%I #23 Aug 04 2024 19:11:08
%S 1,1,3,12,57,295,1620,9261,54459,326952,1994085,12313323,76795452,
%T 482903473,3057550167,19472859840,124645558881,801367484155,
%U 5172056157960,33494998326741,217581335698815,1417270332517704,9254594901658689,60567020813141991
%N G.f.: Sum_{n>=0} [Sum_{k>=0} C(n+k-1,k)^2*x^k]^2 * x^n.
%C Compare g.f. to a g.f. of the Central binomial coefficients (A000984):
%C Sum_{n>=0} [Sum_{k>=0} C(n+k,k)^2*x^k] * x^n.
%H G. C. Greubel, <a href="/A178807/b178807.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: 1 + x * Sum_{n>=0} (Sum_{k=0..n} C(n,k)^2 * x^k)^2 * x^n /(1-x)^(4*n+2).
%F G.f.: 1 + x / AGM(1-7*x+x^2, 1+x+x^2), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean. - _Paul D. Hanna_, Aug 31 2014
%F a(n) ~ phi^(4*n) / (4*Pi*n), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Jan 24 2019
%e G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 57*x^4 + 295*x^5 + 1620*x^6 +...
%e which equals the sum of the series:
%e A(x) = 1 + (1 + x + x^2 + x^3 + x^4 + x^5 +...)^2*x
%e + (1 + 2^2*x + 3^2*x^2 + 4^2*x^3 + 5^2*x^4 + 6^2*x^5 +...)^2*x^2
%e + (1 + 3^2*x + 6^2*x^2 + 10^2*x^3 + 15^2*x^4 + 21^2*x^5 +...)^2*x^3
%e + (1 + 4^2*x + 10^2*x^2 + 20^2*x^3 + 35^2*x^4 + 56^2*x^5 +...)^2*x^4
%e + (1 + 5^2*x + 15^2*x^2 + 35^2*x^3 + 70^2*x^4 + 126^2*x^5 +...)^2*x^5
%e + (1 + 6^2*x + 21^2*x^2 + 56^2*x^3 + 126^2*x^4 + 252^2*x^5 +...)^2*x^6 +...
%e The g.f. can also be expressed as:
%e A(x) = 1 + x/(1-x)^2 + (1+x)^2*x^2/(1-x)^6
%e + (1 + 2^2*x + x^2)^2*x^3/(1-x)^10
%e + (1 + 3^2*x + 3^2*x^2 + x^3)^2*x^4/(1-x)^14
%e + (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)^2*x^5/(1-x)^18
%e + (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5)^2*x^6/(1-x)^22
%e + (1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)^2*x^7/(1-x)^26 +...
%t With[{m=30}, CoefficientList[Series[1 +x*Sum[Sum[Binomial[n,k]^2*x^k, {k,0,n}]^2* x^n/(1-x)^(4*n+2), {n, 0, m}], {x, 0, m}], x]] (* _G. C. Greubel_, Jan 23 2019 *)
%o (PARI) {a(n)=polcoef(sum(m=0,n,sum(k=0,n,binomial(m+k-1,k)^2*x^k)^2*x^m)+x*O(x^n),n)}
%o (PARI) {a(n)=polcoef(1+x*sum(m=0,n,sum(k=0,m,binomial(m,k)^2*x^k)^2*x^m/(1-x+x*O(x^n))^(4*m+2)),n)}
%o (PARI) {a(n)=polcoef( 1 + x / agm(1-7*x+x^2, 1+x+x^2 +x*O(x^n)), n)}
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 26 2010