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.

A375838 Number of rooted chains starting with the cycle (1)(2)(3)...(n) in the permutation poset of [n].

Original entry on oeis.org

1, 1, 2, 9, 83, 1270, 28799, 906899, 37866842, 2024422837, 134850653405, 10950546880152, 1064840930492393, 122158078221727119, 16325324374155336370, 2514183676808883419043, 442023695390488997377405, 87989953715757624724243004, 19688099473681895327628896249, 4919839221134662388853128069571, 1365091729320293490230304687026514
Offset: 0

Views

Author

Rajesh Kumar Mohapatra, Subhashree Sahoo, and Ranjan Kumar Dhani, Sep 10 2024

Keywords

Examples

			Consider the set S = {1, 2, 3}. The a(3) = 1 + 5 + 3 = 9 in the poset of permutations of {1,2,3}:
 |{(1)(2)(3)}| = 1;
 |{(1)(2)(3) < (1)(23), (1)(2)(3) < (2)(13), (1)(2)(3) < (3)(12), (1)(2)(3) < (123), (1)(2)(3) < (132)}|=5;
 |{(1)(2)(3) < (1)(23) < (123), (1)(2)(3) < (2)(13)< (132), (1)(2)(3) < (3)(12) < (123)}| = 3.
		

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember;
          1+add(abs(Stirling1(n, k))*a(k), k=1..n-1)
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Jul 01 2025
  • Mathematica
    T[n_, k_] := T[n, k] = If[k < 0 || k > n, 0, If[(n == 0 && k == 0) || k == 1, 1, Sum[If[r >= 0, Abs[StirlingS1[n, r]]*T[r, k - 1], 0], {r, k - 1, n - 1}]]]; Table[Sum[T[n, k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jul 01 2025, after A375837 *)

Formula

a(n) = Sum_{k=0..n} A375837(n,k).
a(n) = (A375836(n)+1)/2.
Conjecture: a(n) = R(n,0) where R(n,k) = (k+1) * (Sum_{i=0..n-1} R(n-1,i) + Sum_{j=0..k-1} R(n-1,j)) for 0 <= k < n, R(n,n) = 1. - Mikhail Kurkov, Jun 21 2025
a(n) ~ c * n!^2 / (2^n * log(2)^n * n^(1-log(2)/3)), where c = A385521 = 1.59585433050036621247006569740016516964502505848324064247941890934119103861277... - Vaclav Kotesovec, Jul 01 2025
a(n) = 1 + Sum_{k=1..n-1} abs(Stirling1(n,k))*a(k). - Rajesh Kumar Mohapatra, Jul 01 2025