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 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. 6
 2, 16, 104, 128, 3872, 3328, 139904, 167936, 5038592, 2748416, 7886848, 2392064, 6530342912, 39182073856, 235092475904, 16594763776, 8463329656832, 381804347392, 304679869743104, 6647560798208, 10968475319140352, 2861341387784192, 8401385351544832, 5207012459675648 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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. LINKS Paolo Xausa, Table of n, a(n) for n = 1..1000 IBM Research, A Dice Game, Ponder This Challenge, February 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 EXAMPLE 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. MATHEMATICA Array[Numerator[(3^#-1)/((3/2)^#-1)/2] &, 35] (* Paolo Xausa, Mar 13 2024 *) PROG (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<

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Last modified July 15 11:27 EDT 2024. Contains 374332 sequences. (Running on oeis4.)