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

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

Author

Ilya Gutkovskiy, Oct 16 2018

Keywords

• sign
• easy

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000593, A320589, A307548.

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.

Cf. A000041, A007442, A002865, A008284, A095195, A103446, A116608, A218482, A281425, A320568.

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