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

0,2

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.

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

Mathematica

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

Prog

(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

Crossrefs

Cf. A000680.

Sequence in context: A305005 A326267 A337043 * A039718 A307249 A201381

Adjacent sequences: A008266 A008267 A008268 * A008270 A008271 A008272

Keyword

nonn

Author

Lily Yen

Status

approved