A186377 - OEIS

%I #12 Feb 12 2015 04:39:19

%S 1,7,79,1129,18559,333577,6365089,126652183,2598628543,54577439833,

%T 1167481074529,25346459683783,557042221952881,12368307313680871,

%U 277027947337574911,6251808554314780009,142015508983550880703

%N a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + x^p + x^q)^n found at sufficiently large p and q>(n+1)p for some fixed k>0.

%C Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials:

%C . (2 + x^k + x^p + x^q)^n, or

%C . (1 + x^k + 2*x^p + x^q)^n, or

%C . (1 + x^k + x^p + 2*x^q)^n,

%C for all p>(n+1)k and q>(n+1)p and fixed k>0.

%F (1) a(n) = Sum_{k=0..n} C(n,k)^2 *4^(n-k) *Sum_{j=0..k} C(k,j)^2*C(2j,j).

%F Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then

%F (2) A(x) = B(x)^3 * B(2^2*x)

%F where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).

%F Recurrence: (n-1)*n^3*(3*n - 5)*a(n) = 2*(n-1)*(54*n^4 - 174*n^3 + 192*n^2 - 99*n + 20)*a(n-1) - 2*(441*n^5 - 2604*n^4 + 6102*n^3 - 7107*n^2 + 4111*n - 940)*a(n-2) + 2*(n-2)^2*(726*n^3 - 3076*n^2 + 4188*n - 1655)*a(n-3) - 225*(n-3)^2*(n-2)^2*(3*n - 2)*a(n-4). - _Vaclav Kotesovec_, Feb 12 2015

%F a(n) ~ 5^(2*n+2) / (2^(7/2) * Pi^(3/2) * n^(3/2)). - _Vaclav Kotesovec_, Feb 12 2015

%e G.f.: A(x) = 1 + 7*x + 79*x^2/2!^2 + 1129*x^3/3!^2 + 18559*x^4/4!^2 +...

%e The g.f. may be expressed as:

%e A(x) = [Sum_{n>=0} x^n/n!^2]^3 *[Sum_{n>=0} (4x)^n/n!^2] where

%e [Sum_{n>=0} x^n/n!^2]^3 = 1 + 3*x + 15*x^2/2!^2 + 93*x^3/3!^2 + 639*x^4/4!^2 + 4653*x^5/5!^2 +...+ A002893(n)*x^n/n!^2 +...

%t Table[Sum[Binomial[n,k]^2 * 4^(n-k) *Sum[Binomial[k,j]^2 * Binomial[2j,j], {j,0,k}], {k,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Feb 11 2015 *)

%o (PARI) {a(n)=local(V=Vec((1+2*x+x^(n+2)+x^(n^2+2*n+3))^n));V*V~}

%o (PARI) {a(n)=sum(k=0,n,binomial(n,k)^2*4^(n-k)*sum(j=0,k,binomial(k,j)^2*binomial(2*j,j)))}

%o (PARI) {a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)^3*sum(m=0,n,(2^2*x)^m/m!^2),n)}

%Y Cf. A186375, A186376, A186378.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 19 2011