A048144 a(n) = Sum_{k=0..n} (k!)^2 * Stirling_2(n,k)^2.

1, 1, 5, 73, 2069, 95401, 6487445, 610093513, 75796724309, 12020754177001, 2369364111428885, 568128719132038153, 162835627057766030549, 54975855375379966645801, 21593185551426744571090325, 9762238510837560633366673993, 5033241437347149354018370856789

Offset

0,3

Comments

Number of digraphs with loops, with labeled vertices and labeled arcs, with n arcs and with no vertex of indegree 0 or outdegree 0, cf. A121936, A122418, A122399. - Vladeta Jovovic, Sep 06 2006

Chromatic invariant of the complete bipartite graph K_{n+1,n+1}. - Eric W. Weisstein, Jul 11 2011

Generally, for p >= 1, Sum_{k=0..n} (k!*StirlingS2(n,k))^p is asymptotic to n^(p*n+1/2) * sqrt(Pi/(2*p*(1-log(2))^(p-1))) / (exp(p*n) * log(2)^(p*n+1)). - Vaclav Kotesovec, May 10 2014

Links

Alois P. Heinz, Table of n, a(n) for n = 0..100

Eric Weisstein's World of Mathematics, Chromatic Invariant

Eric Weisstein's World of Mathematics, Complete Bipartite Graph

Formula

E.g.f.: Sum_{n>=0} Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*(exp(j*x)-1)^n. a(n) = Sum_{k=0..n} Stirling2(n,k)*k!*A104602(k). - Vladeta Jovovic, Mar 25 2006

a(n) ~ sqrt(Pi/(1-log(2))) * n^(2*n+1/2) / (2*exp(2*n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, May 09 2014

E.g.f.: Sum_{n>=0} (1 - exp(-n*x))^n * exp(-n*x). - Paul D. Hanna, Mar 26 2018

E.g.f.: Sum_{n>=0} (exp(n*x) - 1)^n * exp(-n*(n+1)*x). - Paul D. Hanna, Mar 26 2018

a(n) = A272644(2n,n). - Alois P. Heinz, Oct 17 2024

a(n) = A371761(n, n). - Peter Luschny, Nov 20 2024

a(n) = (n!)^2 * [(x*y)^n] 1 / (exp(x) + exp(y) - exp(x + y)). - Ilya Gutkovskiy, Apr 24 2025

Maple

a := proc(n) local A, j; A := proc(n, k) option remember; if n = 0 then n^k else add(binomial(k + `if`(j>0, 1, 0), j+1) * A(n-1, k-j), j = 0..k) fi end: A(n, n) end:
seq(a(n), n = 0..16);  # Peter Luschny, Nov 20 2024

Mathematica

Table[Sum[(k!)^2*StirlingS2[n, k]^2, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 07 2014 *)

Prog

(PARI) a(n) = sum(k=0, n, k!^2*stirling(n, k, 2)^2); \\ Michel Marcus, Mar 07 2020
(Python)
from functools import cache
from math import comb as binomial
@cache
def A(n, k): return int(k == 0) if n == 0 else sum(binomial(k + int(j > 0), j + 1) * A(n - 1, k - j) for j in range(k + 1))
a = lambda n: A(n, n)
print([a(n) for n in range(17)])  # Peter Luschny, Nov 20 2024

Crossrefs

Cf. A000670, A242280, A023997, A212084, A120732, A104602, A272644, A371761.

Sequence in context: A126748 A276965 A217567 * A144682 A293146 A321189

Adjacent sequences: A048141 A048142 A048143 * A048145 A048146 A048147

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved