A114604 Numerator of partial sums of A005329/A006125.

1, 5, 43, 709, 23003, 1481957, 190305691, 48796386661, 25003673060507, 25613941912987493, 52467767892904362139, 214929296497738201165669, 1760788099067877263041671323, 28849467307107603960961499533157

Offset: 0

Author

Joshua Zucker, Dec 14 2005

Keywords

• easy
• frac
• nonn

Comments

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).

Examples

			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.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..80
• Johann Cigler, Hankel determinants of backward shifts of powers of q, arXiv:2407.05768 [math.CO], 2024. See p. 4.

Crossrefs

Cf. A005329, A006125.

Programs

Mathematica

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

Formula

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

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