A089901 Main diagonal of A089900, also the inverse hyperbinomial of A000312 (offset 1).

1, 3, 18, 159, 1848, 26595, 456048, 9073911, 205437312, 5214027267, 146602156800, 4522866752943, 151895344131072, 5516066815430691, 215373243256915968, 8996883483108522375, 400372897193449586688, 18908951043963993686019

Offset: 0

Author

Paul D. Hanna, Nov 14 2003

Keywords

• nonn

Comments

The n-th row of array A089900 is the n-th binomial transform of the factorials found in row 0: {1!,2!,3!,...,(n+1)!,...}. The hyperbinomial transform of this main diagonal gives: {1,4,27,...,(n+1)^(n+1),...}, which is the next lower diagonal in array A089900.

a(n), for n>=1, is the number of colored labeled mappings from n points to themselves, where each component is one of three colors. - Steven Finch, Nov 28 2021

Links

• G. C. Greubel, Table of n, a(n) for n = 0..380
• Nicholas John Bizzell-Browning, LIE scales: Composing with scales of linear intervallic expansion, Ph. D. Thesis, Brunel Univ. (UK, 2024). See p. 162.
• Steven Finch, Rounds, Color, Parity, Squares, arXiv:2111.14487 [math.CO], 2021.

Crossrefs

Cf. A000312, A089900, A089902, A000312.

Programs

Mathematica

CoefficientList[Series[1/(1+LambertW[-x])^3, {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jul 09 2013 *)
Flatten[{1,Table[Sum[n^(n-k)*Binomial[n,k]*(k+1)!,{k,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Jul 09 2013 *)
a[n_] := (n^(n + 2) + Exp[n] Gamma[n + 2, n]) / (n + 1);
Table[a[n], {n, 0, 17}]  (* Peter Luschny, Nov 29 2021 *)

PARI

/* As (n+1)-th term of the n-th binomial transform of {(n+1)!}: */
{a(n)=if(n<0,0,sum(i=0,n,n^(n-i)*binomial(n,i)*(i+1)!))}
for(n=0,20,print1(a(n),", "))

PARI

/* As (n+1)-th term of inverse hyperbinomial of {(n+1)^(n+1)}: */
{a(n)=if(n<0,0,sum(i=0,n,-(n-i-1)^(n-i-1)*binomial(n,i)*(i+1)^(i+1)))}
for(n=0,20,print1(a(n),", "))

Formula

a(n) = Sum_{k=0..n} n^(n-k) * C(n,k) * (k+1)!.

a(n) = Sum_{k=0..n} -(n-k-1)^(n-k-1) * C(n,k) * (k+1)^(k+1).

E.g.f.: 1 / (1 + LambertW(-x))^3.

E.g.f.: (Sum_{n>=0} (n+1)^(n+1) * x^n/n!) * (Sum_{n>=0} -(n-1)^(n-1) * x^n/n!).

a(n) ~ n^(n+1) * (1 + sqrt(Pi/(2*n))). - Vaclav Kotesovec, Jul 09 2013

a(n) = (n^(n + 2) + exp(n)*Gamma(n + 2, n)) / (n + 1). - Peter Luschny, Nov 29 2021