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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A355607 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of e.g.f. (1 + x)^(x^k).

Original entry on oeis.org

1, 1, 1, 1, 0, 0, 1, 0, 2, 0, 1, 0, 0, -3, 0, 1, 0, 0, 6, 20, 0, 1, 0, 0, 0, -12, -90, 0, 1, 0, 0, 0, 24, 40, 594, 0, 1, 0, 0, 0, 0, -60, 180, -4200, 0, 1, 0, 0, 0, 0, 120, 240, -1512, 34544, 0, 1, 0, 0, 0, 0, 0, -360, -1260, 11760, -316008, 0, 1, 0, 0, 0, 0, 0, 720, 1680, 28224, -38880, 3207240, 0
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2022

Keywords

Examples

			Square array begins:
  1,   1,   1,   1,    1,   1, 1, ...
  1,   0,   0,   0,    0,   0, 0, ...
  0,   2,   0,   0,    0,   0, 0, ...
  0,  -3,   6,   0,    0,   0, 0, ...
  0,  20, -12,  24,    0,   0, 0, ...
  0, -90,  40, -60,  120,   0, 0, ...
  0, 594, 180, 240, -360, 720, 0, ...
		

Crossrefs

Columns k=1..4 give A007113, A007121, (-1)^n * A353229(n), A354625.

Programs

  • PARI
    T(n, k) = n!*sum(j=0, n\(k+1), stirling(n-k*j, j, 1)/(n-k*j)!);

Formula

T(0,k) = 1 and T(n,k) = -(n-1)! * Sum_{j=k+1..n} (-1)^(j-k) * j/(j-k) * T(n-j,k)/(n-j)! for n > 0.
T(n,k) = n! * Sum_{j=0..floor(n/(k+1))} Stirling1(n-k*j,j)/(n-k*j)!.

A354624 Expansion of e.g.f. (1 - x)^(-x^4).

Original entry on oeis.org

1, 0, 0, 0, 0, 120, 360, 1680, 10080, 72576, 2419200, 25660800, 279417600, 3286483200, 41894012160, 794511244800, 13755488947200, 238514695372800, 4269265386946560, 79696849513881600, 1658065431859200000
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2022

Keywords

Crossrefs

Column k=4 of A355609.
Cf. A354625.

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace((1-x)^(-x^4)))
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(-x^4*log(1-x))))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=5, i, j/(j-4)*v[i-j+1]/(i-j)!)); v;
    
  • PARI
    a(n) = n!*sum(k=0, n\5, abs(stirling(n-4*k, k, 1))/(n-4*k)!);

Formula

a(0) = 1; a(n) = (n-1)! * Sum_{k=5..n} k/(k-4) * a(n-k)/(n-k)!.
a(n) = n! * Sum_{k=0..floor(n/5)} |Stirling1(n-4*k,k)|/(n-4*k)!.
a(n) ~ n! * (1 - 4/n - 16*log(n)/n^2). - Vaclav Kotesovec, Jul 21 2022
Showing 1-2 of 2 results.