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A186378 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^2 + x^p)^n found at sufficiently large p for some fixed k>0.

%I A186378 #13 Feb 12 2015 04:43:54
%S A186378 1,7,95,1609,30271,606057,12636689,271026455,5934011839,131978406553,
%T A186378 2971793928145,67586972435495,1549805136840625,35783848365934663,
%U A186378 831089570101489423,19400246240227360809,454864027237803296703
%N A186378 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^2 + x^p)^n found at sufficiently large p for some fixed k>0.
%C A186378 Equivalently, a(n) equals the sum of the squares of the coefficients in the polynomial: (1 + 2*x + x^2 + x^p)^n for all p>2(n+1).
%C A186378 ...
%C A186378 More generally, let B(x) equal the g.f. for the least sum of the squares of the coefficients in (F(x^k) + x^p)^n where F(x) is a finite polynomial in x with degree d and p>(n+1)dk for some fixed k>0,
%C A186378 then B(x) = [Sum_{n>=0} x^n/n!^2]*[Sum_{n>=0} c(n)/n!^2] where c(n) equals the sum of the squares of the coefficients in the polynomial F(x)^n.
%F A186378 (1) a(n) = Sum_{k=0..n} C(n,k)^2*C(4k,2k).
%F A186378 Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
%F A186378 (2) A(x) = [Sum_{n>=0} (4n)!/(2n)!^2 *x^n/n!^2] *[Sum_{n>=0} x^n/n!^2].
%F A186378 Recurrence: n^2*(2*n-1)^2*(1152*n^4 - 8160*n^3 + 21040*n^2 - 23376*n + 9467)*a(n) = 3*(55296*n^8 - 503808*n^7 + 1891456*n^6 - 3812256*n^5 + 4504864*n^4 - 3193428*n^3 + 1326995*n^2 - 296732*n + 27900)*a(n-1) - 3*(451584*n^8 - 4607232*n^7 + 19744768*n^6 - 46227488*n^5 + 64243016*n^4 - 53731348*n^3 + 26049967*n^2 - 6596672*n + 675000)*a(n-2) + (n-2)^2*(2230272*n^6 - 16991232*n^5 + 49582720*n^4 - 69169056*n^3 + 46825856*n^2 - 13847412*n + 1451547)*a(n-3) - 900*(n-3)^2*(n-2)^2*(1152*n^4 - 3552*n^3 + 3472*n^2 - 1168*n + 123)*a(n-4). - _Vaclav Kotesovec_, Feb 12 2015
%F A186378 a(n) ~ 5^(2*n+3/2) / (2^(7/2) * Pi * n). - _Vaclav Kotesovec_, Feb 12 2015
%e A186378 G.f.: A(x) = 1 + 7*x + 95*x^2/2!^2 + 1609*x^3/3!^2 + 30271*x^4/4!^2 +...
%e A186378 The g.f. may be expressed as:
%e A186378 A(x) = C(x) * BesselI(0, 2*sqrt(x)) where
%e A186378 C(x)= 1 + 6*x + 70*x^2/2!^2 + 924*x^3/3!^2 + 12870*x^4/4!^2 +...+ (4n)!/(2n)!^2*x^n/n!^2
%t A186378 Table[Sum[Binomial[n,k]^2 * Binomial[4*k,2*k], {k,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Feb 11 2015 *)
%o A186378 (PARI) {a(n)=local(V=Vec((1+2*x+x^2+x^(2*n+3))^n));V*V~}
%o A186378 (PARI) {a(n)=sum(k=0,n,binomial(n,k)^2*(4*k)!/(2*k)!^2)}
%o A186378 (PARI) {a(n)=n!^2*polcoeff(sum(m=0,n,(4*m)!/(2*m)!^2*x^m/m!^2)*sum(m=0,n,x^m/m!^2+x*O(x^n)),n)}
%Y A186378 Cf. A186375, A186376, A186377, A186391, A186392.
%K A186378 nonn
%O A186378 0,2
%A A186378 _Paul D. Hanna_, Feb 19 2011