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A265920 Numerator of the probability that Alice wins the following game: Alice and Bob take turn (Alice starts first) to gain 1 or 2 chips randomly and independently with 1/2 chance, and the first player that collects at least n chips is the winner. 1
1, 3, 13, 47, 185, 723, 2821, 11127, 43825, 173147, 685181, 2713919, 10762793, 42715619, 169654133, 674238983, 2680944545, 10665068907, 42443750893, 168973210575, 672913173913, 2680539263219, 10680581419493, 42566341729431, 169678604019217, 676501889994363 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

The formula is proved by Wing Hong Tony Wong and Jiao Xu, and then proved by Taoye Zhang and Ju Zhou independently.

The probability that Alice wins the game is a(n)/4^(n-1).

LINKS

Table of n, a(n) for n=1..26.

FORMULA

a(n) = (4^n + Sum_(k=1..n) 4^(n-k)*(binomial(k, n-k)+binomial(k-1, n-k))^2)/8.

MATHEMATICA

Table[(4^n + Sum[4^(n - k)*(Binomial[k, n - k] + Binomial[k - 1, n - k])^2, {k, n}])/8, {n, 100}]

CROSSREFS

A265919 provides the integer sequence of the numerator of the probability that Bob wins the game. The corresponding terms of these two sequences add up to 4^n.

Sequence in context: A228529 A084519 A304628 * A262322 A180278 A193164

Adjacent sequences:  A265917 A265918 A265919 * A265921 A265922 A265923

KEYWORD

nonn

AUTHOR

Wing Hong Tony Wong, Jiao Xu, Dec 18 2015

STATUS

approved

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Last modified July 1 06:01 EDT 2022. Contains 354952 sequences. (Running on oeis4.)