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.

Showing 1-2 of 2 results.

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

Views

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

Cf. A186375, A186376, A186377, A186391, A186392.

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

A186391 a(n) equals the least sum of the squares of the coefficients in (1 + x^k + x^(2k) + x^p)^n found at sufficiently large p for some fixed k>0.

Original entry on oeis.org

1, 4, 32, 340, 4096, 52704, 705956, 9717488, 136443904, 1945097296, 28063032832, 408836809088, 6004266204964, 88779091937488, 1320294416004736, 19733192546306640, 296219343194357760, 4463668854432401280
Offset: 0

Views

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+x+x^2 + x^p)^n for all p>2(n+1).
...
More generally, let B(x) = Sum_{n>=0} b(n)*x^n/n!^2 such that b(n) is the least sum of the squares of the coefficients in (F(x^k) + t*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} (t^2*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 + 4*x + 32*x^2/2!^2 + 340*x^3/3!^2 + 4096*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2] * C(x) where
C(x)= 1 + 3*x + 19*x^2/2!^2 + 141*x^3/3!^2 + 1107*x^4/4!^2 + 8953*x^5/5!^2 + 73789*x^6/6!^2 +...+ A082758(n)*x^n/n!^2 +...

Crossrefs

Cf. A082758, A186392, A186378.

Programs

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

Formula

(1) a(n) = Sum_{k=0..n} C(n,k)^2*A082758(k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = [Sum_{n>=0} x^n/n!^2]*[Sum_{n>=0} A082758(n)*x^n/n!^2]
where A082758(n) is the sum of the squares of the trinomial coefficients in (1+x+x^2)^n.
Recurrence: n^2*(2*n-1)^2*(240*n^5 - 2720*n^4 + 12016*n^3 - 25824*n^2 + 26966*n - 10939)*a(n) = 4*(6000*n^9 - 81920*n^8 + 470480*n^7 - 1486720*n^6 + 2841766*n^5 - 3403995*n^4 + 2557086*n^3 - 1163574*n^2 + 291045*n - 30429)*a(n-1) - 24*(6720*n^9 - 103040*n^8 + 678448*n^7 - 2511512*n^6 + 5740528*n^5 - 8358208*n^4 + 7691502*n^3 - 4262963*n^2 + 1269092*n - 151182)*a(n-2) + 4*(96000*n^9 - 1633280*n^8 + 11993280*n^7 - 49661760*n^6 + 127015344*n^5 - 206357088*n^4 + 210522120*n^3 - 127943860*n^2 + 41074746*n - 5150007)*a(n-3) - 64*(n-3)^2*(4*n - 13)*(4*n - 11)*(240*n^5 - 1520*n^4 + 3536*n^3 - 3696*n^2 + 1686*n - 261)*a(n-4). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 2^(4*n + 1/2) / (sqrt(3) * Pi * n). - Vaclav Kotesovec, Feb 12 2015
Showing 1-2 of 2 results.

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