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.

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A308690 Square array A(n,k), n >= 1, k >= 0, where A(n,k) = Sum_{d|n} d^(k*n/d - k + 1), read by antidiagonals.

Original entry on oeis.org

1, 1, 3, 1, 3, 4, 1, 3, 4, 7, 1, 3, 4, 9, 6, 1, 3, 4, 13, 6, 12, 1, 3, 4, 21, 6, 24, 8, 1, 3, 4, 37, 6, 66, 8, 15, 1, 3, 4, 69, 6, 216, 8, 41, 13, 1, 3, 4, 133, 6, 762, 8, 201, 37, 18, 1, 3, 4, 261, 6, 2784, 8, 1289, 253, 68, 12, 1, 3, 4, 517, 6, 10386, 8, 9225, 2197, 648, 12, 28
Offset: 1

Views

Author

Seiichi Manyama, Jun 17 2019

Keywords

Examples

			Square array begins:
    1,  1,  1,   1,   1,    1,     1, ...
    3,  3,  3,   3,   3,    3,     3, ...
    4,  4,  4,   4,   4,    4,     4, ...
    7,  9, 13,  21,  37,   69,   133, ...
    6,  6,  6,   6,   6,    6,     6, ...
   12, 24, 66, 216, 762, 2784, 10386, ...
    8,  8,  8,   8,   8,    8,     8, ...
		

Crossrefs

Columns k=0..3 give A000203, A055225, A308688, A308689.

Programs

  • Mathematica
    T[n_, k_] := DivisorSum[n, #^(k*n/# - k + 1) &]; Table[T[k, n - k], {n, 1, 12}, {k, 1, n}] // Flatten (* Amiram Eldar, May 09 2021 *)

Formula

L.g.f. of column k: -log(Product_{j>=1} (1 - j^k*x^j)^(1/j^k)).
A(p,k) = p+1 for prime p.

A356561 Expansion of e.g.f. Product_{k>0} 1/(1 - k^3 * x^k)^(1/k^3).

Original entry on oeis.org

1, 1, 4, 18, 204, 1260, 37440, 299880, 11002320, 204860880, 6618628800, 92924647200, 8181137764800, 124123075876800, 7211104918617600, 288085376346768000, 14964000305173920000, 340302035937191328000, 42619767305209750656000
Offset: 0

Views

Author

Seiichi Manyama, Aug 12 2022

Keywords

Crossrefs

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(1/prod(k=1, N, (1-k^3*x^k)^(1/k^3))))
    
  • PARI
    a308689(n) = sumdiv(n, d, d^(3*n/d-2));
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, i, a308689(j)*v[i-j+1]/(i-j)!)); v;

Formula

a(0) = 1; a(n) = (n-1)! * Sum_{k=1..n} A308689(k) * a(n-k)/(n-k)!.
Showing 1-2 of 2 results.