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A005427 Josephus problem: numbers m such that, when m people are arranged on a circle and numbered 1 through m, the final survivor when we remove every 4th person is one of the first three people.
(Formerly M3759)
5
5, 7, 9, 12, 16, 22, 29, 39, 52, 69, 92, 123, 164, 218, 291, 388, 517, 690, 920, 1226, 1635, 2180, 2907, 3876, 5168, 6890, 9187, 12249, 16332, 21776, 29035, 38713, 51618, 68824, 91765, 122353, 163138, 217517, 290023, 386697, 515596, 687461, 916615, 1222153, 1629538, 2172717, 2896956, 3862608, 5150144, 6866859, 9155812, 12207749, 16276999, 21702665, 28936887, 38582516, 51443354 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Is this the same as A072493 with its first 8 terms removed? See also the similar conjecture concerning A005428 and A073941.

From Petros Hadjicostas, Jul 20 2020: (Start)

We describe the counting-off game of Burde (1987) using language from Schuh (1968). Suppose m people are labeled with the numbers 1 through m (say clockwise). (Burde uses the numbers 0 through m-1 probably because he relates this problem to the representation of m in the fractional base k/(k-1) = 4/3. He actually modifies the (4/3)-representation of m to include negative coefficients. See the coefficients f(n;k) below.)

Suppose we start the counting at the person labeled 1, and we remove every 4th person. This sequence gives those numbers m for which the last survivor is one of the first three people.

When m = 5, 9, 12, 16, 218, 517, ... the last survivor is the first person.

When m = 7, 29, 69, 92, 291, 388, ... the last survivor is the second person.

When m = 22, 39, 52, 123, 164, 690, ... the last survivor is the third person.

If we know m = a(n) and the number, say i(n), of the last survivor (when there are a(n) people on the circle), we may find a(n+1) and the number i(n+1) of the new last survivor (when there a(n+1) people on the circle) in the following way:

(a) If 0 = a(n) mod 3, then a(n+1) = (4/3)*a(n), and i(n+1) = i(n).

(b) If 1 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.

(c) If 1 = a(n) mod 3 and i(n) = 2, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1.

(d) If 1 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 2.

(e) If 2 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 2.

(f) If 2 = a(n) mod 3 and i(n) = 2, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.

(g) If 2 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1. (End)

From Petros Hadjicostas, Jul 22 2020: (Start)

In general, for k >= 2, it seems that when m people are placed on a circle, labeled 1 through m, and every k-th person is removed (starting the counting at person 1), we may determine those m for which the last survivor is in {1, 2, ..., k-1} in the following way.

Define the sequence (T(n;k): n >= 1) by T(n;k) = ceiling(Sum_{s=1..n-1} T(s;k)/(k-1)) for n >= 2 starting with T(1; k) = 1. Then the list of those m's for which the last survivor is in {1, 2, ..., k-1} consists of all the numbers T(n;k) >= k (thus, we exclude the cases m = 1, ..., k-1 that may be repeated more than once in the sequence (T(n;k): n >= 1)).

I do not have a general proof of this conjecture though I strongly believe that Schuh's (1968) way of solving the case k = 3 (see pp. 373-375 and 377-379, where he provides two methods of solution) may provide clues for proving the conjecture.

We have T(n; k=2) = A011782(n+1), T(n; k=3) = A073941(n), T(n; k=4) = A072493(n), T(n; k=5) = A120160(n), T(n; k=6) = A120170(n), T(n; k=7) = A120178(n), T(n; k=8) = A120186(n), T(n; k=9) = A120194(n), and T(n; k=10) = A120202(n).

We also have T(n+1;k) = floor((k/(k-1))*T(n;k)) or ceiling((k/(k-1)*T(n;k)).

To identify the last survivor that results when we place T(n; k) people on the circle (with T(n;k) >= k) in the above Josephus problem, we use a modification of Burde's algorithm due to Thériault (2000).

We use the following recursions but we start at T(k;k) (rather than at the smallest n for which T(n;k) >= k). Define the sequence (S(n;k): n >= 1) by S(n;k) = T(n+k-1; k) for n >= 1. (It is easy to prove that S(1;k) = T(k;k) = 1.)

Define also the sequences (j(n;k): n >= 1) and (f(n;k): n >= 1) by j(1;k) = 1, f(1;k) = 0, f(n+1;k) = ((j(n;k) - S(n;k) - 1) mod (k-1)) + 1 - j(n;k) and j(n+1;k) = j(n;k) + f(n+1;k) for n >= 2.

Then for all n s.t. S(n;k) >= k, j(n;k) is the number of the last survivor of the Josephus problem where every k-th person is removed (provided we start the counting at number 1). It will always be the case that j(n;k) is in {1,2,...,k-1}.

We actually have S(n+1; k) = (k*S(n;k) + f(n+1;k))/(k-1) for n >= 1.

Notice that the Burde-Thériault algorithm is a generalization of Schuh's method. (End)

REFERENCES

Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968. [This book is cited in Burde (1987). Table 18, p. 374, is related to a very similar sequence (A073941). Thus, definitely, the counting-off games described in the book are related to a similar counting-off game in Burde (1987).]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

K. Burde, Das Problem der Abzählreime und Zahlentwicklungen mit gebrochenen Basen [The problem of counting rhymes and number expansions with fractional bases], J. Number Theory 26(2) (1987), 192-209.

Nicolas Thériault, Generalizations of the Josephus problem, Utilitas Mathematica, 58 (2000), 161-173 (MR1801309).

Nicolas Thériault, Generalizations of the Josephus problem, Utilitas Mathematica, 58 (2000), 161-173 (MR1801309).

Index entries for sequences related to the Josephus Problem

FORMULA

a(n) = 5 + ceiling(Sum_{k=1..n-1} a(k)/3). - Petros Hadjicostas, Jul 21 2020

EXAMPLE

From Petros Hadjicostas, Jul 22 2020: (Start)

We explain why 5 and 7 are in the sequence but 6 is not.

If we put m = 5 people on the circle, label them 1 through 5, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 3 -> 5 -> 2. Thus, the last survivor is 1, so m = 5 is included in this sequence.

If we put m = 6 people on a circle, label them 1 through 6, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 2 -> 1 -> 3 -> 6. Thus, the last survivor is 5 (not 1, 2, or 3), so m = 6 is not included in this sequence.

If we put m = 7 people on a circle, label them 1 through 7, start the counting at person 1, and remove every 4th person, then the list of people who are removed is 4 -> 1 -> 6 -> 5 -> 7 -> 3. Thus, the last survivor is 2, so m = 7 is included in this sequence.

Strictly speaking, m = 2 and m = 3 should have been included as well (since clearly the last survivor would be 1 or 2 or 3). In addition, m = 4 should have been included as well because the list of people removed is 4 -> 1 -> 3. The case of number 1 does create a problem since there is no survivor. Note that the numbers 1, 2, 3, 4 are all included in A072493. (End)

MATHEMATICA

f[s_] := Append[s, Ceiling[5 + Plus@@(s/3)]]; Nest[f, {5}, 100] (* Vladimir Joseph Stephan Orlovsky, Jan 08 2011 *)

PROG

(PARI) /* Gives an n X 2 matrix w s.t. w[, 1] are the terms of this sequence and w[, 2] are the corresponding numbers of the last survivors (1, 2 or 3). */

lista(nn) = {my(w = matrix(nn, 2)); w[1, 1] = 5; w[1, 2] = 1; for(n=1, nn-1,

if(0 == w[n, 1] % 3, w[n+1, 1] = w[n, 1]*4/3; w[n+1, 2] = w[n, 2]);

if(1 == w[n, 1] % 3 && w[n, 2] == 1, w[n+1, 1] = ceil(w[n, 1]*4/3);  w[n+1, 2] = w[n, 2] + 2);

if(1 == w[n, 1] % 3 && w[n, 2] == 2, w[n+1, 1] = floor(w[n, 1]*4/3); w[n+1, 2] = w[n, 2] - 1);

if(1 == w[n, 1] % 3 && w[n, 2] == 3, w[n+1, 1] = floor(w[n, 1]*4/3); w[n+1, 2] = w[n, 2] - 1);

if(2 == w[n, 1] % 3 && w[n, 2] == 1, w[n+1, 1] = ceil(w[n, 1]*4/3);  w[n+1, 2] = w[n, 2] + 1);

if(2 == w[n, 1] % 3 && w[n, 2] == 2, w[n+1, 1] = ceil(w[n, 1]*4/3);  w[n+1, 2] = w[n, 2] + 1);

if(2 == w[n, 1] % 3 && w[n, 2] == 3, w[n+1, 1] = floor(w[n, 1]*4/3); w[n+1, 2] = w[n, 2] - 2);

); Vec(w[, 1]); } // Petros Hadjicostas, Jul 21 2020

/* Second PARI program for the general case of Josephus problem. We use the Burde-Thériault algorithm, not the formula T(n; k) = ceiling(Sum_{s=1..n-1} T(s; k)/(k-1)). We start with T(k; k) = 1 (and omit all previous 1's). Burde starts with the smallest T(n; k) >= k whose corresponding last survivor is 1. This, however, can be very large. To get the corresponding last survivors, modify the program to get the vector j. */

lista(nn, k) = {my(j=vector(nn)); my(f=vector(nn)); my(N=vector(nn));

j[1]=1; f[1]=0; N[1] = 1;

for(n=1, nn-1, f[n+1] = ((j[n]-N[n]-1) % (k-1)) + 1 - j[n];

j[n+1] = j[n] + f[n+1]; N[n+1] = (k*N[n] + f[n+1])/(k-1); );

for(n=1, nn, if(N[n] > k-1, print1(N[n], ", "))); } \\ Petros Hadjicostas, Jul 23 2020

CROSSREFS

Cf. A005428, A072493.

Similar sequences: A011782 (k = 2), A073941 (k = 3), A072493 (k = 4), A120160 (k = 5), A120170 (k = 6), A120178 (k = 7), A120186 (k = 8), A120194 (k = 9), A120202 (k = 10).

Sequence in context: A336008 A159018 A345980 * A116024 A115913 A200268

Adjacent sequences:  A005424 A005425 A005426 * A005428 A005429 A005430

KEYWORD

nonn

AUTHOR

N. J. A. Sloane, Simon Plouffe

EXTENSIONS

More terms (from the Burde paper, p. 208) from R. J. Mathar, Sep 26 2006

Name edited by Petros Hadjicostas, Jul 20 2020

STATUS

approved

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Last modified August 4 05:02 EDT 2021. Contains 346442 sequences. (Running on oeis4.)