A165556 -id:A165556 - cp's OEIS frontend

A325594 Last remaining numbers after a symmetric variation of the Josephus problem.

1, 1, 3, 4, 3, 6, 1, 3, 9, 1, 11, 5, 11, 7, 9, 14, 5, 12, 7, 12, 11, 14, 9, 22, 5, 20, 7, 28, 3, 30, 1, 11, 25, 9, 27, 5, 35, 7, 33, 3, 41, 1, 43, 5, 43, 7, 41, 19, 33, 17, 35, 13, 43, 15, 41, 27, 33, 25, 35, 29, 35, 31

Offset: 1

Gordon Atkinson, Sep 07 2019

nonn

A variant of the Josephus problem where two numbers are eliminated at every stage, one elimination clockwise, the other counterclockwise. To resolve ambiguities, the usual Josephus problem takes precedence.

Hiroshi Matsui, Toshiyuki Yamauchi, Soh Tatsumi, Takahumi Inoue, Masakazu Naito and Ryohei Miyadera, Interesting Variants of the Josephus Problem, Computer Algebra - Design of Algorithms, Implementations and Applications, Kokyuroku, The Research Institute of Mathematical Science, No. 1652, (2009), 44-54.
Masakazu Naito and Ryohei Miyadera, The Josephus Problem in Both Directions, The Wolfram Demonstrations Project.
Masakazu Naito, Sohtaro Doro, Daisuke Minematsu and Ryohei Miyadera, The Self-Similarity of the Josephus Problem and its Variants, Visual Mathematics, Volume 11, No. 2, 2009.
Index entries for sequences related to the Josephus Problem

Cf. A006257, A165556.

joseboth[m_, mm_] := Block[{t, p, q, u, v, w}, w = mm - 1; t = Range[m]; p = t; q = t; Do[p = RotateLeft[p, w]; u = First[p]; p = Rest[p]; q = Drop[q, Position[q, u][[1]]]; If[Length[p] == 1, Break[],]; q = RotateRight[q, w]; v = Last[q]; q = Drop[q, -1]; p = Drop[p, Position[p, v][[1]]]; If[Length[q] == 1, Break[],], {n, 1, Ceiling[m/2]}]; p[[1]]];

A165556(n) = a(n) mod 2.

A165728 If we divide the sequence into these subsequences, the pattern is obvious. {{1,1}, {0,1}, {1,1}}, {{0,1,0,1}, {1,1,1,1}, {0,1,0,1}}, {{1,1,1,1,1,1,1,1}, {0,1,0,1,0,1,0,1}, {1,1,1,1,1,1,1,1}}, {{0,1,0,1,0,1,0,1,0,1,0,1,0,1,0}, {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1}, {0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1}}, ...

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

Offset: 1

Ryohei Miyadera and Masakazu Naito, Sep 25 2009

nonn

This is a sequence made by a variant of the Josephus Problem mod 2, but we use the last number to eliminate instead the last number that remains.
We put n numbers in a circle, and in this variant two numbers are to be eliminated at the same time.
These two processes of elimination go in different directions. Suppose that there are n numbers. Then the first process of elimination starts with the first number and the 2nd, 4th, 6th numbers, ... are to be eliminated.
The second process starts with the n-th number, and the (n-1)-st, (n-3)-rd, (n-5)-th number, ... are to be eliminated.
We suppose that the first process comes first and the second process second at every stage.
We denote by JI2(n) the position of the last number to be eliminated when we have n numbers.
If we use this sequence JI2(n) for n = 6,7,8,... under mod 2, then we get the above sequence with 1 and 0.
Note that we have to omit the first 5 terms to get this sequence with its beautiful pattern.

			Suppose that there are n = 14 numbers. Then the 2nd, 4th, and 6th numbers will be eliminated by the first process. Similarly 13th, 11th, and 9th numbers will be eliminated by the second process. Now two directions are going to overlap. The first process will eliminate the 8 and 12 and the second process will eliminate 5 and 1. After this the first process will eliminate 3 and 14, and the second process will eliminate 10. The number that remains is 7, and hence the last number to be eliminated is 14. Therefore JI2(14) = 14. JI2(14) = 0 (mod 2).

Hiroshi Matsui, Toshiyuki Yamauchi, Soh Tatsumi, Takahumi Inoue, Masakazu Naito and Ryohei Miyadera, Interesting Variants of the Josephus Problem, Computer Algebra - Design of Algorithms, Implementations and Applications, Kokyuroku, The Research Institute of Mathematical Science, No. 1652, (2009), 44-54.
Masakazu Naito and Ryohei Miyadera, The Josephus Problem in Both Directions, The Wolfram Demonstrations Project.
Masakazu Naito, Daisuke Minematsu and Ryohei Miyadera, The Self-Similarity of the Josephus Problem and its Variants, Visual Mathematics, 11(2) (2009).
Index entries for sequences related to the Josephus Problem

Cf. A113648, A114144, A165556.

last2 = {1, 2, 1, 1, 1, 3, 5, 6, 5, 5}; Table[JI2[n] = last2[[n]], {n, 1, 10}]; JI2[m_] := JI2[m] = Block[{n, h}, h = Mod[m, 8]; n = (m - h)/8; Which[h == 0, 4 JI2[2 n] - 1 - Floor[JI2[2 n]/(n + 1)], h == 1, 8 n + 5 - 4 JI2[2 n], h == 2, 4 JI2[2 n] -3 -Floor[JI2[2 n]/(n + 2)], h == 3, 8 n + 7 - 4 JI2[2 n], h == 4, 8 n + 8 - 4 JI2[2 n + 1] + Floor[JI2[2 n + 1]/(n + 2)], h == 5, 4 JI2[2 n + 1] - 1, h == 6, 8 n + 10 - 4 JI2[2 n + 1] + Floor[JI2[2 n + 1]/(n + 2)], h == 7, 4 JI2[2 n + 1] - 3]]; Table[Mod[JI2[n], 2], {n, 6, 95}]

{JI2(n): n = 1,2,3,4,5,6,7,8} = {1, 2, 1, 1, 1, 3, 5}.
(1) JI2(8*n) = 4*JI2(2*n) - 1 - [JI2(2*n)/(n+1) ].
(2) JI2(8*n+1) = 8*n + 5 - 4*JI2(2*n).
(3) JI2(8*n+2) = 4*JI2(2*n) - 3 - [JI2(2*n)/(n + 2)] .
(4) JI2(8*n+3) = 8*n + 7 - 4*JI2(2*n).
(5) JI2(8*n+4) = 8*n + 8 - 4*JI2(2*n+1) + [JI2(2*n+1)/(n+2)].
(6) JI2(8*n+5) = 4*JI2(2*n+1) - 1.
(7) JI2(8*n+6) = 8*n + 10 - 4*JI2(2*n+1) + [JI2(2*n+1)/(n+2)].
(8) JI2(8*n+7) = 4*JI2(2*n+1) - 3,
Note that recurrence relations are the same as those of A165556, but initial values are different.

A165735 Surviving integers under the double-count Josephus problem (see A054995), modulo 3.

Original entry on oeis.org

1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Ryohei Miyadera and Masakazu Naito, Sep 25 2009

nonn

Old name was: The pattern is obvious. The sequence can be divided into subsequences of {1,1,1,...} and {2,2,2,...}.

Let n be a natural number. We put n numbers in a circle, and we are going to remove every third number. Let J3(n) be the last number that remains. This is the traditional Josephus Problem. Let J3 (mod 3) be the residue of the sequence J3(n) under mod 3. J3 (mod 3) produces the sequence {1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2,...}.

			If we use n = 10, then we put numbers 1,2,3,4,5,6,7,8,9,10 in a circle. We eliminate 3,6,9,2,7,1,8,5,10, and the last number that remains is 4. Therefore J3(10) = 4 and J3(10) = 1 mod 3.

Hiroshi Matsui, Masakazu Naito and Naoyuki Totani, The Period and the Distribution of the Fibonacci-like Sequence Under Various Moduli, Undergraduate Math Journal, Rose-Hulman Institute of Technology, Vol. 10, Issue 1, 2009.
Masakazu Naito and Ryohei Miyadera, The Self-Similarity of the Josephus Problem and its Variants, Visual Mathematics, Volume 11, No.2, 2009.
Wolfram MathWorld, Josephus Problem
Index entries for sequences related to the Josephus Problem

Cf. A010872, A054995, A114144, A113648, A165556.

J3[1] = 1; J3[2] = 2; J3[n_] := J3[n] = Block[{m, t}, t = Mod[n, 3]; m = (n - t)/3; Which[t == 0, J3[2 m] + Floor[(J3[2 m] - 1)/2], t == 1, If[J3[2 m + 1] == 1, 3 m + 1, J3[2 m + 1] + Floor[J3[2 m + 1]/2] - 2], t == 2, J3[2 m + 1] + Floor[J3[2 m + 1]/2] + 1]]; Table[Mod[J3[n], 3], {n, 1, 200}]

(1) J3(1) = 1 and J3(2) = 2.
(2) J3(3m) = J3(2m) + [(J3(2m)-1)/2].
(3a) J3(3m+1) = 3m + 1 (if J3(2m + 1) = 1).
(3b) J3(3m+1) = J3(2m+1) + [J3(2m+1)/2] - 2 (if J3(2m + 1) > 1).
(4) J3(3m+2) = J3(2m+1) + [J3(2m+1)/2] + 1
a(n) = A010872(A054995(n)). - Gordon Atkinson, Aug 21 2019

New name from Gordon Atkinson, Aug 21 2019

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A325594 Last remaining numbers after a symmetric variation of the Josephus problem.

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A165735 Surviving integers under the double-count Josephus problem (see A054995), modulo 3.

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