A337191 A version of the Josephus problem: a(n) is the surviving integer under the skip-eliminate-eliminate version of the elimination process.
1, 1, 1, 4, 4, 1, 7, 4, 1, 7, 4, 10, 7, 13, 10, 16, 13, 1, 16, 4, 19, 7, 22, 10, 25, 13, 1, 16, 4, 19, 7, 22, 10, 25, 13, 28, 16, 31, 19, 34, 22, 37, 25, 40, 28, 43, 31, 46, 34, 49, 37, 52, 40, 1, 43, 4, 46, 7, 49, 10, 52, 13, 55, 16, 58, 19, 61, 22, 64, 25, 67
Offset: 1
Keywords
Examples
Consider 4 people in a circle in order 1,2,3,4. In the first round, person 1 is skipped and persons 2 and 3 are eliminated. Now people are in order 4,1. In the second round, person 4 is skipped and person 1 is eliminated. Person 4 is freed. Thus, a(4) = 4. - _Tanya Khovanova_, Apr 14 2025
Links
Programs
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Mathematica
nxt[{n_,a_,b_}]:={n+1,b,If[Mod[a+3,n+1]!=0,Mod[a+3,n+1],n+1]}; NestList[nxt,{2,1,1},70][[;;,2]] (* Harvey P. Dale, Jul 27 2024 *) -
PARI
a(n) = if (n <= 2, 1, my(x = (a(n-2) + 3) % n); if (x, x, n)); \\ Michel Marcus, Aug 20 2020
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PARI
a(n) = if (n<=1, return(1)); my(v=vector(n, i, i), w); while (#v > 3, if (#v <=3, w = [], w = vector(#v-3, k, v[k+3])); w = concat(w, Vec(v, 1)); v = w;); v[1]; \\ Michel Marcus, Mar 25 2025
Formula
a(1) = 1, a(2) = 1, a(n) = (a(n-2) + 3) (mod n) if (a(n-2) + 3) (mod n) is not 0; a(n) = n if (a(n-2) + 3) (mod n)=0.
Any number n can be written as either 2*(3^k) + 2m (where 0 <= m < 3^k, k = 0,1,2,...) or 3^k + 2m (where 0 <= m < 3^k, k = 0,1,2,...), in either case a(n) = 3m + 1.
Extensions
More terms from Michel Marcus, Aug 20 2020
Title corrected by Tanya Khovanova, Apr 14 2025
Comments