A107231 a(n) = C(n+2,2)*C(n,floor(n/2)).

1, 3, 12, 30, 90, 210, 560, 1260, 3150, 6930, 16632, 36036, 84084, 180180, 411840, 875160, 1969110, 4157010, 9237800, 19399380, 42678636, 89237148, 194699232, 405623400, 878850700, 1825305300, 3931426800, 8143669800, 17450721000

Offset

0,2

Comments

Third column of A107230. Related to the generalized pentagonal numbers A001318. The sequence 0,0,1,3,12,... is an inverse Chebyshev transform of 0,0,1,3,8,... (see A034828). This transform maps a g.f. g(x) to (1/sqrt(1-4x^2))g(c(x^2)). Thus A001318, as first differences of A034828, can be expressed in terms of A107231.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Formula

G.f.: (1+x)*(1-sqrt(1-4*x^2))^3*(sqrt(1-4*x^2)-4*x^2+1)^2/(8*x^4*(1-4*x^2)^(5/2)*(sqrt(1-4*x^2)+2*x-1)^2).

a(n) = Sum_{k=0..floor((n+2)/2)} binomial(n+2, k)*A034828(n+2-2*k). [corrected by Jason Yuen, Sep 02 2024]

Conjecture: n*a(n) +(n-4)*a(n-1) +2*(-2*n-5)*a(n-2) -4*n*a(n-3)=0. - R. J. Mathar, Nov 24 2012

G.f.: (1+x)/((1+2*x)^(3/2)*(1-2*x)^(5/2)). - Vladimir Reshetnikov, Aug 01 2018

Sum_{n>=0} 1/a(n) = Pi^2/9 - 2*Pi/sqrt(3) + 4. - Amiram Eldar, Sep 03 2024

Mathematica

Table[Binomial[n + 2, 2]*Binomial[n, Floor[n/2]], {n, 0, 50}] (* G. C. Greubel, Jun 13 2017 *)

Prog

(PARI) for(n=0, 50, print1(binomial(n+2, 2)*binomial(n, n\2), ", ")) \\ G. C. Greubel, Jun 13 2017

Crossrefs

Cf. A000217, A001405.

Cf. A001318, A034828, A107230.

Sequence in context: A296289 A089143 A073952 * A363913 A352157 A293656

Adjacent sequences: A107228 A107229 A107230 * A107232 A107233 A107234

Keyword

easy,nonn

Author

Paul Barry, May 13 2005

Status

approved