A008269 Number of strings on n symbols in Stockhausen problem.

1, 2, 9, 112, 2921, 126966, 8204497, 735944084, 87394386417, 13265365173706, 2504688393449081, 575664638548522392, 158222202503521622809, 51242608446417388426622, 19312113111034490954560641, 8379247307752508262094697596, 4146836850351947542340780899937

Offset: 0

Author

Lily Yen

Keywords

• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 0..235
• R. C. Read, Combinatorial problems in theory of music, Discrete Math. 167 (1997), 543-551.
• Ronald C. Read, Lily Yen, A note on the Stockhausen problem, J. Comb. Theory, Ser. A 76, No. 1, 1-10.

Crossrefs

Cf. A000680.

Programs

Mathematica

Table[HypergeometricPFQ[{1,1/2,-n},{},-2],{n,0,20}] (* Vaclav Kotesovec, Feb 18 2015 *)

PARI

for(n=0,14,print1(2^(-n)*round(intnum(x=0,999,(2+x^2)^n*exp(-x))),", ")) \\ Gerald McGarvey, Oct 12 2007

PARI

a(n) = sum(i=0, n, binomial(n,i) * (2*i)!/2^i); \\ Michel Marcus, May 13 2022

Formula

a(n) = (2*n^2-5*n+4)*a(n-1) + (-4*n^2+15*n-14)*a(n-2) + (2*n^2-10*n+12)*a(n-3).

a(n) = hypergeom([1, 1/2, -n], [], -2). - Vladeta Jovovic, Apr 08 2007

a(n) = (1/2^n) * Integral_{x>=0} (2+x^2)^n*exp(-x) dx. - Gerald McGarvey, Oct 12 2007

a(n) ~ sqrt(Pi) * 2^(n+1) * n^(2*n+1/2) / exp(2*n). - Vaclav Kotesovec, Feb 18 2015

a(n) = Sum_{i=0..n} binomial(n,i) * A000680(i). - Håvar Andre Melheim Salbu, May 13 2022