A281425 a(n) = [q^n] (1 - q)^n / Product_{j=1..n} (1 - q^j).

1, 0, 1, -1, 2, -4, 9, -21, 49, -112, 249, -539, 1143, -2396, 5013, -10550, 22420, -48086, 103703, -223806, 481388, -1029507, 2187944, -4625058, 9742223, -20490753, 43111808, -90840465, 191773014, -405523635, 858378825, -1817304609, 3845492204, -8129023694, 17162802918, -36191083386

Offset: 0

Author

Ilya Gutkovskiy, Oct 05 2017

Keywords

• sign

Comments

a(n) is n-th term of the Euler transform of -n + 1, 1, 1, 1, ...

Inverse zero-based binomial transform of A000041. The version for strict partitions is A380412, or A293467 up to sign. - Gus Wiseman, Feb 06 2025

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..3000
• A. M. Odlyzko, Differences of the partition function, Acta Arithmetica 49.3 (1988): 237-254.
• Dennis Stanton and Doron Zeilberger, The Odlyzko conjecture and O’Hara’s unimodality proof, Proceedings of the American Mathematical Society 107.1 (1989): 39-42.

Crossrefs

Cf. A128566, A292463, A292541, A292613.

Cf. A000041, A002865, A053445, A275638, A275639, A275640, A275641, A275642, A275643, A275644.

Cf. A218481, A294466, A095051.

Cf. A266232, A294467, A293467, A294468.

Row n=0 of A175804 (row n=1 is A320590 except first term).

Cf. A000009, A008284, A087897, A116608, A129519, A225486, A320591, A377056, A378622.

Programs

Maple

b:= proc(n, k) option remember; `if`(k=0,
      combinat[numbpart](n), b(n, k-1)-b(n-1, k-1))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..35);  # Alois P. Heinz, Dec 21 2024

Mathematica

Table[SeriesCoefficient[(1 - q)^n / Product[(1 - q^j), {j, 1, n}], {q, 0, n}], {n, 0, 35}]
Table[SeriesCoefficient[(1 - q)^n QPochhammer[q^(1 + n), q]/QPochhammer[q, q], {q, 0, n}], {n, 0, 35}]
Table[SeriesCoefficient[1/QFactorial[n, q], {q, 0, n}], {n, 0, 35}]
Table[Differences[PartitionsP[Range[0, n]], n], {n, 0, 35}] // Flatten
Table[Sum[(-1)^j*Binomial[n, j]*PartitionsP[n-j], {j, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Oct 06 2017 *)

Formula

a(n) = [q^n] 1/((1 + q)*(1 + q + q^2)*...*(1 + q + ... + q^(n-1))).

a(n) = Sum_{j=0..n} (-1)^j * binomial(n, j) * A000041(n-j). - Vaclav Kotesovec, Oct 06 2017

a(n) ~ (-1)^n * 2^(n - 3/2) * exp(Pi*sqrt(n/12) + Pi^2/96) / (sqrt(3)*n). - Vaclav Kotesovec, May 07 2018