1,4

This variation of the Josephus problem is related to under-down-down card dealing. - Tanya Khovanova, Apr 14 2025

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Eric Huang, Tanya Khovanova, Timur Kilybayev, Ryan Li, Brandon Ni, Leone Seidel, Samarth Sharma, Nathan Sheffield, Vivek Varanasi, Alice Yin, Boya Yun, and William Zelevinsky, Card Dealing Math, arXiv:2509.11395 [math.NT], 2025. See p. 17.

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.

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

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

(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

Cf. A006257, A225381, A321298, A378635, A381048, A381049, A382528.

Sequence in context: A247252 A016495 A335826 * A341863 A047213 A128213

Adjacent sequences: A337188 A337189 A337190 * A337192 A337193 A337194

Robert W. Vallin, Aug 18 2020

More terms from Michel Marcus, Aug 20 2020

Title corrected by Tanya Khovanova, Apr 14 2025

approved