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.

A022571 Expansion of Product_{m>=1} (1+x^m)^6.

Original entry on oeis.org

1, 6, 21, 62, 162, 384, 855, 1806, 3648, 7110, 13434, 24702, 44361, 78006, 134592, 228302, 381300, 627840, 1020394, 1638528, 2601849, 4088780, 6363354, 9813504, 15005458, 22760262, 34261248, 51204222, 76005906, 112092438, 164296989, 239404860, 346898496, 499971968, 716906394
Offset: 0

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References

  • A. P. Prudnikov, Yu. A. Brychkov and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", New York, Gordon and Breach Science Publishers, 1986-1992, p. 755, Eq. 6.2.2.2. MR0874986 (88f:00013)

Crossrefs

Cf. A000009.
Column k=6 of A286335.

Programs

  • Magma
    Coefficients(&*[(1+x^m)^6:m in [1..40]])[1..40] where x is PolynomialRing(Integers()).1; // G. C. Greubel, Feb 26 2018
  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+q^m)^6,{m,1,nmax}],{q,0,nmax}],q] (* Vaclav Kotesovec, Mar 05 2015 *)
  • PARI
    a(n)=if(n<0, 0, polcoeff( prod(k=1,n, 1+x^k, 1+x*O(x^n))^6, n)) /* Michael Somos, Jul 09 2005 */
    

Formula

Euler transform of period 2 sequence [6, 0, ...]. - Michael Somos, Jul 09 2005
Expansion of q^(-1/4)(eta(q^2)/eta(q))^6 in powers of q. - Michael Somos, Jul 09 2005
Expansion of q^(-1/4)(1/2)k^(1/2)/k' in powers of q. - Michael Somos, Jul 09 2005
Given g.f. A(x), then B(x)=(x*A(x^4))^4 satisfies 0=f(B(x), B(x^2)) where f(u, v)=(4096uv+48u+1)v-u^2 . - Michael Somos, Jul 09 2005
Given g.f. A(x), then B(x)=x*A(x^4) satisfies 0=f(B(x), B(x^3)) where f(u, v)=(u^2-v^2)^2 -uv(1+8uv)^2 . - Michael Somos, Jul 09 2005
G.f.: Product_{k>0} (1+x^k)^6.
a(n) ~ exp(Pi * sqrt(2*n)) / (16 * 2^(1/4) * n^(3/4)). - Vaclav Kotesovec, Mar 05 2015
a(0) = 1, a(n) = (6/n)*Sum_{k=1..n} A000593(k)*a(n-k) for n > 0. - Seiichi Manyama, Apr 03 2017
G.f.: exp(6*Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k))). - Ilya Gutkovskiy, Feb 06 2018