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A121689 G.f.: Sum_{n>=0} x^n * (1+x)^(n^2). 10

%I #23 Sep 12 2021 13:50:00

%S 1,1,2,5,16,57,231,1023,4926,25483,140601,822422,5074015,32881868,

%T 223027542,1578435549,11625317128,88894615929,704269188135,

%U 5770209550496,48810504348082,425650324975153,3821377057170313

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

%H Seiichi Manyama, <a href="/A121689/b121689.txt">Table of n, a(n) for n = 0..613</a>

%H MathOverflow, <a href="http://mathoverflow.net/questions/154528/asymptotic-behaviour-of-sequence">Asymptotic behaviour of sequence</a>

%F a(n) = Sum_{k=0..n} C(k^2,n-k).

%F From _Paul D. Hanna_, Apr 24 2010: (Start)

%F Let q = (1+x), then g.f. A(x) equals the continued fraction:

%F 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 due to an identity of a partial elliptic theta function.

%F (End)

%F 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 G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 16*x^4 + 57*x^5 + 231*x^6 + ...

%e where

%e 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 Table[Sum[Binomial[k^2,n-k],{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Mar 06 2014 *)

%o (PARI) a(n)=sum(k=0,n,binomial(k^2,n-k))

%o (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 Cf. A217285.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Aug 15 2006

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