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A008269
Number of strings on n symbols in Stockhausen problem.
1
1, 2, 9, 112, 2921, 126966, 8204497, 735944084, 87394386417, 13265365173706, 2504688393449081, 575664638548522392, 158222202503521622809, 51242608446417388426622, 19312113111034490954560641, 8379247307752508262094697596, 4146836850351947542340780899937
OFFSET
0,2
LINKS
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
KEYWORD
nonn
AUTHOR
STATUS
approved