A121689 G.f.: Sum_{n>=0} x^n * (1+x)^(n^2).
1, 1, 2, 5, 16, 57, 231, 1023, 4926, 25483, 140601, 822422, 5074015, 32881868, 223027542, 1578435549, 11625317128, 88894615929, 704269188135, 5770209550496, 48810504348082, 425650324975153, 3821377057170313
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 16*x^4 + 57*x^5 + 231*x^6 + ... where 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) + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..613
- MathOverflow, Asymptotic behaviour of sequence
Crossrefs
Cf. A217285.
Programs
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Mathematica
Table[Sum[Binomial[k^2,n-k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Mar 06 2014 *) -
PARI
a(n)=sum(k=0,n,binomial(k^2,n-k))
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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
Formula
a(n) = Sum_{k=0..n} C(k^2,n-k).
From Paul D. Hanna, Apr 24 2010: (Start)
Let q = (1+x), then g.f. A(x) equals the continued fraction:
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- ...)))))))))
due to an identity of a partial elliptic theta function.
(End)
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