A178325 - OEIS

%I #32 Jan 10 2023 07:23:59

%S 1,1,2,6,21,83,363,1730,8889,48829,284858,1755325,11374092,77208275,

%T 547261631,4039201624,30967330941,246084049137,2023030659970,

%U 17175765057532,150367445873108,1355528352031358,12566899017130088

%N G.f.: A(x) = Sum_{n>=0} x^n/(1-x)^(n^2).

%C Equals the row sums of triangle A214398.

%C a(n) is the number of weak compositions of n such that if the first part is equal to k then there are a total of k^2 + 1 parts.  A weak composition is an ordered partition of the integer n into nonnegative parts. a(3) = 6 because we have: 1+2, 2+0+0+0+1, 2+0+0+1+0, 2+0+1+0+0, 2+1+0+0+0, 3+0+0+0+0+0+0+0+0+0. - _Geoffrey Critzer_, Oct 09 2013

%H Paul D. Hanna, <a href="/A178325/b178325.txt">Table of n, a(n) for n = 0..250</a>

%F a(n) = Sum_{k=0..n} C((n-k)^2 + k-1, k).

%F G.f.: A(x) = Sum_{n>=0} (x-x^2)^n*Product_{k=1..n} ((1-x)^(4*k-3) - x)/((1-x)^(4*k-1) - x) due to a q-series identity.

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

%F . A(x) = 1/(1- q*x/(1- q*(q^2-1)*x/(1- q^5*x/(1- q^3*(q^4-1)*x/(1- q^9*x/(1- q^5*(q^6-1)*x/(1- q^13*x/(1- q^7*(q^8-1)*x/(1- ...)))))))))

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

%F log(a(n)) ~ n*(log(n) - 2) * (1 + log(4*n) - log((log(n) - 2)*log(n))) / log(n). - _Vaclav Kotesovec_, Jan 10 2023

%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 83*x^5 + 363*x^6 +...

%e A(x) = 1 + (x-x^2)*((1-x)-x)/((1-x)^3-x) + (x-x^2)^2*((1-x)-x)*((1-x)^5-x)/(((1-x)^3-x)*((1-x)^7-x)) + (x-x^2)^3*((1-x)-x)*((1-x)^5-x)*((1-x)^9-x)/(((1-x)^3-x)*((1-x)^7-x)*((1-x)^11-x)) +...

%t nn=22;CoefficientList[Series[Sum[x^k/(1-x)^(k^2),{k,0,nn}],{x,0,nn}],x]  (* _Geoffrey Critzer_, Oct 09 2013 *)

%o (PARI) {a(n)=sum(k=0,n,binomial((n-k)^2+k-1,k))}

%o (PARI) {a(n)=polcoeff(sum(m=0,n,x^m/(1-x+x*O(x^n))^(m^2)),n)}

%o (PARI) {a(n)=polcoeff(sum(m=0,n,(x-x^2)^m*prod(k=1,m,((1-x)^(4*k-3)-x)/((1-x)^(4*k-1)-x +x*O(x^n)))),n)}

%Y Cf. A214398, A230050, A227934, A227935.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Dec 21 2010