A370823 a(n) is the numerator of the ratio of winning probabilities P_A/P_B of winning in a 2-player game with a ratio of odds for A and B in a single round of 2:1. To win the game it is necessary to win n rounds in a row.

2, 16, 104, 128, 3872, 3328, 139904, 167936, 5038592, 2748416, 7886848, 2392064, 6530342912, 39182073856, 235092475904, 16594763776, 8463329656832, 381804347392, 304679869743104, 6647560798208, 10968475319140352, 2861341387784192, 8401385351544832, 5207012459675648

Offset: 1

Author

Hugo Pfoertner, Mar 02 2024

Keywords

• nonn
• frac

Comments

Such a game can be implemented, for instance, by rolling a single die per round, with A winning the round if the numbers are 1 to 4 and B winning if the numbers are 5 and 6.

Examples

			a(n)/A370824(n) for n = 1..11: 2/1, 16/5, 104/19, 128/13, 3872/211, 3328/95, 139904/2059, 167936/1261, 5038592/19171, 2748416/5275, 7886848/7613.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..1000
• IBM Research, A Dice Game, Ponder This Challenge, February 2024.

Crossrefs

A370824 are the corresponding denominators.

Programs

Mathematica

Array[Numerator[(3^#-1)/((3/2)^#-1)/2] &, 35] (* Paolo Xausa, Mar 13 2024 *)

PARI

a370823_4(n, A=2/3, B=1/3) = my (an=A^n, bn=B^n); (1-A) * an * (1-bn) / ((1-B) * bn * (1-an));
\\ or by determination of the eigenvalues of the Markov matrix
a370823_4(n, na=2, nb=1) = { if (n==1, na/nb, my (ntot=na+nb, A=na/ntot, B=nb/ntot, M=matrix(2*n+1)); M[1,2]=A; M[1,3]=B; for (rp=1, n-1, my (rb=2*rp+1, ra=rb-1);  M[ra,3]=B; M[rb,2]=A; M[ra,ra+2]=A; M[rb,rb+2]=B); M[2*n,2*n]=M[2*n+1,2*n+1]=1; my (ME=mateigen(M)); ME[1,2]/ME[1,3])};
numerator(a370823_4(n))

Python

from math import gcd
def A370823(n): return (a:=3**n-1<Chai Wah Wu, Mar 07 2024

Formula

See the solution page of the "Ponder This" challenge for the formula derived from the Markov matrix representing the rules of the game.

Numerator of 2^(n-1)*(3^n-1)/(3^n-2^n). - Chai Wah Wu, Mar 07 2024