A292613 -id:A292613 - cp's OEIS frontend

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

Ilya Gutkovskiy, Oct 05 2017

sign

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

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.

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.

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

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 *)

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

A292508 Number A(n,k) of partitions of n with k kinds of 1; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 2, 1, 4, 7, 7, 5, 2, 1, 5, 11, 14, 12, 7, 4, 1, 6, 16, 25, 26, 19, 11, 4, 1, 7, 22, 41, 51, 45, 30, 15, 7, 1, 8, 29, 63, 92, 96, 75, 45, 22, 8, 1, 9, 37, 92, 155, 188, 171, 120, 67, 30, 12, 1, 10, 46, 129, 247, 343, 359, 291, 187, 97, 42, 14

Offset: 0

Alois P. Heinz, Sep 17 2017

Partial sum operator applied to column k gives column k+1.

A(n,k) is also defined for k < 0. All given formulas and programs can be applied also if k is negative.

			Square array A(n,k) begins:
  1,  1,  1,   1,   1,    1,    1,    1,     1, ...
  0,  1,  2,   3,   4,    5,    6,    7,     8, ...
  1,  2,  4,   7,  11,   16,   22,   29,    37, ...
  1,  3,  7,  14,  25,   41,   63,   92,   129, ...
  2,  5, 12,  26,  51,   92,  155,  247,   376, ...
  2,  7, 19,  45,  96,  188,  343,  590,   966, ...
  4, 11, 30,  75, 171,  359,  702, 1292,  2258, ...
  4, 15, 45, 120, 291,  650, 1352, 2644,  4902, ...
  7, 22, 67, 187, 478, 1128, 2480, 5124, 10026, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-10 give: A002865, A000041, A000070, A014153, A014160, A014161, A120477, A320753, A320754, A320755, A320756.
Rows n=0-4 give: A000012, A001477, A000124, A004006(k+1), A027927(k+3).
Main diagonal gives A292463.
A(n,n+1) gives A292613.
Cf. A292622, A292741, A292745.

A:= proc(n, k) option remember; `if`(n=0, 1, add(
      (numtheory[sigma](j)+k-1)*A(n-j, k), j=1..n)/n)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..14);
# second Maple program:
A:= proc(n, k) option remember; `if`(n=0, 1, `if`(k<1,
      A(n, k+1)-A(n-1, k+1), `if`(k=1, combinat[numbpart](n),
      A(n-1, k)+A(n, k-1))))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..14);
# third Maple program:
b:= proc(n, i, k) option remember; `if`(n=0 or i<2,
      binomial(k+n-1, n), add(b(n-i*j, i-1, k), j=0..n/i))
    end:
A:= (n, k)-> b(n$2, k):
seq(seq(A(n, d-n), n=0..d), d=0..14);

b[n_, i_, k_] := b[n, i, k] = If[n == 0 || i < 2, Binomial[k + n - 1, n], Sum[b[n - i*j, i - 1, k], {j, 0, n/i}]];
A[n_, k_] := b[n, n, k];
Table[A[n, d - n], {d, 0, 14}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 17 2018, translated from 3rd Maple program *)

G.f. of column k: 1/(1-x)^k * 1/Product_{j>1} (1-x^j).
Column k is Euler transform of k,1,1,1,... .
For fixed k>=0, A(n,k) ~ 2^((k-5)/2) * 3^((k-2)/2) * n^((k-3)/2) * exp(Pi*sqrt(2*n/3)) / Pi^(k-1). - Vaclav Kotesovec, Oct 24 2018

A292424 a(n) = [x^n] Product_{k=1..n} 1/((1 - x)^k * (1 - x^k)).

Original entry on oeis.org

1, 2, 11, 92, 1080, 16490, 311238, 7007796, 183431836, 5474465390, 183502419505, 6825981504602, 279041903645153, 12434720809043056, 599929817745490600, 31155278025923406979, 1732781419647450834768, 102761486514549541577999, 6473124665688520200808139

Offset: 0

Vaclav Kotesovec, Sep 20 2017

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..368

Cf. A292613.

Table[SeriesCoefficient[Product[1/((1-x)^k * (1-x^k)), {k, 1, n}], {x, 0, n}], {n, 0, 20}]
Table[SeriesCoefficient[Product[1/(1-x^k), {k, 1, n}] / (1-x)^(n*(n+1)/2), {x, 0, n}], {n, 0, 20}]

{a(n)= polcoef(prod(k=1, n, 1/((1-x)^k*(1-x^k) +x*O(x^n))), n)};
for(n=0,20, print1(a(n), ", ")) \\ G. C. Greubel, Feb 02 2019

a(n) ~ exp(n+2) * n^(n-1/2) / (sqrt(Pi) * 2^(n+1/2)).

A304781 a(n) = [x^n] (1/(1 - x)^n)*Product_{k>=1} (1 + x^k).

Original entry on oeis.org

1, 2, 6, 21, 75, 274, 1016, 3807, 14377, 54627, 208584, 799669, 3076167, 11867511, 45897145, 177888715, 690770763, 2686879415, 10466761637, 40828165464, 159453481037, 623427464093, 2439907421914, 9557831470082, 37472409664888, 147028505564603, 577302980976146

Offset: 0

Ilya Gutkovskiy, May 18 2018

nonn

Number of partitions of n into odd parts with n + 1 kinds of 1.

Index entries for sequences related to partitions

Cf. A000009, A036469, A095944, A128566, A128593, A292463, A292613, A293467.

Table[SeriesCoefficient[1/(1 - x)^n Product[(1 + x^k), {k, 1, n}], {x, 0, n}], {n, 0, 26}]
Table[SeriesCoefficient[1/(1 - x)^n Product[1/(1 - x^(2 k - 1)), {k, 1, n}], {x, 0, n}], {n, 0, 26}]
Table[SeriesCoefficient[1/(1 - x)^n Exp[Sum[(-1)^(k + 1) x^k/(k (1 - x^k)), {k, 1, n}]], {x, 0, n}], {n, 0, 26}]
Table[SeriesCoefficient[QPochhammer[-1, x]/(2 (1 - x)^n), {x, 0, n}], {n, 0, 26}]

a(n) = [x^n] (1/(1 - x)^n)*Product_{k>=1} 1/(1 - x^(2*k-1)).
a(n) = [x^n] (1/(1 - x)^n)*exp(Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k))).
a(n) ~ QPochhammer[-1, 1/2] * 4^(n-1) / sqrt(Pi*n). - Vaclav Kotesovec, May 18 2018

cp's OEIS Frontend

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

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A292508 Number A(n,k) of partitions of n with k kinds of 1; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A292424 a(n) = [x^n] Product_{k=1..n} 1/((1 - x)^k * (1 - x^k)).

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A304781 a(n) = [x^n] (1/(1 - x)^n)*Product_{k>=1} (1 + x^k).

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