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A037962
a(n) = n*(15*n^3 + 30*n^2 + 5*n - 2)*(n+4)!/5760.
4
0, 1, 62, 1806, 40824, 834120, 16435440, 322494480, 6411968640, 130456085760, 2731586457600, 59056027430400, 1320663933388800, 30575780537702400, 733062897120153600, 18198613875746304000
OFFSET
0,3
COMMENTS
For n>=1, a(n) is equal to the number of surjections from {1,2,...,n+4} onto {1,2,...,n}. - Aleksandar M. Janjic and Milan Janjic, Feb 24 2007
REFERENCES
Identity (1.20) in H. W. Gould, Combinatorial Identities, Morgantown, 1972, page 3.
FORMULA
From G. C. Greubel, Jun 20 2022: (Start)
a(n) = (-1)^n * Sum_{j=0..n} (-1)^j * binomial(n, j)*j^(n+4).
a(n) = n!*StirlingS2(n+4, n).
a(n) = A131689(n+4, n).
a(n) = A019538(n+4, n).
E.g.f.: x*(1 + 22*x + 58*x^2 + 24*x^3)/(1-x)^9. (End)
MATHEMATICA
Table[(n+4)!n(15n^3+30n^2+5n-2)/5760, {n, 0, 20}] (* Harvey P. Dale, Nov 16 2020 *)
Table[n!*StirlingS2[n+4, n], {n, 0, 30}] (* G. C. Greubel, Jun 20 2022 *)
PROG
(Magma) [Factorial(n)*StirlingSecond(n+4, n): n in [0..30]]; // G. C. Greubel, Jun 20 2022
(SageMath) [factorial(n)*stirling_number2(n+4, n) for n in (0..30)] # G. C. Greubel, Jun 20 2022
CROSSREFS
KEYWORD
nonn
STATUS
approved