacem regina ferebat: Minguét
Populeam jirgam mater Regina ferebat: Badcock
Populeam virgam mater Regina ferebat: van Etten; Hunt; Schnippel; Ozanam 1725; Les Amusemens; Hooper; Jackson; Manuel des Sorciers; Boy's Own Book; The Sociable; Le Vallois; Gaidoz; Lucas
Populeam virgam mater regina reserrat: Agostini's version of Pacioli;
Populeam virgam Mater Regina tenebat: Hyde from Wit's Interpreter; Murphy; Schnippel/Bolte
Mort tu ne failliras pas en me liurant le trespas: van Etten
Mort, tu ne falliras pas En me livrant au trépas: Manuel des Sorciers;
Mort, tu ne falliras pas. En me livrant le trépas: Schnippel/Bolte; Ozanam 1725; Les Amusemens; The Sociable; Le Vallois (without the first comma); Ducret; Lucas
On tu ne dai la pace ei la rendea: Schnippel/Bolte
Gott schuf den Mann in Amalek, der (or den) Israel bezwang: Schnippel
Gott schlug den Mann in Amalek, den Israel bezwang: Schnippel/Bolte
So du etwan bist gfalln hart, Stehe widr, Gnade erwart: Schnippel/Bolte
Non dum pena minas a te declina degeas: Schnippel/Bolte
Nove la pinta dà e certi mantena: Benedetto da Firenze
From member's aid and art, Never will fame depart: Schnippel/Bolte
From numbers, aid and art / Never will fame depart: Wingate/Kersey
From numbers, aid, and art, Never will fame depart: Ingleby; Jackson; Rational Recreations
From number's aid and art, Never will fame depart: Gaidoz
From numbers aid and art / Never will fame depart: The Sociable
I have only one example of a mnemonic for 15 & 15 counted by 10s.
Rex Paphicum Gente Bonadat Signa Serena: Hunt
See 5.AD for the general problem of stacking a deck to produce a desired effect.
Josephus. De Bello Judaico. c80. Book III, chap. 8, sect. 7. (Translated by Whiston or by Thackeray (Loeb Classical Library, Heinemann, London, 1927, vol. 2, pp. 685 687.)) (Many later authors cite Hegesippus which is a later version of Josephus.) This says that Josephus happened to survive "by chance or God's providence".
H. St. J. Thackeray. Josephus, the Man and the Historian. Jewish Institute Press, NY, 1929, p. 14. Comments on the Slavonic text, which says that Josephus "counted the numbers with cunning and thereby misled them all" but gives no indication how.
Ahrens. MUS II. 1918. Chap. XV: Das Josephsspiel, pp. 118169. This is the most extended and thorough discussion of this problem and its history. I have used it as the basis of this section. He gives a rather complex method, based on work of Busche, Schubert and Tait, for determining the last man, or any other man in the sequence of counting out, which I never worked through, but which is clearly explained under Richards (1999/9).
Gerard Murphy. The puzzle of the thirty counters. Béaloideas  The Journal of the Folklore of Ireland Society XII (1942) 3 28. In this work and the material cited (mostly ??NYS), the problem of 15 and 15 counted by 9s is shown to have the medieval name Ludus Sancti Petri = St. Peters Spiel = St. Peter's Lake (lake being an Old English word for game [or AngloSaxon for 'to play']) = Sankt Peter Lek or Sankt Päders Lek (in Swedish). Murphy cites Schnippel & Bolte to assert that it was known to the Arabs in the 14C  cf below. He says the usual European version has Christians and Jews on a ship, with St. Peter present and suggesting the counting out process. [I had forgotten that such versions occur in Ahrens, MUS II 130.] However, Murphy was unable to consult MUS, so his background is not as complete as it might be.
Murphy demonstrates that the problem was recently well known in both Scots and Irish Gaelic in a form where a woman has to choose between two groups of warriors seated in a circle, with emotional reasons for her preference. The solution is given in a vernacular mnemonic, using actual numbers as in early Latin forms, while later Latin and vernacular forms used vowel mnemonics. He gives an Irish reconstruction, with English translation, based on several 18C MSS whose texts he estimates as 13C to 17C, probably 16C. This is titled: Goid Fhinn Agus Dubháin Anso (Here is the Thieving of Fionn and Dubhán). One MS has the Latin subtitle: Populeam virgam Mater Regina tenebat, which is a common Latin vowel mnemonic. One of Murphy's sources says this refers to the Queenly Mary appearing to the ship's captain and holding a poplar rod.
Murphy also gives an extended Irish story (3pp) built around the problem: Ceann Dubhrann na Ndumhchann Bán (Ceann Dubhrann of the White Sandhills). The Gaelic names Fionn and Dubhán are derived from 'fionn' and 'dub' meaning 'white' and 'black'. Murphy gives a contemporary Irish version on board a ship with a white captain and a black wife and a crew of 15 and 15, with half having to go overboard due to lack of food. He sketches numerous other Irish and Scots version with varying combinations of details, but using essentially the same verse mnemonic.
Murphy cites a study by Manitius of a 9C MS (Bib. Nat. Paris, No. 13029) where the problem begins "Quadam nocte niger dub nomine, candiduus alter" (One night a black man named Dub and another [named] White). They have to choose between the blacks and the whites to keep watch. Cf. Codex Einsidelensis No. 326 below. The MS ends with the prose line "These two Irish soldiers, one named 'Find' the other 'Dub', were engaged in hunting. 'Find' means "white", 'dub' "black"." The 12C Rouen MS No. 1409 attributes the problem to a Clemens Scottus, which Murphy interprets as Clement the Irishman. The 12C MS Bib. Nat. Paris No. 8091 attributes it to a Thomas Scottus.
Murphy concludes that the problem has an Irish origin, c800. He gives what he believes to be the earliest Latin form, basically Bib. Nat. Paris No. 13029, and opines there must have been an Irish predecessor.
Codex Einsidelensis No. 326. 10C. F. 88'. Latin verse. Published by Th. Mommsen, Handschriftliches. Zur lateinischen Anthologie. Rheinischen Museum für Philologie (NS) 9 (1854) 296 301, with material of interest on pp. 298299. Latin given in: M. Curtze, Bibliotheca Math. (2) 9 (1895) 34 35. Latin & German in MUS II 123 125. Begins: "Quadam nocte niger dux nomine, candidus alter". 15 white & 15 black soldiers, half to keep watch, counted off by 9. The colours refer to clothing, not skin!
Codex lat. Monacensis 14836. 11C. F. 80' gives rules for 15 and 15 counted by 9 (though this value is not specified) and mentions counting by 8 and other values. No mention of what is being counted. Quoted and discussed by: M. Curtze; Zur Geschichte der Josephspiels; Bibliotheca Math. (2) 8 (1894) 116 and in: Die Handschrift No. 14836 der Königl. Hof und Staats bibliothek zu München; AGM 7 (1895) 105 & 111 112 (Supplement to Zeitsch. für Math. und Physik 40 (1895)).
Codex Bernensis 704. 12C. Published by: Hermann Hage; Carmina medii aevi maximam partem inedita; Ex Bibliothecis Helveticus collecta; Bern, 1877; no. 85, pp. 145 146. ??NYS. Latin in: Curtze, op. cit. at Codex Einsidelensis, pp. 35 36; and in: MUS II 127. Jews & Christians.
Ahrens, MUS II 118 147, gives many further references from 10 13C. Originals ??NYS.
Meermanische Codex, 10C. Mentions counting by other values.
Leiden Miscellancodex, 12C
Basel Miscellancodex, 13C
Michinori Fujiwara (11061159). This work is lost, but has been conjectured to contain a form of the problem  see under Kenkō, c1331, and Yoshida, 1634.
Rabbi Abraham ben Ezra. Ta'hbula (or Tachbûla), c1150. ??NYS  described in: Moritz Steinschneider; Abraham ibn Ezra (Abraham Judaeus, Avenare); Zur Geschichte der mathematischen Wissenschaft im XII Jahrhundert; Zeitschr. für Math. und Physik 25 (1880): Supp: AGM 3, Part II (1880). The material is Art. 20, pp. 123 124. 15 students and 15 goodfornothings on a ship, counted by 9s. This seems to be the first extant example on board a ship. Verse mnemonic, which Steinschneider says is not original. Steinschneider cites further sources. Smith & Mikami, p. 84, say ben Ezra died in 1067  ??
Reinmar von Zweter. Meisterlied: "Ander driu, wie man juden und cristen ûz zelt". 13C. (In MUS II 128.) Jews and Christians on a ship, counts by 10.
Kenkō, also known as Kenkō Yoshida or Urabe no Kaneyoshi or Yoshida no Kaneyoshi (12831350). Tsurezuregusa. c1331. Translated by Donald Keene as: Essays in Idleness The Tsurezuregusa of Kenkō; Columbia Univ. Press, NY, 1967. (Kenkō was the author's monastic name. His lay name was Urabe no Kaneyoshi. He lived for a long time at Yoshida in Kyoto.)
This book is one of the classics of Japanese literature, consisting of 243 essays, ranging from single sentences to several pages. The most common themes of these relate to the impermanence of life and the vanity of man.
In Japanese, the Josephus problem is called Mamakodate or Mamakodate San (or Mamako tate no koto  cf Matŭoka, 1808) or Mamagodate (Scheme to benefit the stepchildren or Stepchild disposition). It is said to have been in the lost work of Michinori Fujiwara (1106 1159), qv. The word Mamagodate first occurs in essay 137 of Kenkō, pp. 115121 in Keene's version (including a doublepage illustration which doesn't depict the problem), whose beginning is characteristic of Kenkō's style: "Are we to look at cherry blossoms only in full bloom, the moon only when it is cloudless? To long for the moon while looking on the rain, to lower the blinds and be unaware of the passing of the spring  these are even more deeply moving." The passage of interest is toward the end, on p. 120 of Keene: "When you make a mamagodate^{1} with backgammon counters, at first you cannot tell which of the stones arranged before you will be taken away. Your count then falls on a certain stone and you remove it. The others seem to have escaped, but as you renew the count you will thin out the pieces one by one, until none is left. Death is like that." The footnote refers to counting 15 and 15 by 10s, so that 14 white stones are eliminated, then the counting is reversed and all the black stones are eliminated. "The Japanese name mamagodate (stepchild disposition) derives from the story of a man with fifteen children by one wife and fifteen by another; his estate was disposed of by means of the game, one stepchild in the end inheriting all." Kenkō's text clearly shows he was familiar with the process of counting to the last man and the use of the name indicates that he was familiar with the version mentioned in the footnote, though its earliest explicit appearance in Japan is in Yoshida, 1634, qv. My thanks to Takao Hayashi for the reference to Keene.
Thomas Hyde. Mandragorias seu Historia Shahiludii, .... (= Vol. 1 of De Ludis Orientalibus, see 4.B.5 for vol. 2.) From the Sheldonian Theatre (i.e. OUP), Oxford, 1694. Prolegomena curiosa. The initial material and the Prolegomena are unpaged but the folios of the Prolegomena are marked (a), (a 1), .... The material is on (e 1).v  (e 2).v, which are pages 3436 if one starts counting from the beginning of the Prolegomena. Cited by Bland (loc. cit. in 5.F.1 under Persian MS 211, p. 31); Ahrens (MUS II 136) & Murray 280. Several citations are to ii.23, which may be to the 1767 reprint of Hyde's works.
Hyde asserts that the problem of the ship with 15 Moslems and 15 Christians on a ship, counted by 9s, was given by alSafadi (Şalâhaddîn aş Şafadî [NOTE: Ş, ş denote S, s, with an underdot and the h should have an underdot.] = alSâphadi = AlSáphadi) (d. 1363) in his Lâmiyato ’l Agjam (variously printed in the text). This must be his Sharh [the h should have an underdot] Lâmîyat al ‘Ajam of c1350. Hyde gives an Arabic mnemonic using the first five letters of the Arabic alphabet, which he transliterates as: Dahbagja Ababgja Baba [= Da h b a gj a A ba b gj a Ba b a]. He says the problem occurs in an English book called Wit's Interpreter (??NYS) (8^{o}S.87.Art) where the mnemonic Populeam virgam mater Regina tenebat is given. He then says that the problem is also described in 'Megjdium & Abulphedam'  p. 43 of his main text identifies Abulpheda as a prince born in 672 AH  ??
Shihâbaddîn Abû’l ‘Abbâs Ahmad [the h should have an underdot] ibn Yahya [the h should have an underdot] ibn Abî Hajala [the H should have an underdot] at Tilimsâni alH anbalî [the H should have an underdot]. Kitâb ’anmûdhaj al qitâl fi la‘b ash shaţranj [NOTE: ţ denotes a t with an underdot] (Book of the examples of warfare in the game of chess). Copied by Muhammed ibn ‘Ali ibn Muhammed al Arzagî in 1446.
This is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray. Murray 280 says "Man. 36 45 relate to as Safadî's problem of the ship (see Hyde, ii.23)", described by Murray as 15 Christians and 15 Muslims counted by n. Bland has "the wellknown problem of the Ship, first as described by Safadi, and then in other varieties. (Hyde, p. 23.)"
Murray 620 says the problem is of Muslim origin and says it appears in the c1530 Italian version of the Bonus Socius collection which Murray denotes It. (See 5.F.1 under Bonus Socius.) Murray refers to 15 & 15 counted by 9s, but it is not clear if this refers to this particular MS.
Murray 622 cites MS Sloan 3281 in the BM, 14C, as giving the Latin mnemonic solution.
Bartoli. Memoriale. c1420. F. 100r (= Sesiano p. 135). Il giuocho de' Cristiani contra Saracin. 15 Christians and 15 Saracens  the text ends in the middle of the statement of the problem.
AR. c1450. Prob. 80, pp. 52, 181 182 & 229. 15 Christians and 15 Jews. Gives only mnemonics for counting by 10, 9, 8, 6 or 12.
Codex lat. Monacensis 14908. c1460. F. 76 gives mnemonics for 15 Jews and 15 Christians counted by 6, 8, 9, 10, 12. Quoted and discussed by Curtze, opp. cit. under Cod. lat. Mon. 14836, above. [In the first paper, the codex number is misprinted as 14809.]
Benedetto da Firenze. c1465. Pp. 142 143. 15 Christians and 15 Jews on a boat counted by 9s. Vowel mnemonic: Nove la pinta dà e certi mantena. Diagrammatic picture on p. 143.
Murray 643 says the MS Lasa version, c1475, of the Civis Bononiae collection (described in 5.F.1 under Civis Bononiae) has "16 diagrams of the 'ship' puzzle under different conditions".
Chuquet. 1484. Prob. 146. English in FHM 228230, with reproduction of the original on p. 229. 15 Jews and 15 Christians on a ship, counting by 9s. Says one can have 18 or 24 of each and can count by 10s, etc. The reproduction on FHM 229 shows a circle marked out, with Populeam virgam matre regina tenebat written in the middle. The commentary says this "problem is comparatively rare in fifteenth century texts", which doesn't seem like a fair assessment to me.
Calandri. Aritmetica. c1485. Ff. 102v103v, pp. 205 207. Coloured plate opp. p. 192 of the text volume. (Tropfke 654 gives this in B&W.) Franciscans and Camoldensians on a boat: 15 & 15 counted by 9s.
Pacioli. De Viribus. c1500. Probs. 56 60.
Ff. 99r  102r. LVI. (Capitolo) de giudei Chri'ani in diversi modi et regole. a farne quanti se vole etc (Of Jews and Christians in diverse methods and rules, to make as many as one wants, etc.). = Peirani 140143. Does 2 & 30 by 9s  there is a diagram for this in the margin of f. 100r, but it is not in the transcription and Peirani says another diagram is lacking. Pacioli suggests counting the passengers on shore and doing the counting out with coins or pebbles in case one will need to know the arrangement in a hurry. He also says one might count by 8s, 7s, 6s, 13s, etc., with any number of Christians and Jews.
Ff. 102r  102v. [Unnumbered.] de .18. Giudei et .2. Chri'ani. = Peirani 144. 2 & 18 by 7s.
F. 102v. LVII. C(apitolo). de .30. Giudei et .2. contando per .7. ch' toca va in aqua (Of 30 Jews and 2 counting by 7 with the touched going in the water). = Peirani 144.
Ff. 102v  103r. LVIII. C(apitolo). de .15. Giudei et .15. Chri'ani per .9. in aqua (Of 15 Jews and 15 Christians by 9 in the water.) = Peirani 144145. 15 & 15 by 9s. In order to remember the arrangement, he says to see the next section.
Ff. 103r. LIX. C(apitolo). Quater quinque. duo. unus. tres unus. et unus. bis duo. ter unus. duo duobus un' (4, 5, 2, 1, 3, 1, 1, 2, 2, 3, 1, 2, 2, 1). = Peirani 145. Pattern for 15 & 15 counted by 9s.
Ff. 103r  103v. LX. C(apitolo). si da unaltro verso viz. Populea. irga. mater regina. reserra. ['unaltro' is in the margin with a mark showing where it is to go.] Vowel mnemonic for 15 & 15 counted by 9s, which he explains in detail. Agostini says this is intended to be: Populeam virgam mater regina reserrat but both Pacioli's heading and his discussion have Populea irgam mater regina reserra. The Index has LVIIILX under one heading which only refers to 'versi memorevili'.
Elias Levita der Deutsche. Ha Harkabah. Rome, 1518. ??NYS. Attributes to ben Ezra, c1150??. Smith & Mikami, p. 84, say this seems to be the first printed version of the problem.
Cardan. Practica Arithmetice. 1539. Chap. LXI, section 18, ff. T.iiii.r T.v.r, but the material of interest is just a few sentences on f. T.iv.v (p. 113). Very brief description of 15 white and 15 black as 'ludus Josephus', saying one can work out any numbers with some pebbles. MUS says this is first to relate the problem to Josephus as the last man, but he doesn't give any numerical details.
Hans Sachs (14941576). Meisterleid: 'Historia Die XV Christen und XV Türcken, so auff dem meer furen'. (MUS II 132 133 gives text.)
Tartaglia. General Trattato, 1556, art. 203, pp. 264v 265r. 15 whites and 15 blacks (or Turks and Christians) counted out by 3, 4, ..., 12. No reference to Josephus.
Buteo. Logistica. 1559. Prob. 89, pp. 303304. 15 Christians and 15 Jews on a ship counted by 10s. [Mentioned in H&S 52.]
Simon Jacob von Coburg. Ein new und Wolgegründt Rechenbuch .... 1565 or 1612 (in quarto, not to be confused with octavo versions of 1565 and 1613 which do not contain the problem), f. 250v. ??NYS  described in MUS II 133134. 12 drinkers deciding who shall pay the bill. Ahrens doesn't specify the counting number. Ahrens & Bolte (below) say this is the earliest example, after Cardan, of finding the last man. Ahrens describes numerous later examples of this type from 1693 on.
Prévost. Clever and Pleasant Inventions. (1584), 1998. Pp. 183185. This seems like a version of the Josephus problem but isn't. Place ten counters in a circle and then ten on top of them. Start anywhere and count off five and remove the top counter. He says to count five again  starting on the place where the counter was removed, so we now would say he is counting four  and remove the top counter. Continue in this way, counting the places where a top counter has been removed and you manage to remove all the top counters. In fact this is impossible, but after removing five counters, you subtly start counting from the next position rather than where the top counter was removed! Hence you remove the top counters in the order 5, 9, 3, 7, 1, 6, 10, 4, 8, 2. Your audience will not observe this and hence cannot reproduce the effect.
The mathematical description is simpler if one counts by fours, removing 4, 8, 2, 6, 10, 5, 9, 3, 7, 1. The first five values are the values of 4a (mod 10) for a = 1, 2, 3, 4, 5. Because GCD (4, 10) = 2, this sequence repeats with period 5. Your trick shifts from the even values to the odd values and then you can count out the five odd values.
Bachet. Problemes. 1612. Préface, 1624: A.5.v  A.7.r; 1884: 89 & prob. XX, 1612: 103 106; prob. XXIII, 1624: 174177; 1884: 118 121. Turks & Christians  discusses Josephus as last man.
van Etten. 1624. Prob. 7 (7), pp. 7 9 (16 19). 15 Turks and 15 Christians counted by 9s. Mnemonics: Populeam virgam mater Regina ferebat; Mort tu ne failliras pas en me liurant le trespas. Discusses other cases, Roman decimation and Josephus as 40 counted by 3s. In the 1630 edition, 40 is changed to 41. Henrion's Notte, pp. 9 10, refers to Bachet's prob. 23 and mentions the correction of 40 to 41.
Hunt. 1631 (1651). Pp. 266269 (258261). 15 Christians & 15 Turks counted by 9s; mnemonic: Populeam virga mater regina ferebat. Then does the same counting by 10s and gives the mnemonic: Rex Paphicum Gente Bonadat Signa Serena.
Yoshida (Shichibei) Kōyū (= Mitsuyoshi Yoshida) (15981672). Jinkō ki. Additional problems in the 2nd ed., 1634. Op. cit. in 5.D.1. ??NYS. Shimodaira (see the entry in 5.D.1) discusses the Josephus problem on pp. 1214. He gives some of the information on the Japanese names and on Michinori (11061159) and Kenkō, c1331, which is presented under them. I have a transcription of (some of?) Yoshida into modern Japanese which includes this material as prob. 3 on pp. 6667.
15 children (in black) and 15 stepchildren (in white) counted by 10s. When 14 stepchildren are eliminated, the last stepchild says the arrangement was unfair and requests the counting to go the other way from him (so that he is number 1 in the counting). His stepmother agrees and thereby eliminates all her own children.
This is discussed in Smith & Mikami, pp. 8084. They quote a slightly later version by Seki Kōwa (16421708) where the stepmother simply reverses the order due to overconfidence. (On p. 121, they identify the source as Sandatsu Kempu, a MS of Kōwa.) This is also discussed in MUS II 139 140, where it says that the change in counting was an error on the stepmother's part. Needham, p. 62, gives a picture from the 1634 ed. of Yoshida, but this is different than the picture in my modern transcription. I have a photocopy from an 1801 ed. Dean, 1997, gives the picture, discusses this and provides some additional details, citing the Heibonsha encyclopaedia for the version with the intelligent stepchild. Dean, 1997, also gives an illustration from a 1767 version called Shinpen Jinkoki, cf at 1767.
Ahmed elQalyubi (d. 1659). Naouadir (or Nauadir), c1650?, published at Boulaq (a suburb of Cairo), 1892, hist. 176, p. 82. ??NYS  described in Basset (18861887 below) and MUS II 136. 15 Moslems and 15 infidels on a ship counted by 9s.
Muramatsu Kudayū Mosei. Mantoku Jinkō ri. 1665. ??NYS  described in MUS II 139 and Smith & Mikami, pp. 8084. Smith & Mikami, p. 81, and Dean, 1997, give Muramatsu's schematic diagrams. The top diagram is for the classic 15 and 15 counted by 10s. The second has 32 people counted by 10s to the last two, though the first 15 are coloured black and the second 15 are coloured white, with the last two drawn as squares marked by dice patterns for 5 and 6.
Wingate/Kersey. 1678?. Prob. 3, pp. 531532. 15 & 15 counted by 9s or 10s or any other. Christians and Turks. From numbers, aid and art / Never will fame depart. Discusses Josephus.
Thomas Hyde. Historia Nerdiludii, hoc est dicere, Trunculorum; .... (= Vol. 2 of De Ludis Orientalibus, see above for vol. 1.) From the Sheldonian Theatre (i.e. OUP), Oxford, 1694. De Ludo Charîgj seu Charîtch, pp. 225226. Says it is an Arabic game, i.e. Ludus Exeundi & Eliminadi. The description is very vague, but it seems to involve counting out in a circle. The diagram shows a circle of 21 and the text mentions counting by ten or by five. No reference to any other version of the process. ??need to read the Latin more carefully.
Emil Schnippel (& Johannes Bolte). Das St. PetersSpiel (with a Nachtrag by Bolte). Zeitschrift für Volkskunde 39 (1929) 190192 (& 192194). Schnippel describes the appearance of solutions of the St. PetersSpiel = Sankt Päders Lek = Saint Peter's Lake on 1718C rune calendars from Sweden, which mystified academics until identified by G. Stephens in 1866. He gives the vowelmnemonic: Populeam virgam mater regina ferebat. The rune marks are X for Χριστιαvoι (Xristianoi) and I for ’Ioυδαîoι (Ioudaioi). He gives the German vowelmnemonic: Gott schuf den Mann in Amalek, der (or den) Israel bezwang. He cites other writers (??NYS) who describe a 1497 MS with 15 & 15 Christians and Jews counted by 10s, and versions counted by 7s and 11s, and a version with 17 & 15 Christians and Jews counted by 12s. He cites: a 1604 reference to Josephus but without specific numbers; a 1703 version with 15 & 15 French and Germans; and a 1782 version with 30 deserters, 15 to be pardoned.
[Nigel Pennick; Mazes and Labyrinths; Robert Hale, London, 1990, p. 37, says that in Finland, stone labyrinths are sometimes called "Pietarinleikki (St Peter's Game). The latter name refers to a traditional numerical sequence which appears to be related to the lunar cycle. It is known from rock carvings and ancient Scandinavian calendars and as an antisemitic folktale." Can anyone provide details of a connection to the lunar cycle or its appearance in rock carvings??]
Bolte's Nachtrag cites Gaidoz et al. (below at 18861887) and MUS and an article by himself in Euphorion 3 (1896) 351362, ??NYS  cited MUS II 132. He sketches the history as given by Ahrens. Mentions the Japanese versions and reproduces Matuoka's picture. He adds three citations including a 1908 Indian version with 15 honest men and 15 thieves counted by 9s to the last man (??). He gives vowelmnemonics in Latin, French, German, English and Italian as follows.
Non dum pena minas a te declina degeas.
Populeam virgam mater regina ferebat.
Mort, tu ne falliras pas. En me livrant le trépas.
So du etwan bist gfalln hart, Stehe widr, Gnade erwart.
Gott schlug den Mann in Amalek, den Israel bezwang.
From member's (sic) aid and art, Never will fame depart.
On tu ne dai la pace ei la rendea.
Ozanam. Murphy, note 4, says the problem is not in the 1694 ed.  but see below which could explain why Murphy didn't find it here.
Ozanam. 1696. Preface to vol. 2  first and second of unnumbered pages, which are pp. 269 270. 1708: Author's Preface  second and third of unnumbered pp. Discusses Josephus, citing Bachet.
Ozanam. 1725. Prob. 45, 1725: 246 250. Prob. 17, 1778: 168171; 1803: 168171; 1814: 148150. Prob. 16, 1840: 7677. 15 Turks and 15 Christians counted by 9s. Gives two verse mnemonics: Mort, tu ne failliras pas, En me livrant le trépas; Populeam virgam mater Regina ferebat. Discusses decimation. Quotes Bachet on Josephus and asserts Hegesippus says Josephus used the method and suggests 41 counted by 3s (however, Hegesippus doesn't say this!).
Kanchusen. Wakoku Chiekurabe. 1727. Pp. 8 & 35. 8 and 8 counted by 8s. This is pointing out the remarkable fact that one can count out either set first by starting at different points, in different directions. See: Dudeney, 1899 & 1905; Shaw, 1944?
Minguet. 1733. Pp. 152154 (1755: 110111; 1822: 169171; 1864: 146148). 15 & 15 by 9s, whites and blacks. Populea virga pacem regina ferebat.
Alberti. 1747. 'Modo di disporre 30 cose ...', pp. 132 134 (77 78). 15 Christians and 15 Turks or Jews, counted by 3, 8, 9, 10.
Les Amusemens. 1749. Prob. 16, p. 138: Tiré de Josephe l'Historien. 15 and 15 counted by 9s. French and Latin mnemonics: Mort tu ne failliras pas En me livrant le trépas; Populeam Virgam Mater Regina ferebat.
Shinpen Jinkoki (New Edition of the Jonkoki), more correctly entitled Sanpo Shinan Guruma (A Mathematical Compass). 1767. BL ORB 30/3411. ??NYS  illustration reproduced in Dean, 1997.
Fujita Sadasuke. Sandatsu Kaigi. 1774. ??NYS  cited in a draft version of Dean, 1997, as a Japanese commentary on the problem.
Hooper. Rational Recreations. Op. cit. in 4.A.1. 1774. Vol. 1, recreation XIII, pp. 4243. 30 deserters of whom 15 are to be punished, counted by 9s. Populeam virgam mater regina ferebat.
L. Euler. Observationes circa novum et singulare progressionum genus. (Novi Comment. Acad. Sci. Petropol. 20 (1775 (1776)) 123 139.) = Opera Omnia (1) 7, (1923) 246 261. Gets the recurrence for the last man: L(n) L(n1) + k (mod n).
Miyake Kenryū. Shojutsu Sangaku Zuye. 1795. ??NYS. (Described in MUS II 142 143.) First(?) to modify Yoshida's problem (1634?) so that the last stepchild sees his imminent fate and asks for the count to restart with him. Smith, History II 543 and Smith & Mikami, p. 82, give a poorish picture from this. Dean, 1997, is a better picture.
Matuoka (= Matŭoka ??= Matsuoka Nōichi). 1808. ??NYS  translated by Le Vallois, with reproductions of the pictures, cf below. Ahrens, MUS II 140 142, discusses this, based on Le Vallois and reproduces the main picture from Le Vallois. Gives Miyake's version. Le Vallois gives the title as: Mamako tate no koto (Problème des beauxfils (i.e. stepsons)). There is a diagram showing the countingout processes.
Ingleby. Ingleby's Whole Art of Legerdemain, containing all the Tricks and Deceptions, (Never before published) As performed by the Emperor of Conjurors, at the Minor Theatre, with copious explanations; Also, several new and astonishing Philosophical and Mathematical Experiments, with Preliminary Observations, Including directions for practicing the Slight of Hand. T. Hughes & C. Chaple, London, nd [1815]. Trick L. The Turks and Christians, pp. 104106. 15 & 15 counted by 9s. "This ingenious trick, which is scarcely known, ...." "From numbers, aid, and art, / Never will fame depart."
Sanpo Chie Bukuro (A Bag of Mathematical Wisdom). 1818. BL ORB 30/3411. ??NYS  illustration reproduced and discussed in Dean, 1997. Here a man and a woman are studying a set of 29 black and white go stones and the text describes the problem and how to arrange the children.
Badcock. Philosophical Recreations, or, Winter Amusements. [1820]. Pp. 8385, no. 130: Thirty soldiers having deserted, so to place them in a ring, that you may save any fifteen you please, and it shall seem the effect of chance. 15 & 15 by 9s. Populeam jirgam mater Regina ferebat. (jirgam must be a misprint of virgam.) Says Josephus and 'thirty or forty of his soldiers' hid in a cave and Josephus arranged to be one of the last.
Jackson. Rational Amusement. 1821. Arithmetical Puzzles.
No. 16, pp. 45 & 5456. 15 Turks and 15 Christians counted by 9s. Solution gives: From numbers, aid, and art, Never will fame depart and Populeam virgam mater regina ferebat.
No. 39, pp. 910 & 6162. Decimation of a troop of 40 to be counted by 10s  where to place the four ringleaders so they will be the four to be shot.
Rational Recreations. 1824. Feat 27, p. 103. 15 Turks and 15 Christians counted by 9s. Gives: From numbers, aid, and art, / Never will fame depart.
Manuel des Sorciers. 1825. ??NX Pp. 7980, art. 40. 15 & 15 by 9s. Populeam virgam mater regina ferebat. Mort, tu ne falliras pas En me livrant au trépas. Says one can also do 18 & 6 by 8s, etc. Cf Gaidoz, below, col. 429.
Endless Amusement II. 1826? P. 117 (misprinted 711 in 1826?): Predestination illustrated. 30 and 10 counted by 12s.
Boy's Own Book.
The slighted lady. 1828: 411 412; 18282: 417418; 1829 (US): 210211; 1855: 565 566; 1868: 670. 13 counted down to last person by 9s. Before 1868, it gives the survivor for 2, 3, ..., 13, counted out by 9s.
The partial reprieve. 1828: 417 418; 18282: 422; 1855: 571; 1868: 672673; 1881: 214. 30 criminals counted by 9s to eliminate 15. Populeam virgam mater regina ferebat.
Nuts to Crack XIV (1845), no. 72. 21 counted by 7s to the last man.
Magician's Own Book. 1857.
The fortunate ninth, pp. 221222. 15 oranges and 15 apples, counted by 9s. English mnemonics based on vowel coding.
Another decimation of fruit, p. 224225. 30 apples and 10 oranges, counted by 12s in order to get the oranges first.
The Sociable. 1858.
Prob. 31: The puzzle of the Christians and the Turks, pp. 296 & 312314. From numbers aid and art / Never will fame depart. Mort, tu ne faillras pas / en me livrant le trepas. Populeam Virgam Mater regina ferebat. Then considers counting out 10 from 40, counting by 12s. Discusses Josephus, citing Hegesippus, and suggests counting by 3s. = Book of 500 Puzzles, 1859, prob. 31, pp. 14 & 3032.
Prob. 39: The landlord tricked, pp. 298 & 316. 21 counted by 7s to the last man. = Book of 500 Puzzles, 1859, prob. 39, pp. 16 & 34. = Wehman; New Book of 200 Puzzles; 1908, p. 51.
The Secret Out. 1859. The Circle of Fourteen Cards, p. 87. This appears to be counting out all 14 cards by 6s (it says by 7s, but it takes the counted out card as one for the next stage), but it's not clear what the object is. This seems to be a corruption of an earlier version??
Indoor & Outdoor. c1859. Part II, prob. 19: The landlord tricked, p. 136. Identical to The Sociable.
Boy's Own Conjuring Book. 1860.
The fortunate ninth, pp. 190 191. Identical to Magician's Own Book.
Another decimation of fruit, p. 194. Identical to Magician's Own Book.
Vinot. 1860. Art. XXVI: De l'historien Josèphe, pp. 5556. Gives the Josephus story and does it as counting from 41 by 3s to the last man.
The Secret Out (UK). c1860. A delicate distribution, p. 12. Count 13 by 9s to the last person (different context than Leske). Mentions counting 12 by 9s.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 56430, pp. 254 & 396: Aus 12 Dreizehn machen. Count 13 by 9s to the last person.
M. le Capitaine Le Vallois. Les Sciences exactes chez les Japonais. With comments by Louis de Zélinski & M. Sédillot. Congrès International des Orientalists (= International Congress of Orientalists). CompteRendu de la première session, Paris, 1873. Maisonneuve et Cie., Paris, 1874. T. 1, pp. 289 299, with comments on pp. 299303. The material of interest is on pp. 294298. Gives a translation of Matŭoka, 1808, and reproduces the pictures, cf above. Discusses Bachet, Ozanam, Mort tu ne failliras pas En me livrant le trépas, Populeam virgam mater Regina tenebat, Josephus (saying Josephus arranged to be last).
Hanky Panky. 1872. The landlord tricked, pp. 129130. Identical to The Sociable, prob. 39.
Kamp. Op. cit. in 5.B. 1877. No. 7, pp. 323 324. 28 counted by 9s until one is left. Footnote seems to refer to 24 counted by 9s.
Mittenzwey. 1880. Prob. 282285, pp. 5052 & 100101; 1895?: 311314, pp. 5456 & 102103; 1917: 311 314, pp. 49 50 & 97 98.
282 (311): 15 Negroes and 17 Europeans on a ship, counted by 12s.
283 (312): 7 students and a crafty Jew who wishes to make two of the students, A & B, pay the bill since they had been rude to him. Initially he is not included. Starting with A and counting clockwise by 3s, A & B are left. Starting with B and counting anti clockwise by 3s, A & B are left. Then the Jew is included and starts with himself, counting anticlockwise by 3s and again A & B are left.
284 (313): 14 counted by 10s to the last man.
285 (314): 21 counted by 8s to the last man.
Cassell's. 1881. P. 103: To reward the favourites, and show no favouritism. = Manson, 1911, p. 256. 15 & 15 counted by 9s.
Henri Gaidoz, Israël Lévi & René Basset. Le jeu de Saint Pierre  Amusement arithmétique. This is a series of five notes in Mélusine 3 (1886 87).
Gaidoz. Part I. Col. 273 274. Gives classical version with St. Peter, 15 Christians & 15 Jews counted by 9s. He then gives two versions from Ceylon. One version is called Yonmaruma  The massacre of the Moors  and has 15 Portuguese & 15 Moors with a Singhalese verse mnemonic. The second version involves the Portuguese siege of Kandy in 1821, again 15 & 15 by 9s, but different versions have the Portuguese winning or losing. These versions come from: The Orientalist 2 (1885) 177, ??NYS. The editor of The Orientalist added a version learned from an Irish soldier with the vowel mnemonic: From number's aid and art, Never will fame depart. Gaidoz says he cannot venture a source for the puzzle.
Gaidoz. Part II. Col. 307 308. Comments on correspondence generated by Part I which provided: 'Populeam virgam mater regina ferebat'; the version with the Virgin instead of St. Peter; a version with negroes and whites and a negro captain; a version with French and English; references to Josephus, Bachet and Ozanam.
Lévi. Part III. Col. 332. Says ibn Ezra's "Tahboula" (Stratagem), c1150, is devoted to this game. Cites Schwenter (1623); Steinschneider's 1880 article discussed above at Ezra; Steinschneider's Catalog librorum hebr. Biblioth. Bodleianae, col. 687  all ??NYS. Previously Steinschneider opined the game derived from Jahia ibn al Batrik's Secret of Secrets (8C), but Lévi says that that is a different amusement involving 9.
Gaidoz. Part IV. Col. 429. Cites: Le Manuel des Sorciers, Paris, 2nd ed., 1802, p. 70 for a version with French and English. ??NYS, but see the 1825 ed above.
Basset. Part V. Col. 528. Describes elQalyubi, c1650?  cf above.
Robert Harrison. UK Patent 15,105  An Improved Puzzle or Game. Applied: 25 Sep 1889; accepted: 2 Nov 1889. 2pp + 1p diagrams. 12 whites and 12 blacks on a boat with a lifeboat that will hold 12, counted by 6s, called The Captain's Dilemma.
É. Ducret. Récréations Mathématiques. Op. cit. in 4.A.1. 1892?
Pp. 105106: Une dame pas contente. 13 counted by 2s to last person.
Pp. 118119: Stratagéme de Joséphe. 41 counted by 3s to last two, claimed to be the method used by Josephus.
Pp. 120121: Les marauders punis. 15 & 15 counted by 9s. Officers and soldiers to be executed.
Pp. 121122: Les naufrages. Same numbers, with Turks and Christians on a boat. Mort, tu ne failliras pas, En me livrant au trépas.
P. 122: Les Élections perfectionnées. 36 counted by 10s  want the first six chosen.
Hoffmann. 1893. Chap. 4, pp. 156157 & 210211 = HoffmannHordern, pp. 134135, with photo.
No. 54: Tenth man out. 15 whites and 15 blacks on a ship, counted by 10s, but first 15 get to go into the lifeboats. Photo on p. 135 shows L'Equipage Decime, with box and instructions, by Watilliaux, 18741895.
No. 55: Ninth man out. Same, counted by 9s. Hoffmann cites Bachet and gives a Latin mnemonic. Photo on p. 135 shows La Question des Boches, with box having instructions on base, 19141918.
É. Lucas. Problem 32. Intermed. des Math. 1 (1894) 9. n^{2} persons, counted by n until n 1 are left. "Problème dit de Caligula".
E. Cesarò. Solution to 32. Ibid., pp. 30 31.
J. Franel. Deuxième réponse [to Problem 32]. Ibid., p. 31. Cites: Busche, CR 103, pp. 118, ??NYS.
Adrien Akar. Troisième réponse [to Problem 32]. Ibid., pp. 189 190.
E. Lemoine. Problem 330. Ibid, pp. 184 185. Asks for last man of n counted by p.
Adrien Akar; H. Delannoy; J. Franel; C. Moreau. Independent solvers of Lemoine's problem. Ibid., 2 (1895) 120 122 & 229 230. Akar refers to Josephus, Bachet, etc. Moreau has the clearest form of the recurrence.
Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 7: The tenth man out. Almost identical to Hoffmann, no. 54. No solution.
Lucas. L'Arithmétique Amusante. 1895. Pp. 1218.
Le stratagème de Josèphe, pp. 1217. Prob. VI. 15 Christians and 15 Turks, counted by 9s. Vowel mnemonics: Mort, tu ne falliras pas, En me livrant le trépas!; Populeam virgam mater Regina ferebat. Discusses and quotes Bachet's 1624 Préface which gives the Josephus story and the idea of counting 41 by 3s.
Prob. VII: Le procédé de Caligula, pp. 1718. 6 and 30 counted by 10s so as to count the 6 first.
H. Schubert. Zwölf Geduldspiele. 1895. P. 125. ??NYS  cited by Ahrens; Mathematische Spiele; Encyklopadie article, op. cit. in 3.B; 1904.
E. Busche. Ueber die Schubert'sche Lösung eines Bachet'schen Problems. Math. Annalen 47 (1896) 105 112.
Clark. Mental Nuts. 1897, no. 13; 1904, no. 22. The ship's crew. 1897 has the usual 15 and 15 counted by 9s, starting with the captain, involving whites and blacks on a ship and half being thrown overboard. 1904 has 14 whites and 15 blacks and the captain must discharge 15 at a port. He joins the crew and starts counting from himself and wants to discharge the 15 blacks.
P. G. Tait. On the generalization of Josephus' problem. Proc. Roy. Soc. Edin. 22 (1898) 165 168. = Collected Scientific Papers, vol. II, pp. 432 435. Says the Josephus passage is "very obscure, ..., but it obviously suggests deliberate fraud of some kind on Josephus' part." Develops a way of computing the last man.
Les Bourgeois Punis. Puzzle from c1900, shown in S&B, p. 133. 8 and 2 counted by ?? to leave the 2 at the end.
Dudeney. A batch of puzzles. Royal Magazine 1:3 (Jan 1899) & 1:4 (Feb 1899) 368372. The prisoners of Omdurman. 8 Europeans followed by 8 Abyssinians in a ring. Start counting with the first European. Determine the countingout number to eliminate the Abyssinians in sequence. Doing it in reverse sequence works for any multiple of 16, 15, 14, ..., 9. The LCM is 720720. But doing it in forward sequence can be done with 360361. Since the pattern is symmetric in the two types of people, a change of initial position, but keeping the same direction, will count out the others first.
Dudeney. "The Captain" puzzle corner. The Captain 3:2 (May 1900) 97 & 179 & 3:4 (Jul 1900) 303. The "blacks" and "whites" puzzle or The twelve schoolboys. 6 consecutive "blacks" and 6 consecutive "whites" in a circle. What is the smallest number to count out by which will count out the "whites" first? You can start anywhere and in any direction. Answer is 322 and one starts counting on the fourth "white" in the direction of counting.
Sreeramula Rajeswara Sarma. Mathematical literature in Telugu: An overview. Sree Venkateswara University Oriental Journal 28 (1985) 7790. Telugu is one of the Dravidian languages of south India, spoken in the area north of Madras, and is the state language of Andhra Pradesh. On pp. 8387 & 90, he reports finding examples in Telugu in the notebooks of the schoolmaster Panakalu Rayudu (18831928) who was a collector of material from many sources. Unfortunately there is no indication of where Rayudu obtained these examples and Sarma knows of no Indian versions. He has 15 thieves and 15 brahmins counted by 9s, then 30 thieves and 30 brahmins counted by 12s. Solutions are given in some literary form. The second problem is new to me. In his notes, Sarma cites the German mnemonic Gott schuf den Mann in Amalek, der (or den) Israel bezwang given by Schnippel, and that he has learned that the problem occurs in the Peddabālaśikşa [NOTE: ş denotes an s with an underdot.], a work which is unknown to me.
H. D. Northrop. Popular Pastimes. 1901. No. 8: the landlord tricked, pp. 6768 & 72. = The Sociable, no. 39.
Ahrens. Mathematische Spiele. Encyklopadie article, op. cit. in 3.B. 1904. Pp. 1088 1089 discusses Schubert's work and its later developments.
Benson. 1904. The black and white puzzle, pp. 225 226. As in Hoffmann, no. 54, but first 15 get thrown overboard. Solution is linear rather than circular.
Dudeney. Tit Bits (14 Oct & 28 Oct 1905). ??NYS  described by Ball; MRE, 5th ed., 1911, pp. 25 27. 5 and 5 arranged so that one method eliminates one group while another method eliminates the other group. Determine the two starting points and counts. Ball doesn't give these values, but seems to imply that both counts go in the same direction, and this is the case in the examples given below. Ball asks if the starting points can ever be the same for two groups of C? He gives solutions for C = 2 (counted by 4 & 3), 3 (counted by 7 & 8), 4 (counted by 9 & 5). He believes this question is new. Note on p. 27 gives the solution for C = 5, but with different starting points. See MRE, 10th ed., 1920, for a general solution with the same starting points. See: Kanchusen, 1727; Dudeney, 1899; Shaw, 1944?
Pearson. 1907. Part II, no. 62, pp. 126 & 203. 15 Christians, including St. Peter, who does the counting, and 15 Jews, counted by 9s.
Ball. MRE, 5th ed. 1911. See under Dudeney, 1905.
Loyd. Cyclopedia. 1914. Christians and Turks, pp. 198 & 365. = MPSL2, prob. 42, pp. 30 31 & 134. Like Dudeney's 1905 version with a different arrangement of 5 and 5.
Williams. Home Entertainments. 1914. A decimation problem, pp. 122124. 15 whites & 15 blacks counted by 10s. Half have to go over because of shortage of provisions. Simple circular picture with man counting in middle.
Ball. MRE, 10th ed., 1920, pp. 2627. See under Dudeney, 1905, for the previous version. Incorporates the solution for the case C = 5 into the text and adds a general solution due to Swinden.
See: Will Blyth; Money Magic; 1926 for a related problem.
Collins. Book of Puzzles. 1927. Sailors don't care puzzle, pp. 7071. 15 whites & 15 blacks counted by 10s. Captain has to throw half over because of shortage of provisions. Diagram of 15 circles in a row above a picture with 15 circles in a row below, but normally numbered  it would seem natural in this problem to have the lower row numbered backward to simulate a circle.
William P. Keasby. The Big Trick and Puzzle Book. Whitman Publishing, Racine, Wisconsin, 1929. Counting out problem, pp. 71 & 95. 15 & 15 shown in a circle with the starting point and direction given. Determine the counting number.
Rudin. 1936. Nos. 102103, pp. 3738 & 98.
No. 102. 13 counted by 9s until last man.
No. 103. 17 and 15 counted by 12s to eliminate the 15 first.
Ern Shaw. The Pocket Brains Trust  No. 2. Op. cit. in 5.E. c1944. Prob. 50: Poser with pennies. Pattern of 5 Hs and 5 Ts given  determine the count to count the Hs first, which turns out to be 11. Though not mentioned, the pattern of Hs is equivalent to that for Ts, so one can count out the Ts first by starting at a different point in the opposite direction. The pattern is the same as Dudeney (1905).
Robert Gibbings. Lovely is the Lee. Dent, London, 1945. Pp. 111114. He says he was shown the puzzle by an old man on the Aran Islands. Cites Murphy, but his version is different than anything in Murphy. ??NYS  information sent by Michael Behrend in an email of 12 Jun 2000.
Le Rob Alasdair MacFhraing (= Robert A. Rankin, who tells me that the Gaelic particle 'Le' means 'by' and is not part of the name). Àireamh muinntir Fhinn is Dhubhain, agus sgeul Iosephuis is an dà fhichead Iudhaich (The numbering of Fionn's men and Dubhan's men, and the story of Josephus and the forty Jews) (in Scots Gaelic with English summary). Proc. Roy. Irish Acad. Sect. A. 52:7 (1948) 8793. A more detailed description than in the Summary appears in Rankin's review in: Math. Reviews 10, #509b (= A995 in: Reviews in Number Theory). I had assumed that this was a development from Murphy's article, but Rankin writes that he had not heard of Murphy's article until I wrote for a reprint of Rankin's article in 1991. He gives a Scottish Gaelic version which is clearly a variant of those studied by Murphy. He then studies the problem of determining the last man, citing Tait. For counting out by 2s, the rule is simple. [There is a story that the the number of mathematicians fluent in Scots Gaelic is so small and the author's name is so obscured that the journal sent the paper to Rankin to referee, not knowing he was actually the author. The story continues that the referee made a number of suggestions for improvement which the author gratefully accepted. However, Rankin told me that he was not the referee. But he did review it for Math. Reviews!]
Joseph Leeming. Games with Playing Cards Plus Tricks and Stunts. Op. cit. in 6.BE, 1949. ??NYS  but two abridged versions have appeared which contain the material  see 6.BE.
24 Stunts with Cards, 7th & 19th stunts. A surprising card deal & All in order. Dover: pp. 124 & 131. Gramercy: pp. 11 & 18. Both stunts involve dealing cards by putting one out face up, then the next is put at the bottom of the deck, then the next is dealt face up, .... This process is the same as counting out by 2s. The object is to produce the cards in a particular order. The first has 8 cards and wants an alternation of face cards and nonface cards. The second has the 13 cards of a suit and wants them produced in order. This is the only example that I can recall of the use of the Josephus idea as a card trick, though other forms of counting out are common, e.g. by counting 1, 2, 3, ..., or by spelling the words one, two, three, ....
N. S. Mendelsohn, proposer; Roger Lessard, solver. Problem E 898  Discarding cards. AMM 57 (1950) 3435 & 488489. Basically counting out by 2s. Determine the position of the last card discarded. No mention of Josephus, though the editor asks what happens if every rth card is discarded and gets the recurrence f(N) r + f(N 1) (mod N).
W. J. Robinson. Note 2876: The Josephus problem. MG 44 (No. 347) (Feb 1960) 47 52. Analyses what sequences of persons can be removed by varying the count. Applies to Dudeney's problem.
Barnard. 50 Observer BrainTwisters. 1962. Prob. 36: Circle of fate, pp. 41 42, 64 65 & 95. Princess counts out from 17 suitors by 3s. She sees that her favourite will be the next one out, so she reverses direction and then the favourite is the survivor.
F. Jakóbczyk. On the generalized Josephus problem. Glasgow Math. J. 14 (1973) 168173. Gives a method of determining when the ith man is removed and which is the kth to be removed. Somewhat similar to Rankin's method.
D. Woodhouse The extended Josephus problem. Revista Matematica HispanoAmericana 33 (1973) 207218. Gets recurrences for the last person, but unnecessarily complicates the process by considering the starting point. By combining the recurrences, he gets an n fold iteration for the result, but this doesn't really clarify anything. Only cites Josephus.
Israel N. Herstein & Irving Kaplansky. Matters Mathematical. 1974; slightly revised 2nd ed., Chelsea, NY, 1978. Chap. 3, section 5: The Josephus permutation, pp. 121128. They study the permutation where f(i) = number of ith man eliminated, but restrict to the case where one counts by 2s, which has considerable structure. Gives a substantial bibliography, mostly included here.
Sandy L. Zabell. Letter [on the history of the Josephus problem]. Fibonacci Quarterly 14 (1976) 48 & 51. Sketches the history.
I. M. Richards. The Josephus problem. MS 24 (1991/92) 97104. Studies the case of counting out by 3s. Shows the 'Tait numbers', i.e. n such that L(n) = 1 or 2, are given by [η(3/2)^{i} + 1/3], where η = 1.216703..., and obtains a formula for L(n). Presumably this could be extended to the general case??
Michael Dean. Josephus and the Mamakodate san (Scheme to benefit the stepchildren). International Netsuke Society Journal 17:2 (Summer 1997) 4153. There are inro boxes from late 17C Japan which have pictures of the 15 children and 15 stepchildren problem. These initially mystified the art historians, but eventually they discovered the Josephus problem and its Japanese forms, but only as far back as Bachet. Dean gives a brief history for the benefit of art collectors, with references to a number of Japanese sources (some of which I have not seen)  see above at 1767, 1795, 1818  and some photos of the inro boxes (including a fine late 17C example from the collection of Michael and Hiroko Dean) and other material.
Ian M. Richards. The Josephus Problem and Ahrens arrays. MS 31:2 (1998/9) 3033. He has finally obtained a copy of Ahrens' work, but from the first edition, and states the result clearly. For n persons, labelled 1, 2, ..., n, counted out by k, if we want to locate the eth man counted out, form a sequence starting with 1 + k(ne) and then form each next term by multiplying a term by k/(k1) and rounding the result up to an integer. (I.e. x_{n+1} = x_{n} * k/(k1).) Then the position number of the eth person eliminated is the difference between kn + 1 and the largest term in the sequence less than kn + 1. The sequence is giving the points where L(n, k) is zero in some sense. Note that when e = n, so we are looking for the last person, then the sequence starts at 1, which is because we start counting with the first person as one. kn + 1 is the total amount counted in counting n people by k, For other values of e, the change of the starting point of the sequence compensates for the fact that one only counts k(ne) + 1 to eliminate the eth person. Ahrens then examined the sequences obtained, with rational multipliers, and found some nice properties which Richards states. Richards generalises to arbitrary multipliers and finds connections with Beatty sequences, a^{n}.
Ian M. Richards. Towards an analytic solution of the Josephus problem. Unpublished preprint sent to me on 21 Mar 1999, 12pp. (Available from the author, 3 Empress Avenue, Penzance, Cornwall, TR18 2UQ.) Gets formulae for the case k = 4 which give the result with a maximum error of ±1.
David Singmaster. Adjacent survivors in the Josephus Problem. Nov 2003, 5pp, but may be extended. This was inspired by the first example in Pacioli's De Viribus, which has 2 'good guys' and 30 'bad guys' arranged in a circle and every 9th person is thrown overboard. I was struck by the fact that the two survivors were adjacent in the original circle as clearly marked in the marginal diagram. Offhand it seems an unlikely result, but one soon observes that this remains true as the counting out takes place. That is, if the two survivors in counting off N by Ks are adjacent, then this is also true for counting off n by Ks for 3 n < N. This paper investigates the maximal N for which counting out by Ks leaves two last survivors who were originally adjacent.
