cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

1, 7, 95, 1609, 30271, 606057, 12636689, 271026455, 5934011839, 131978406553, 2971793928145, 67586972435495, 1549805136840625, 35783848365934663, 831089570101489423, 19400246240227360809, 454864027237803296703
Offset: 0

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Author

Paul D. Hanna, Feb 19 2011

Keywords

Comments

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).
...
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,
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.

Examples

			G.f.: A(x) = 1 + 7*x + 95*x^2/2!^2 + 1609*x^3/3!^2 + 30271*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = C(x) * BesselI(0, 2*sqrt(x)) where
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
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]^2 * Binomial[4*k,2*k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)
  • PARI
    {a(n)=local(V=Vec((1+2*x+x^2+x^(2*n+3))^n));V*V~}
    
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*(4*k)!/(2*k)!^2)}
    
  • 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)}

Formula

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(4k,2k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = [Sum_{n>=0} (4n)!/(2n)!^2 *x^n/n!^2] *[Sum_{n>=0} x^n/n!^2].
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
a(n) ~ 5^(2*n+3/2) / (2^(7/2) * Pi * n). - Vaclav Kotesovec, Feb 12 2015