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 A370827 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 3:2. To win the game it is necessary to win n rounds in a row. 2
 3, 63, 1053, 16443, 250533, 28431, 56859813, 853737003, 12811093893, 17472421953, 2883131020773, 25013333547, 648727335888453, 9730949220408843, 145964473398624933, 128792265384372219, 32842036136344638213, 3703989419737954191, 7389459197057088616293, 10076535434903231752353 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS If you play the game with odds of 3:2 with one dice, you can decide what happens to the unused number. In the present sequence, this number is ignored without effect. However, if one defines the occurrence of the 6th number as a draw, which interrupts consecutive wins by both players, then the sequence pair A370825/A370826 results. The addition of a draw increases the odds ratio in favor of the player who has the higher chance of winning a single round. The relative advantage is small with 2 games to be won, i.e., 2/(63/32)-1 = 1/63, with 3 rounds 1/27, with 4 rounds 965/16443, but increases with a higher number of consecutive rounds to be won, and reaches asymptotically 1/8. LINKS Table of n, a(n) for n=1..20. FORMULA a(n)/A370828(n) = (2/3) * (3/5)^n * ((5/2)^n - 1) / (1 - (3/5)^n). EXAMPLE a(n)/A370828(n) for n = 1..8: 3/2, 63/32, 1053/392, 16443/4352, 250533/46112, 28431/3584, 56859813/4860032, 853737003/49160192. PROG (PARI) a370827(n) = numerator((2/3) * (3/5)^n * ((5/2)^n - 1) / (1 - (3/5)^n)) (Python) from math import gcd def A370827(n): return (a:=3**(n-1)*(5**n-(1<

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Last modified June 21 20:04 EDT 2024. Contains 373559 sequences. (Running on oeis4.)