A032434 -id:A032434 - cp's OEIS frontend

A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.

0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Offset: 0

N. J. A. Sloane

Write the numbers 1 through n in a circle, start at 1 and cross off every other number until only one number is left.
A version of the children's game "One potato, two potato, ...".
a(n)/A062383(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006
Iterating a(n), a(a(n)), ... eventually leads to 2^A000120(n) - 1. - Franklin T. Adams-Watters, Apr 09 2010
By inspection, the solution to the Josephus Problem is a sequence of odd numbers (from 1) starting at each power of 2.  This yields a direct closed form expression (see formula below). - Gregory Pat Scandalis, Oct 15 2013
Also zero together with a triangle read by rows in which row n lists the first 2^(n-1) odd numbers (see A005408), n >= 1. Row lengths give A011782. Right border gives A000225. Row sums give A000302, n >= 1. See example. - Omar E. Pol, Oct 16 2013
For n > 0: a(n) = n + 1 - A080079(n). - Reinhard Zumkeller, Apr 14 2014
In binary, a(n) = ROL(n), where ROL = rotate left = remove the leftmost digit and append it to the right. For example, n = 41 = 101001_2 => a(n) = (0)10011_2 = 19. This also explains FTAW's comment above. - M. F. Hasler, Nov 02 2016
In the under-down Australian card deck separation: top card on bottom of a deck of n cards, next card separated on the table, etc., until one card is left. The position a(n), for n >= 1, from top will be the left over card. See, e.g., the Behrends reference, pp. 156-164. For the down-under case see 2*A053645(n), for n >= 3, n not a power of 2. If n >= 2 is a power of 2 the botton card survives. - Wolfdieter Lang, Jul 28 2020

			From _Omar E. Pol_, Jun 09 2009: (Start)
Written as an irregular triangle the sequence begins:
  0;
  1;
  1,3;
  1,3,5,7;
  1,3,5,7,9,11,13,15;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,
   43,45,47,49,51,53,55,57,59,61,63;
...
(End)
From _Omar E. Pol_, Nov 03 2018: (Start)
An illustration of initial terms, where a(n) is the area (or number of cells) in the n-th region of the structure:
   n   a(n)       Diagram
   0    0     _
   1    1    |_|_ _
   2    1      |_| |
   3    3      |_ _|_ _ _ _
   4    1          |_| | | |
   5    3          |_ _| | |
   6    5          |_ _ _| |
   7    7          |_ _ _ _|
(End)

Erhard Behrends, Der mathematische Zauberstab, Rowolth Taschenbuch Verlag, rororo 62902, 4. Auflage, 2019, pp. 156-164. [English version: The Math Behind the Magic, AMS, 2019.]
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 10.
M. S. Petković, "Josephus problem", Famous Puzzles of Great Mathematicians, page 179, Amer. Math. Soc. (AMS), 2009.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Paul Weisenhorn, Josephus und seine Folgen, MNU, 59(2006), pp. 18-19.

Iain Fox, Table of n, a(n) for n = 0..100000 (terms 0..1000 from T. D. Noe, terms 1001..10000 from Indranil Ghosh).
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197, ex. 34.
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Paul Barry, Conjectures and results on some generalized Rueppel sequences, arXiv:2107.00442 [math.CO], 2021.
Daniel Erman and Brady Haran, The Josephus Problem, Numberphile video (2016)
Chris Groër, The Mathematics of Survival: From Antiquity to the Playground, Amer. Math. Monthly, 110 (No. 9, 2003), 812-825.
Alasdair MacFhraing, Aireamh Muinntir Fhinn Is Dhubhain, Agus Sgeul Josephuis Is An Da Fhichead Iudhaich, [Gaelic with English summary], Proc. Royal Irish Acad., Vol. LII, Sect. A., No. 7, 1948, 87-93.
Yuri Nikolayevsky and Ioannis Tsartsaflis, Cohomology of N-graded Lie algebras of maximal class over Z_2, arXiv:1512.87676 [math.RA], (2016), pages 2, 6.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Josephus Problem
Wikipedia, Josephus problem
Index entries for sequences related to the Josephus Problem

Cf. A005428, A035327, A038572, A049940, A049964, A053644, A053645, A088147, A088442.
Second column, and main diagonal, of triangle A032434.
Cf. A181281 (with s=5), A054995 (with s=3).
Column k=2 of A360099.

Require Import ZArith.
Fixpoint a (n : positive) : Z :=
match n with
| xH => 1
| xI n' => (2*(a n') + 1)%Z
| xO n' => (2*(a n') - 1)%Z
end.
(* Stefan Haan, Aug 27 2023 *)

a006257 n = a006257_list !! n
a006257_list =
   0 : 1 : (map (+ 1) $ zipWith mod (map (+ 1) $ tail a006257_list) [2..])
-- Reinhard Zumkeller, Oct 06 2011

[0] cat [2*(n-2^Floor(Log(2,n)))+1: n in [1..100]]; // Vincenzo Librandi, Jan 14 2016

a(0):=0: for n from 1 to 100 do a(n):=(a(n-1)+1) mod n +1: end do:
seq(a(i),i=0..100); # Paul Weisenhorn, Oct 10 2010; corrected by Robert Israel, Jan 13 2016
A006257 := proc(n)
    convert(n,base,2) ;
    ListTools[Rotate](%,-1) ;
    add( op(i,%)*2^(i-1),i=1..nops(%)) ;
end proc: # R. J. Mathar, May 20 2016
A006257 := n -> 2*n  - Bits:-Iff(n, n):
seq(A006257(n), n=0..78); # Peter Luschny, Sep 24 2019

Table[ FromDigits[ RotateLeft[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v, Sep 21 2003 *)
Flatten@Table[Range[1, 2^n - 1, 2], {n, 0, 5}] (* Birkas Gyorgy, Feb 07 2011 *)
m = 5; Range[2^m - 1] + 1 - Flatten@Table[Reverse@Range[2^n], {n, 0, m - 1}] (* Birkas Gyorgy, Feb 07 2011 *)

a(n)=sum(k=1,n,if(bitxor(n,k)Paul D. Hanna

a(n)=if(n, 2*n-2^logint(2*n,2)+1, 0) \\ Charles R Greathouse IV, Oct 29 2016

import math
def A006257(n):
     return 0 if n==0 else 2*(n-2**int(math.log(n,2)))+1 # Indranil Ghosh, Jan 11 2017

def A006257(n): return bool(n&(m:=1<Chai Wah Wu, Jan 22 2023
(C#)
static long cs_A006257(this long n) => n == 0 ? 0 : 1 + (1 + (n - 1).cs_A006257()) % n; // Frank Hollstein, Feb 24 2021

To get a(n), write n in binary, rotate left 1 place.
a(n) = 2*A053645(n) + 1 = 2(n-msb(n))+1. - Marc LeBrun, Jul 11 2001. [Here "msb" = "most significant bit", A053644.]
G.f.: 1 + 2/(1-x) * ((3*x-1)/(2-2*x) - Sum_{k>=1} 2^(k-1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = number of positive integers k < n such that n XOR k < n. a(n) = n - A035327(n). - Paul D. Hanna, Jan 21 2006
a(n) = n for n = 2^k - 1. - Zak Seidov, Dec 14 2006
a(n) = n - A035327(n). - K. Spage, Oct 22 2009
a(2^m+k) = 1+2*k; with 0 <= m and 0 <= k < 2^m; n = 2^m+k; m = floor(log_2(n)); k = n-2^m; a(n) = ((a(n-1)+1) mod n) + 1; a(1) = 1. E.g., n=27; m=4; k=11; a(27) = 1 + 2*11 = 23. - Paul Weisenhorn, Oct 10 2010
a(n) = 2*(n - 2^floor(log_2(n))) + 1 (see comment above). - Gregory Pat Scandalis, Oct 15 2013
a(n) = 0 if n = 0 and a(n) = 2*a(floor(n/2)) - (-1)^(n mod 2) if n > 0. - Marek A. Suchenek, Mar 31 2016
G.f. A(x) satisfies: A(x) = 2*A(x^2)*(1 + x) + x/(1 + x). - Ilya Gutkovskiy, Aug 31 2019
For n > 0: a(n) = 2 * A062050(n) - 1. - Frank Hollstein, Oct 25 2021

More terms from Robert G. Wilson v, Sep 21 2003

A321298 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in the Josephus elimination process for n people and a count of 2, 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 2, 1, 3, 2, 4, 3, 1, 2, 4, 1, 5, 3, 2, 4, 6, 3, 1, 5, 2, 4, 6, 1, 5, 3, 7, 2, 4, 6, 8, 3, 7, 5, 1, 2, 4, 6, 8, 1, 5, 9, 7, 3, 2, 4, 6, 8, 10, 3, 7, 1, 9, 5, 2, 4, 6, 8, 10, 1, 5, 9, 3, 11, 7, 2, 4, 6, 8, 10, 12, 3, 7, 11, 5, 1, 9, 2, 4, 6, 8, 10, 12, 1, 5, 9, 13, 7, 3, 11, 2, 4, 6, 8, 10, 12, 14

Offset: 1

Zeph L. Turner, Nov 02 2018

In the Josephus elimination process for n and k, the numbers 1 through n are written in a circle. A pointer starts at position 1. Each turn, the pointer skips (k-1) non-eliminated number(s) going around the circle and eliminates the k-th number, until no numbers remain. This sequence represents the triangle J(n, i), where n is the number of people in the circle, i is the turn number, and k is fixed at 2 (every other number is eliminated).

			Triangle begins:
  1;
  2, 1;
  2, 1, 3;
  2, 4, 3, 1;
  2, 4, 1, 5,  3;
  2, 4, 6, 3,  1,  5;
  2, 4, 6, 1,  5,  3, 7;
  2, 4, 6, 8,  3,  7, 5, 1;
  2, 4, 6, 8,  1,  5, 9, 7,  3;
  2, 4, 6, 8, 10,  3, 7, 1,  9,  5;
  2, 4, 6, 8, 10,  1, 5, 9,  3, 11, 7;
  2, 4, 6, 8, 10, 12, 3, 7, 11,  5, 1, 9;
  2, 4, 6, 8, 10, 12, 1, 5,  9, 13, 7, 3, 11;
  ...
For n = 5, to get the entries in 5th row from left to right, start with (^1, 2, 3, 4, 5) and the pointer at position 1, indicated by the caret. 1 is skipped and 2 is eliminated to get (1, ^3, 4, 5). (The pointer moves ahead to the next "live" number.) On the next turn, 3 is skipped and 4 is eliminated to get (1, 3, ^5). Then 1, 5, and 3 are eliminated in that order (going through (^3, 5) and (^3)). This gives row 5 of the triangle and entries a(11) through a(15) in this sequence.

Index entries for sequences related to the Josephus Problem

The right border of this triangle is A006257.

Cf. A032434, A054995, A181281, A225381, A378635 (row permutation inverses).

Table[Rest@ Nest[Append[#1, {Delete[#2, #3 + 1], #2[[#3 + 1]], #3}] & @@ {#, #[[-1, 1]], Mod[#[[-1, -1]] + 1, Length@ #[[-1, 1]]]} &, {{Range@ n, 0, 0}}, n][[All, 2]], {n, 14}] // Flatten (* Michael De Vlieger, Nov 13 2018 *)

def A321298(n,k):
    if 2*k<=n: return 2*k
    n2,r=divmod(n,2)
    if r==0: return 2*A321298(n2,k-n2)-1
    if k==n2+1: return 1
    return 2*A321298(n2,k-n2-1)+1 # Pontus von Brömssen, Sep 18 2022

From Pontus von Brömssen, Sep 18 2022: (Start)
The terms are uniquely determined by the following recursive formulas:
  T(n,k) = 2*k if k <= n/2;
  T(2*n,k) = 2*T(n,k-n)-1 if k > n;
  T(2*n+1,k) = 2*T(n,k-n-1)+1 if k > n+1;
  T(2*n+1,n+1) = 1.
(End)
From Pontus von Brömssen, Dec 11 2024: (Start)
The terms are also uniquely determined by the following recursive formulas:
  T(1,1) = 1;
  T(n,1) = 2 if n > 1;
  T(n,k) = T(n-1,k-1)+2 if k > 1 and T(n-1,k-1) != n-1;
  T(n,k) = 1 if k > 1 and T(n-1,k-1) = n-1.
T(n,A225381(n)) = 1.
T(n,A225381(n+1)-1) = n.
(End)

Name clarified by Pontus von Brömssen, Sep 18 2022

A054995 A version of Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, delete the integer two places clockwise from i. Repeat, counting two places from the next undeleted integer, until only one integer remains.

Original entry on oeis.org

1, 2, 2, 1, 4, 1, 4, 7, 1, 4, 7, 10, 13, 2, 5, 8, 11, 14, 17, 20, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 1, 4, 7, 10, 13, 16, 19, 22, 25

Offset: 1

John W. Layman, May 30 2000

nonn

If one counts only one place (rather than two) at each stage to determine the element to be deleted, the Josephus survivors (A006257) are obtained.

			a(5) = 4 because the elimination process gives (1^,2,3,4,5) -> (1,2,4^,5) -> (2^,4,5) -> (2^,4) -> (4), where ^ denotes the counting reference position.
a(13) = 13 => a(14) = (a(13) + 2) mod 14 + 1 = 2. - _Paul Weisenhorn_, Oct 10 2010

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..10000
R. Baumann, Das Josephus-Problem, LOG IN, Heft Nr. 165, pp. 70-71, 2010 (in German).
Ph. Dumas, Algebraic aspects of B-regular series. [Broken link]
Philippe Dumas, Algebraic aspects of B-regular series, Research Report, RR-1931, INRIA, 1993.
Ph. Dumas, Algebraic aspects of B-regular series, in: International Colloquium on Automata, Languages and Programming, ICALP 1993 (A. Lingas, R. Karlsson, S. Carlsson, eds.), pp. 457-468, Lecture Notes in Computer Science, vol. 700, Springer, Berlin, 1993.
L. Halbeisen and N. Hungerbühler, The Josephus Problem, J. Théor. Nombres Bordeaux 9 (1997), no. 2, 303-318.
Alasdair MacFhraing, Aireamh Muinntir Fhinn Is Dhubhain, Agus Sgeul Josephuis Is An Da Fhichead Iudhaich, [Gaelic with English summary], Proc. Royal Irish Acad., Vol. LII, Sect. A., No. 7, 1948, 87-93.
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
Index entries for sequences related to the Josephus Problem

Cf. A032434, A005427, A005428, A006257, A007495, A000960, A056530.

Cf. A181281 (with s=5). - Paul Weisenhorn, Oct 10 2010

(* First do *) Needs["Combinatorica`"] (* then *) f[n_] := Last@ InversePermutation@ Josephus[n, 3]; Array[f, 70] (* Robert G. Wilson v, Jul 31 2010 *)
Table[Nest[Rest@RotateLeft[#, 2] &, Range[n], n - 1], {n, 72}] // Flatten (* Arkadiusz Wesolowski, Jan 14 2013 *)

a(n) = 3*n + 1 - floor(K(3)*(3/2)^(ceiling(log((2*n+1)/K(3))/log(3/2)))) where K(3) = (3/2)*K = 1.622270502884767... (K is the constant described in A061419); a(n) = 3n + 1 - A061419(k+1) where A061419(k+1) is the least integer such that A061419(k+1) > 2n.
a(1) = 1 and, for n > 1, a(n) = (a(n-1) + 3) mod n, if this value is nonzero, n otherwise.
a(n) = (a(n-1) + 2) mod n + 1. - Paul Weisenhorn, Oct 10 2010

A007495 Josephus problem: survivors.

Original entry on oeis.org

1, 1, 2, 2, 2, 4, 5, 4, 8, 8, 7, 11, 8, 13, 4, 11, 12, 8, 12, 2, 13, 7, 22, 2, 8, 13, 26, 4, 26, 29, 17, 27, 26, 7, 33, 20, 16, 22, 29, 4, 13, 22, 25, 14, 22, 37, 18, 46, 42, 46, 9, 41, 12, 7, 26, 42, 24, 5, 44, 53, 52, 58, 29, 22, 12, 48, 27, 30, 58, 52, 49, 57, 13, 14, 32, 24, 75, 8, 67

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

If, in a circle of k persons, every n-th person is removed, the survivor is t(k,n) + 1. So the recurrence generates a sequence of survivors. See the formula. For more details see the "Proof of the formula". - Gerhard Kirchner, Oct 23 2016

The recurrence formula looks like a simple congruential generator for pseudo-random numbers. Is a(n) pseudo-random? It seems so, see: "Stochastic aspects". I used the formula for extending a(n) up to n=2^20. - Gerhard Kirchner, Nov 10 2016

			From _Gerhard Kirchner_, Oct 23 2016: (Start)
If n = 4 we have that:
  t(1,4) = 0.
  t(2,4) = (0+4) mod 2 = 0.
  t(3,4) = (0+4) mod 3 = 1.
  t(4,4) = (1+4) mod 4 = 1.
So a(4) = 1 + 1 = 2. (End)

Friend H. Kierstead, Jr., Computer Challenge Corner, J. Rec. Math., 10 (1977), see p. 124.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 1..5000 (first 1000 terms from T. D. Noe)
Gerhard Kirchner, Proof of the formula.
Gerhard Kirchner, Stochastic aspects.
Eric Weisstein's World of Mathematics, Josephus Problem.
Robert G. Wilson v, Notes, n.d..
Index entries for sequences related to the Josephus Problem

Cf. A032434, A054995.

(* First do *) Needs["Combinatorica`"] (* then *) f[n_] := Last@ InversePermutation@ Josephus[n, n]; Array[f, 80] (* Robert G. Wilson v, Jul 31 2010 *)
t[k_, n_] := t[k, n] = Mod[t[k-1, n]+n, k]; t[1, ] = 0; a[n] := t[n, n]+1; Array[a, 1000] (* Jean-François Alcover, Oct 23 2016, after Gerhard Kirchner *)

Let t(k,n) = (t(k-1,n) + n) mod k and t(1,n) = 0; then a(n) = t(n,n) + 1. - Gerhard Kirchner, Oct 23 2016

More terms from Robert G. Wilson v, Jul 31 2010

A088333 A version of Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, delete the integer 3 places clockwise from i. Repeat, counting 3 places from the next undeleted integer, until only one integer remains.

Original entry on oeis.org

1, 1, 2, 2, 1, 5, 2, 6, 1, 5, 9, 1, 5, 9, 13, 1, 5, 9, 13, 17, 21, 3, 7, 11, 15, 19, 23, 27, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42

Offset: 1

N. J. A. Sloane, Nov 13 2003

If one counts only one place (resp. two places) at each stage to determine the element to be deleted, we get A006257 (resp. A054995).

See A054995 for references and links.

Index entries for sequences related to the Josephus Problem

Cf. A006257, A054995, A032434, A005427, A005428, A006257, A007495, A000960, A056530.

It is tempting (in view of A054995) to conjecture that a(1)=1 and, for n>1, a(n) = (a(n-1)+4) mod n. The conjecture is false; counterexample: a(21)=21; a(20)=17; (a(20)+4)mod 21=0; corrected formula: a(n)=(a(n-1)+3) mod n +1;
The conjecture is true. After removing the 4th number, we are reduced to the n-1 case, but starting with 5 instead of 1. - David Wasserman, Aug 08 2005
a(n) = A032434(n,4) if n>=4. - R. J. Mathar, May 04 2007

More terms from David Wasserman, Aug 08 2005

A198789 Array T(n,k) read by antidiagonals: Last survivor positions in Josephus problem for n numbers and a count of k, n >= 1, k >= 1.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 2, 3, 4, 1, 1, 2, 1, 5, 1, 2, 2, 1, 3, 6, 1, 1, 1, 2, 4, 5, 7, 1, 2, 1, 2, 1, 1, 7, 8, 1, 1, 3, 3, 2, 5, 4, 1, 9, 1, 2, 3, 2, 4, 1, 2, 7, 3, 10, 1, 1, 2, 3, 4, 4, 6, 6, 1, 5, 11, 1, 2, 2, 3, 1, 5, 3, 3, 1, 4, 7, 12, 1, 1, 1, 4, 2, 3, 5, 1, 8, 5, 7, 9, 13

Offset: 1

William Rex Marshall, Nov 21 2011

Arrange 1, 2, 3, ..., n clockwise in a circle. Starting the count at 1, delete every k-th integer clockwise until only one remains, which is T(n,k).
The main diagonal (1, 1, 2, 2, 2, 4, 5, 4, ...) is A007495.
Concatenation of consecutive rows (up to the main diagonal) gives A032434.
The periods of the rows, (1, 2, 6, 12, 60, 60, 420, 840, ...), is given by A003418.

			.n\k  1  2  3  4  5  6  7  8  9 10
----------------------------------
.1 |  1  1  1  1  1  1  1  1  1  1
.2 |  2  1  2  1  2  1  2  1  2  1
.3 |  3  3  2  2  1  1  3  3  2  2
.4 |  4  1  1  2  2  3  2  3  3  4
.5 |  5  3  4  1  2  4  4  1  2  4
.6 |  6  5  1  5  1  4  5  3  5  2
.7 |  7  7  4  2  6  3  5  4  7  5
.8 |  8  1  7  6  3  1  4  4  8  7
.9 |  9  3  1  1  8  7  2  3  8  8
10 | 10  5  4  5  3  3  9  1  7  8

William Rex Marshall, First 141 antidiagonals of array, flattened
Index entries for sequences related to the Josephus Problem

Cf. A000027 (k = 1), A006257 (k = 2), A054995 (k = 3), A088333 (k = 4), A181281 (k = 5), A360268 (k = 6), A178853 (k = 7), A109630 (k = 8).

Cf. A003418, A007495 (main diagonal), A032434, A198788, A198790.

T[n_, k_] := T[n, k] = If[n == 1, 1, Mod[T[n-1, k]+k-1, n]+1];
Table[T[n-k+1, k], {n, 1, 13}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Mar 04 2023 *)

T(1,k) = 1; for n > 1: T(n,k) = ((T(n-1,k) + k - 1) mod n) + 1.

A321781 Least q > 1 letting Josephus survive if he finds himself at position j in the circle of m persons, but is allowed to name the elimination parameter q such that every q-th person is executed, written as triangle T(m,j), m > 1, j <= m.

Original entry on oeis.org

0, 2, 3, 5, 3, 2, 2, 4, 6, 10, 4, 5, 2, 3, 11, 3, 10, 8, 6, 2, 27, 11, 4, 6, 3, 7, 5, 2, 2, 19, 5, 7, 12, 4, 3, 9, 3, 7, 2, 42, 35, 11, 6, 5, 21, 8, 19, 5, 3, 2, 15, 9, 10, 7, 12, 16, 26, 24, 40, 7, 36, 2, 5, 4, 14, 12, 4, 9, 6, 26, 8, 11, 18, 13, 2, 3, 12, 7, 21, 10, 15, 11, 4, 5, 23, 13, 6, 12, 2, 18, 3

Offset: 1

Hugo Pfoertner, Nov 18 2018

Exercise 23 associated with Chapter 1.3 in "Concrete Mathematics" about the Josephus Problem asks: "Suppose that Josephus finds himself in a given position j, but he has a chance to name the elimination parameter q such that every qth person is executed. Can he always save himself?"

T(1,1) is set to 0 to complete the triangle. q > 1 serves to avoid the obviously merciless choice of q = 1 in the case of Josephus being located at position m.

			The triangle begins:
   0
   2  3
   5  3  2
   2  4  6 10
   4  5  2  3 11
   3 10  8  6  2 27
  11  4  6  3  7  5  2
   2 19  5  7 12  4  3  9
   3  7  2 42 35 11  6  5 21
   8 19  5  3  2 15  9 10  7 12
  16 26 24 40  7 36  2  5  4 14 12
   4  9  6 26  8 11 18 13  2  3 12  7
   ...
3 persons:
  q = 2: 111 -> 101 -> 001. Position 3 survives, therefore T(3,3) = 2;
  q = 3: 111 -> 110 -> 010. Position 2 survives, therefore T(3,2) = 3;
  q = 4: 111 -> 011 -> 010. Position 2 survives, already covered by q = 3;
  q = 5: 111 -> 101 -> 100. Position 1 survives, therefore T(3,1) = 5.

Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994, page 20.

Index entries for sequences related to the Josephus Problem

The first column of the table is A187788.

Cf. A003418, A032434, A321793, A321794.

A128982 If in a line of n persons every n-th person is eliminated until only one person is left, which position P should one assume in the original lineup to avoid being eliminated?

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 2, 6, 6, 10, 2, 12, 2, 6, 8, 16, 2, 18, 2, 16, 18, 22, 2, 22, 12, 16, 8, 28, 2, 30, 2, 28, 18, 22, 12, 36, 2, 6, 8, 40, 2, 42, 2, 30, 42, 46, 2, 42, 14, 40, 30, 52, 2, 36, 24, 52, 54, 58, 2, 60, 2, 6, 30, 48, 24, 66, 2, 30, 18, 70, 2, 72, 2, 6, 20, 60, 18, 78, 2, 72, 78

Offset: 1

Harri Aaltonen, Apr 30 2007

nonn

The difference between this, A007495 and the diagonal of A032434 is that for each of the n-1 elimination processes, counting from 1 to n starts at the lowest position in the line that is still occupied, not right after the most recently eliminated position. Wrapping around when n exceeds the number of residual occupied positions still occurs in circular fashion as in the original Josephus problem. - R. J. Mathar, May 07 2007

			Elimination at n=6: 1,2,3,4,5,6 -> 1,2,3,4,5 -> 2,3,4,5 -> 2,4,5 -> 2,4 -> 2. After the 3 is eliminated, counting does not start at 4 but again at 2.

Hagen von Eitzen, Table of n, a(n) for n = 1..10000
Index entries for sequences related to the Josephus Problem

Cf. A007495, A032434.

int a(int n) { if (n<3) return 1; int L=1, R=n-1, M, t, s, q, r; while (R>L+1) { s = M = (L+R)/2; t= n-1; while (s && sHagen von Eitzen, Nov 08 2022 */

A128982 := proc(n) local l ; l := [seq(i,i=1..n)] ; for i from 1 to n-1 do rm := ((n-1) mod nops(l))+1 ; l := subsop(rm=NULL,l) ; od ; RETURN(op(1,l)) ; end: for n from 1 to 85 do printf("%d, ",A128982(n)) ; od ; # R. J. Mathar, May 07 2007

a[n_] := Module[{l = Range[n]}, Do[l = Delete[l, Mod[n-1, Length[l]]+1], {n-1}]; If[l == {}, Nothing, l[[1]]]];
a /@ Range[0, 100] (* Jean-François Alcover, Apr 01 2020 *)

If n is prime then P = n - 1. If n is prime + 1 then P = 2.

This is a version of the Josephus problem. Several other versions are already in the OEIS. - N. J. A. Sloane, May 01 2007

Corrected and extended by R. J. Mathar, May 07 2007

A198788 Array T(k,n) read by descending antidiagonals: Last survivor positions in Josephus problem for n numbers and a count of k, n >= 1, k >= 1.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 4, 3, 2, 1, 5, 1, 2, 1, 1, 6, 3, 1, 2, 2, 1, 7, 5, 4, 2, 1, 1, 1, 8, 7, 1, 1, 2, 1, 2, 1, 9, 1, 4, 5, 2, 3, 3, 1, 1, 10, 3, 7, 2, 1, 4, 2, 3, 2, 1, 11, 5, 1, 6, 6, 4, 4, 3, 2, 1, 1, 12, 7, 4, 1, 3, 3, 5, 1, 3, 2, 2, 1, 13, 9, 7, 5, 8, 1, 5, 3

Offset: 1

William Rex Marshall, Nov 21 2011

Arrange 1, 2, 3, ... n clockwise in a circle. Starting the count at 1, delete every k-th integer clockwise until only one remains, which is T(k,n).
The main diagonal of the array (1, 1, 2, 2, 2, 4, 5, 4, ...) is A007495.
Consecutive columns down to the main diagonal (1, 2, 1, 3, 3, 2, 4, 1, 1, 2, ...) is A032434.
Period lengths of columns (1, 2, 6, 12, 60, 60, 420, 840, ...) is A003418.

			.k\n  1  2  3  4  5  6  7  8  9 10
----------------------------------
.1 |  1  2  3  4  5  6  7  8  9 10  A000027
.2 |  1  1  3  1  3  5  7  1  3  5  A006257
.3 |  1  2  2  1  4  1  4  7  1  4  A054995
.4 |  1  1  2  2  1  5  2  6  1  5  A088333
.5 |  1  2  1  2  2  1  6  3  8  3  A181281
.6 |  1  1  1  3  4  4  3  1  7  3
.7 |  1  2  3  2  4  5  5  4  2  9  A178853
.8 |  1  1  3  3  1  3  4  4  3  1  A109630
.9 |  1  2  2  3  2  5  7  8  8  7
10 |  1  1  2  4  4  2  5  7  8  8

William Rex Marshall, First 141 antidiagonals of array, flattened
Index entries for sequences related to the Josephus Problem

Cf. A000027 (k = 1), A006257 (k = 2), A054995 (k = 3), A088333 (k = 4), A181281 (k = 5), A178853 (k = 7), A109630 (k = 8).

Cf. A003418, A007495 (main diagonal), A032434, A198789, A198790.

T(k,1) = 1;

for n > 1: T(k,n) = ((T(k,n-1) + k - 1) mod n) + 1.

A032435 Triangle of second-to-last man to survive in Josephus problem of n men in a circle with every k-th killed, with 1 <= k <= n and n >= 2.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 2, 4, 3, 2, 1, 2, 5, 1, 1, 5, 1, 4, 6, 3, 1, 2, 1, 3, 4, 7, 1, 4, 6, 3, 1, 3, 4, 8, 3, 1, 1, 2, 7, 1, 3, 7, 9, 5, 4, 5, 3, 3, 8, 1, 6, 4, 10, 7, 2, 9, 1, 9, 4, 1, 4, 3, 4, 11, 1, 5, 1, 1, 3, 11, 5, 1, 1, 3, 2, 12, 3, 8, 5, 6, 9, 5, 4, 10, 2, 1, 1, 7, 13, 5, 2, 9, 2, 1, 12, 7, 5

Offset: 2

N. J. A. Sloane

			Triangle T(n,k) (with rows n >= 2 and columns k = 2..n) begins
   1, 1;
   2, 1, 1;
   3, 1, 1, 2;
   4, 3, 2, 1, 2;
   5, 1, 1, 5, 1, 4;
   6, 3, 1, 2, 1, 3,  4;
   7, 1, 4, 6, 3, 1,  3, 4;
   8, 3, 1, 1, 2, 7,  1, 3, 7;
   9, 5, 4, 5, 3, 3,  8, 1, 6, 4;
  10, 7, 2, 9, 1, 9,  4, 1, 4, 3, 4;
  11, 1, 5, 1, 1, 3, 11, 5, 1, 1, 3, 2;
  ...

W. W. R. Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th ed., New York: Dover, pp. 32-36, 1987.
M. Kraitchik, "Josephus' Problem", Sec. 3.13 in Mathematical Recreations, New York: W. W. Norton, pp. 93-94, 1942.
Eric W. Weisstein, The CRC Concise Encyclopedia in Mathematics, 2nd ed., Chapman and Hall/CRC, 2002. [The first 8 rows of the triangle appear on p. 1595 of this book under the topic "Josephus Problem".]

W. W. R. Ball, Mathematical Recreations and Essays, 4th ed., New York: The MacMillan Company, 1905 (see "Decimation" on pp. 19-20).
Sean A. Irvine, A032435 and A032436 Josephus problem data mismatch, message in seqfan, June 2020.
F. Jakóbczyk, On the generalized Josephus problem, Glasow Math. J. 14(2) (1973), 168-173. [It contains algorithms that allow the identification of the original position of the second-to-last person to survive in Josephus problem.]
M. Kraitchik, "Josephus' Problem", Sec. 3.13 in Mathematical Recreations, New York: W. W. Norton, pp. 93-94, 1942. [Available only in the USA through the Hathi Trust Digital Library.]
Eric Weisstein's World of Mathematics, Josephus Problem. [It contains a new, apparently corrected, triangle.]
Wikipedia, Josephus problem.
Index entries for sequences related to the Josephus Problem

Cf. A032434, A032436.

cp's OEIS Frontend

A006257 Josephus problem: a(2*n) = 2*a(n)-1, a(2*n+1) = 2*a(n)+1.

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A321298 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in the Josephus elimination process for n people and a count of 2, 1 <= k <= n.

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A007495 Josephus problem: survivors.

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A198789 Array T(n,k) read by antidiagonals: Last survivor positions in Josephus problem for n numbers and a count of k, n >= 1, k >= 1.

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A321781 Least q > 1 letting Josephus survive if he finds himself at position j in the circle of m persons, but is allowed to name the elimination parameter q such that every q-th person is executed, written as triangle T(m,j), m > 1, j <= m.

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A006257 Josephus problem: a(2n) = 2a(n)-1, a(2n+1) = 2a(n)+1.