A292167 -id:A292167 - cp's OEIS frontend

A292190 Sum of n-th powers of products of terms in all partitions of n into distinct parts.

1, 1, 4, 35, 337, 11925, 371081, 49032439, 3545396034, 3416952655320, 749189363202730, 598250899004413536, 2383502427069445040595, 1729793152213690218766715, 131751643363739706679145099315, 271212858254426215135033141804302

Offset: 0

Seiichi Manyama, Sep 11 2017

nonn

			5 = 4 + 1 = 3 + 2. So a(5) = 5^5 + (4*1)^5 + (3*2)^5 = 11925.

Alois P. Heinz, Table of n, a(n) for n = 0..98

Main diagonal of A292189.

Cf. A265949, A292167, A292194, A292305, A292306.

b:= proc(n, i, k) option remember; (m->
      `if`(mn, 0, i^k*b(n-i, i-1, k)))))(i*(i+1)/2)
    end:
a:= n-> b(n$3):
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 11 2017

nmax = 15; Table[SeriesCoefficient[Product[(1 + k^n*x^k), {k, 1, nmax}], {x, 0, n}], {n, 0, nmax}] (* Vaclav Kotesovec, Sep 12 2017 *)

{a(n) = polcoeff(prod(k=1, n, 1+k^n*x^k+x*O(x^n)), n)}

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

A292166 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of Product_{j>=1} (1 - j^k*x^j).

Original entry on oeis.org

1, 1, -1, 1, -1, -1, 1, -1, -2, 0, 1, -1, -4, -1, 0, 1, -1, -8, -5, -1, 1, 1, -1, -16, -19, -7, 5, 0, 1, -1, -32, -65, -37, 27, 1, 1, 1, -1, -64, -211, -175, 155, 17, 13, 0, 1, -1, -128, -665, -781, 927, 205, 167, 4, 0, 1, -1, -256, -2059, -3367, 5675, 2129, 2089, 110, 0, 0

Offset: 0

Seiichi Manyama, Sep 10 2017

			Square array begins:
    1,  1,  1,   1,    1, ...
   -1, -1, -1,  -1,   -1, ...
   -1, -2, -4,  -8,  -16, ...
    0, -1, -5, -19,  -65, ...
    0, -1, -7, -37, -175, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..2 give A010815, A022661, A292164.
Rows n=0..2 give A000012, (-1)*A000012, (-1)*A000079.
Main diagonal gives A292167.

A[n_, k_] := A[n, k] = If[n == 0, 1, -(1/n)*Sum[Sum[d^(1+k*j/d), {d, Divisors[j]}]*A[n-j, k], {j, 1, n}]];
Table[A[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 04 2022 *)

A(0,k) = 1 and A(n,k) = -(1/n) * Sum_{j=1..n} (Sum_{d|j} d^(1+k*j/d)) * A(n-j,k) for n > 0. - Seiichi Manyama, Nov 02 2017

cp's OEIS Frontend

A292190 Sum of n-th powers of products of terms in all partitions of n into distinct parts.

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