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A121689 G.f.: Sum_{n>=0} x^n * (1+x)^(n^2).

%I A121689 #23 Sep 12 2021 13:50:00
%S A121689 1,1,2,5,16,57,231,1023,4926,25483,140601,822422,5074015,32881868,
%T A121689 223027542,1578435549,11625317128,88894615929,704269188135,
%U A121689 5770209550496,48810504348082,425650324975153,3821377057170313
%N A121689 G.f.: Sum_{n>=0} x^n * (1+x)^(n^2).
%H A121689 Seiichi Manyama, <a href="/A121689/b121689.txt">Table of n, a(n) for n = 0..613</a>
%H A121689 MathOverflow, <a href="http://mathoverflow.net/questions/154528/asymptotic-behaviour-of-sequence">Asymptotic behaviour of sequence</a>
%F A121689 a(n) = Sum_{k=0..n} C(k^2,n-k).
%F A121689 From _Paul D. Hanna_, Apr 24 2010: (Start)
%F A121689 Let q = (1+x), then g.f. A(x) equals the continued fraction:
%F A121689 A(x) = 1/(1- q*x/(1- (q^3-q)*x/(1- q^5*x/(1- (q^7-q^3)*x/(1- q^9*x/(1- (q^11-q^5)*x/(1- q^13*x/(1- (q^15-q^7)*x/(1- ...)))))))))
%F A121689 due to an identity of a partial elliptic theta function.
%F A121689 (End)
%F A121689 G.f.: Sum_{n>=0} x^n * (1+x)^n * Product_{k=1..n} (1 - x*(1+x)^(4*k-3)) / (1 - x*(1+x)^(4*k-1)) due to a q-series identity. - _Paul D. Hanna_, May 08 2010
%e A121689 G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 16*x^4 + 57*x^5 + 231*x^6 + ...
%e A121689 where
%e A121689 A(x) = 1 + x*(1+x) + x^2*(1+x)^4 + x^3*(1+x)^9 + x^4*(1+x)^16 + x^5*(1+x)^25 + x^6*(1+x)^36 + x^7*(1+x)^49 + x^8*(1+x)^64 + ... + x^n*(1+x)^(n^2) + ...
%t A121689 Table[Sum[Binomial[k^2,n-k],{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Mar 06 2014 *)
%o A121689 (PARI) a(n)=sum(k=0,n,binomial(k^2,n-k))
%o A121689 (PARI) {a(n)=polcoeff(sum(m=0,n,x^m*(1+x)^m*prod(k=1,m,(1-x*(1+x)^(4*k-3))/(1-x*(1+x)^(4*k-1) + x*O(x^n)))),n)} \\ _Paul D. Hanna_, May 08 2010
%Y A121689 Cf. A217285.
%K A121689 nonn
%O A121689 0,3
%A A121689 _Paul D. Hanna_, Aug 15 2006