A265919 - OEIS

%I #21 Nov 29 2018 16:00:25

%S 0,1,3,17,71,301,1275,5257,21711,88997,363395,1480385,6014423,

%T 24393245,98781323,399502841,1614022751,6514800277,26275725843,

%U 105904696369,426598453863,1717507247885,6911604624923,27802402448233,111796372691439,449398016848261

%N Numerator of the probability that Bob 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.

%C The formula is proved by _Wing Hong Tony Wong_ and _Jiao Xu_, and then proved by Taoye Zhang and Ju Zhou independently.

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

%H Michael De Vlieger, <a href="/A265919/b265919.txt">Table of n, a(n) for n = 1..1662</a>

%H Tony W. H. Wong, Jiao Xu, <a href="https://www.emis.de/journals/JIS/VOL21/Wong2/wong8.html">A Probabilistic Take-Away Game</a>, J. Int. Seq., Vol. 21 (2018), Article 18.6.3.

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

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

%Y A265920 provides the integer sequence of the numerator of the probability that Alice wins the game. The corresponding terms of these two sequences add up to 4^n.

%K nonn

%O 1,3

%A _Wing Hong Tony Wong_, _Jiao Xu_, Dec 18 2015