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.

A320591 Expansion of Product_{k>=1} (1 + x^k/(1 + x)^k).

Original entry on oeis.org

1, 1, 0, 1, -2, 4, -7, 11, -16, 23, -36, 65, -129, 256, -473, 772, -1028, 835, 776, -5755, 17562, -41750, 86678, -165145, 299949, -541837, 1020029, -2068203, 4509512, -10252952, 23465297, -52762788, 115160832, -243018459, 496094524, -982431070, 1894710043, -3574095362
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 16 2018

Keywords

Comments

After the first term, this is the second term of the n-th differences of A000009, or column n=1 of A378622. - Gus Wiseman, Feb 03 2025

Crossrefs

The version for non-strict partitions is A320590, row n=1 of A175804.
Column n=1 (except first term) of A378622. See also A293467, A377285, A378970, A378971, A380412 (column n=0).
A000009 counts strict integer partitions, differences A087897, A378972.
A266232 gives zero-based binomial transform of strict partitions, differences A129519.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R! ( (&*[(1 + x^k/(1 + x)^k): k in [1..(m+2)]]) )); // G. C. Greubel, Oct 29 2018
  • Maple
    seq(coeff(series(mul((1+x^k/(1+x)^k),k=1..n),x,n+1), x, n), n = 0 .. 37); # Muniru A Asiru, Oct 16 2018
  • Mathematica
    nmax = 37; CoefficientList[Series[Product[(1 + x^k/(1 + x)^k), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 37; CoefficientList[Series[Exp[Sum[Sum[(-1)^(k/d + 1) d, {d, Divisors[k]}] x^k/(k (1 + x)^k), {k, 1, nmax}]], {x, 0, nmax}], x]
    Prepend[Table[Differences[PartitionsQ/@Range[0,k+1],k][[2]],{k,0,30}],1] (* Gus Wiseman, Jan 29 2025 *)
  • PARI
    m=50; x='x+O('x^m); Vec(prod(k=1, m+2, (1 + x^k/(1 + x)^k))) \\ G. C. Greubel, Oct 29 2018
    

Formula

G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k/(k*((1 + x)^k - x^k))).
G.f.: exp(Sum_{k>=1} A000593(k)*x^k/(k*(1 + x)^k)).
From Peter Bala, Dec 22 2020: (Start)
O.g.f.: Sum_{n >= 0} x^(n*(n+1)/2)/Product_{k = 1..n} ((1 + x)^k - x^k). Cf. A307548.
Conjectural o.g.f.: (1/2) * Sum_{n >= 0} x^(n*(n-1)/2)*(1 + x)^n/( Product_{k = 1..n} ( (1 + x)^k - x^k ) ). (End)
a(n+1) = Sum_{k=0..n} (-1)^(n-k) binomial(n,k) A000009(k+1). - Gus Wiseman, Feb 03 2025

A309575 Expansion of Product_{k>=1} (1 - (x*(1 + x))^k).

Original entry on oeis.org

1, -1, -2, -2, -1, 1, 5, 11, 17, 26, 36, 35, 20, -5, -65, -221, -510, -897, -1379, -2157, -3498, -5225, -6500, -6425, -4775, -1463, 5951, 25905, 74833, 173129, 334719, 563200, 876876, 1363232, 2208921, 3621969, 5631470, 7896109, 9725768, 10374574, 9340382, 6104500, -1413334
Offset: 0

Views

Author

Seiichi Manyama, Sep 21 2019

Keywords

Crossrefs

Convolution inverse of A238441.

Programs

  • Mathematica
    nmax = 40; CoefficientList[Series[Product[(1 - (x*(1+x))^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 22 2019 *)
  • PARI
    N=66; x='x+O('x^N); Vec(prod(k=1, N, 1-(x*(1+x))^k))
    
  • PARI
    N=66; x='x+O('x^N); Vec(exp(-sum(k=1, N, sigma(k)*(x*(1+x))^k/k)))

Formula

G.f.: exp(-Sum_{k>=1} sigma(k)*(x*(1+x))^k/k).
Showing 1-2 of 2 results.