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.

A112291 Denominator of sum{k=1 to n} 1/S(n,k), where S(n,k) is a Stirling number of the second kind.

Original entry on oeis.org

1, 1, 3, 42, 150, 36270, 270900, 9440379900, 3332912051700, 2004302168707167000, 1424191116445997823000, 3936008766237071969447818200, 10888542544398564939894000, 3606055788316324023953497288103040
Offset: 1

Views

Author

Leroy Quet, Sep 01 2005

Keywords

Examples

			a(4) = 42, the denominator of 1/1 + 1/7 + 1/6 + 1 = 97/42.
The first few fractions are: 1, 2, 7/3, 97/42, 331/150, 77089/36270, 562609/270900.
		

Crossrefs

Cf. A112290.

Programs

  • Maple
    with(combinat): a:=n->denom(sum(1/stirling2(n,k),k=1..n)): seq(a(n),n=1..15); # Emeric Deutsch, Sep 02 2005
  • Mathematica
    f[n_] := Sum[1/StirlingS2[n, k], {k, n}]; Table[Denominator[f[n]], {n, 15}] (* Ray Chandler, Sep 02 2005 *)

Extensions

Extended by Emeric Deutsch and Ray Chandler, Sep 02 2005

A354478 a(n) is the numerator of Sum_{k=1..n} 1 / Stirling1(n,k).

Original entry on oeis.org

1, 0, 7, 25, 3991, 3923773, 4901627, 527165212865, 9823031039961293027, 123877274974851473572937, 443645907754951021537851199, 246932542361393897304051461727006396307, 1474846779473982897350113519971401527250089, 46578509609937575127608478711343978511593638945099881
Offset: 1

Views

Author

Ilya Gutkovskiy, Jun 02 2022

Keywords

Comments

Conjecture: a(n)/A354479(n) tends to 1 as n tends to infinity. For comparison: A112290(n)/A112291(n) tends to 2 as n tends to infinity. - Vaclav Kotesovec, Jun 02 2022

Examples

			1, 0, 7/6, 25/33, 3991/4200, 3923773/4192200, 4901627/5115600, 527165212865/545250747888, ...
		

Crossrefs

Cf. A008275, A046825, A112288, A112290, A354479 (denominators).

Programs

  • Mathematica
    Table[Sum[1/StirlingS1[n, k], {k, 1, n}], {n, 1, 14}] // Numerator
  • PARI
    a(n) = numerator(sum(k=1, n, 1/stirling(n, k, 1))); \\ Michel Marcus, Jun 02 2022
Showing 1-2 of 2 results.