A114604 - OEIS

%I #14 Jul 15 2024 10:17:58

%S 1,5,43,709,23003,1481957,190305691,48796386661,25003673060507,

%T 25613941912987493,52467767892904362139,214929296497738201165669,

%U 1760788099067877263041671323,28849467307107603960961499533157

%N Numerator of partial sums of A005329/A006125.

%C To win a game, you must flip n+1 heads in a row, where n is the total number of tails flipped so far. The probability of having won before n+1 tails (that is, winning by flipping n+1 or fewer heads in a row) is a(n)/A006125(n). The probability of winning for the first time after n tails (that is, by flipping n+1 heads in a row) is A005329(n)/A006125(n).

%H Michael De Vlieger, <a href="/A114604/b114604.txt">Table of n, a(n) for n = 0..80</a>

%H Johann Cigler, <a href="https://arxiv.org/abs/2407.05768">Hankel determinants of backward shifts of powers of q</a>, arXiv:2407.05768 [math.CO], 2024. See p. 4.

%F a(n) = numerator(Sum_{k=0..n} A005329(k)/A006125(k)).

%F a(n) = a(n-1) * 2^(n+1) + A005329(n).

%e a(3) = 43 because 1/2 + 1/8 + 3/64 = 43/64, or because a(2) * 2^(2+1) + A005329(2) = 5 * 8 + 3 = 43.

%t Nest[Append[#1, #1[[-1]]*2^(#2 + 1) + Product[2^i - 1, {i, #2}]] & @@ {#, Length[#]} &, {1}, 13] (* _Michael De Vlieger_, Jul 15 2024 *)

%Y Cf. A005329, A006125.

%K easy,frac,nonn

%O 0,2

%A _Joshua Zucker_, Dec 14 2005