%I #25 Sep 04 2024 10:56:32

%S 1,3,12,30,90,210,560,1260,3150,6930,16632,36036,84084,180180,411840,

%T 875160,1969110,4157010,9237800,19399380,42678636,89237148,194699232,

%U 405623400,878850700,1825305300,3931426800,8143669800,17450721000

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

%C 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.

%H G. C. Greubel, <a href="/A107231/b107231.txt">Table of n, a(n) for n = 0..1000</a>

%F 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).

%F 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]

%F 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

%F G.f.: (1+x)/((1+2*x)^(3/2)*(1-2*x)^(5/2)). - _Vladimir Reshetnikov_, Aug 01 2018

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

%t Table[Binomial[n + 2, 2]*Binomial[n, Floor[n/2]], {n,0,50}] (* _G. C. Greubel_, Jun 13 2017 *)

%o (PARI) for(n=0,50, print1(binomial(n+2,2)*binomial(n,n\2), ", ")) \\ _G. C. Greubel_, Jun 13 2017

%Y Cf. A000217, A001405.

%Y Cf. A001318, A034828, A107230.

%K easy,nonn

%O 0,2

%A _Paul Barry_, May 13 2005