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A037961 a(n) = n^2*(n+1)*(n+3)!/48.

%I A037961 #35 Sep 08 2022 08:44:53
%S A037961 0,1,30,540,8400,126000,1905120,29635200,479001600,8083152000,
%T A037961 142702560000,2637143308800,50999300352000,1031319184896000,
%U A037961 21785854970880000,480178027929600000,11029155770400768000
%N A037961 a(n) = n^2*(n+1)*(n+3)!/48.
%C A037961 For n>=1, a(n) is equal to the number of surjections from {1,2,...,n+3} onto {1,2,...,n}. - Aleksandar M. Janjic and _Milan Janjic_, Feb 24 2007
%D A037961 Identity (1.19) in H. W. Gould, Combinatorial Identities, Morgantown, 1972; page 3.
%H A037961 Vincenzo Librandi, <a href="/A037961/b037961.txt">Table of n, a(n) for n = 0..400</a>
%H A037961 H. W. Gould, ed. J. Quaintance, <a href="http://www.math.wvu.edu/~gould/Vol.4.PDF">Combinatorial Identities</a>, May 2010 (section 10, p.45)
%H A037961 Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>
%F A037961 (n-1)^2*a(n) - n*(n+3)*(n+1)*a(n-1) = 0. - _R. J. Mathar_, Jul 26 2015
%F A037961 E.g.f.: x*(1 + 8*x + 6*x^2)/(1 - x)^7. - _Ilya Gutkovskiy_, Feb 20 2017
%F A037961 a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*k^(n+3). - _Peter Bala_, Mar 28 2017
%F A037961 From _G. C. Greubel_, Jun 20 2022: (Start)
%F A037961 a(n) = n!*StirlingS2(n+3, n).
%F A037961 a(n) = A131689(n+3, n).
%F A037961 a(n) = A019538(n+3, n). (End)
%t A037961 Table[n!*StirlingS2[n+3, n], {n,0,30}] (* _G. C. Greubel_, Jun 20 2022 *)
%o A037961 (PARI) a(n)=(n+3)!*n^2*(n+1)/48 \\ _Charles R Greathouse IV_, Nov 02 2011
%o A037961 (Magma) [Factorial(n+3)*n^2*(n+1)/48: n in [0..20]]; // _Vincenzo Librandi_, Nov 18 2011
%o A037961 (SageMath) [factorial(n)*stirling_number2(n+3, n) for n in (0..30)] # _G. C. Greubel_, Jun 20 2022
%Y A037961 Cf. A000142, A001286, A001297, A019538, A037960, A131689.
%K A037961 nonn,easy
%O A037961 0,3
%A A037961 _N. J. A. Sloane_