Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the 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