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.

Showing 1-2 of 2 results.

A343579 a(n) = Sum_{k=0..floor(n/2)} |Stirling1(n - k, k)|.

Original entry on oeis.org

1, 0, 1, 1, 3, 9, 36, 176, 1030, 7039, 55098, 486346, 4780445, 51787405, 613045468, 7873065045, 109021348618, 1619197654575, 25675094145535, 432908683794379, 7733991639921585, 145933532935469016, 2900112108790279902, 60543749629794205640, 1324677739541613767983
Offset: 0

Views

Author

Peter Luschny, Apr 20 2021

Keywords

Comments

Equals antidiagonal sums of the triangle of unsigned Stirling numbers of the first kind (A132393).

Crossrefs

Variant: A237653.

Programs

  • Mathematica
    Table[Sum[Abs[StirlingS1[n - k, k]], {k, 0, Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Apr 09 2022 *)
  • PARI
    a(n) = sum(k=0, n\2, abs(stirling(n-k, k, 1))); \\ Michel Marcus, Apr 22 2021
    
  • PARI
    my(N=40, x='x+O('x^N)); Vec(sum(k=0, N, x^k*prod(j=0, k-1, j+x))) \\ Seiichi Manyama, Apr 08 2022

Formula

a(n) ~ n! / n^2. - Vaclav Kotesovec, Apr 09 2022

A331327 T(n, k) = [x^k] Pochhammer(x, n-k) for n >= 0, 0 <= k <= floor(n/2). Irregular triangle read by rows.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 2, 1, 0, 6, 3, 0, 24, 11, 1, 0, 120, 50, 6, 0, 720, 274, 35, 1, 0, 5040, 1764, 225, 10, 0, 40320, 13068, 1624, 85, 1, 0, 362880, 109584, 13132, 735, 15, 0, 3628800, 1026576, 118124, 6769, 175, 1, 0, 39916800, 10628640, 1172700, 67284, 1960, 21
Offset: 0

Views

Author

Peter Luschny, Jan 25 2020

Keywords

Examples

			Triangle starts:
[ 0] 1;
[ 1] 0;
[ 2] 0, 1;
[ 3] 0, 1;
[ 4] 0, 2,     1;
[ 5] 0, 6,     3;
[ 6] 0, 24,    11,    1;
[ 7] 0, 120,   50,    6;
[ 8] 0, 720,   274,   35,   1;
[ 9] 0, 5040,  1764,  225,  10;
[10] 0, 40320, 13068, 1624, 85, 1;
		

Crossrefs

Row sums are: 1, 0, A237653.
Cf. A132393.

Programs

  • Maple
    A331327row := n -> seq(coeff(expand(pochhammer(x, n-k)), x, k), k=0..n/2):
    seq(A331327row(n), n=0..13);
  • Mathematica
    T[n_, k_] := Abs[StirlingS1[n - k, k]];
    Table[T[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten

Formula

Rows are the antidiagonals of A132393.
Showing 1-2 of 2 results.