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A028245 a(n) = 5^(n-1) - 4*4^(n-1) + 6*3^(n-1) - 4*2^(n-1) + 1 (essentially Stirling numbers of second kind).

%I A028245 #32 Sep 08 2022 08:44:50
%S A028245 0,0,0,0,24,360,3360,25200,166824,1020600,5921520,33105600,180204024,
%T A028245 961800840,5058406080,26308573200,135666039624,694994293080,
%U A028245 3542142833040,17980946172000,90990301641624
%N A028245 a(n) = 5^(n-1) - 4*4^(n-1) + 6*3^(n-1) - 4*2^(n-1) + 1 (essentially Stirling numbers of second kind).
%C A028245 For n>=2, a(n) is equal to the number of functions f: {1,2,...,n-1}->{1,2,3,4,5} such that Im(f) contains 4 fixed elements. - Aleksandar M. Janjic and _Milan Janjic_, Mar 08 2007
%H A028245 Seiichi Manyama, <a href="/A028245/b028245.txt">Table of n, a(n) for n = 1..1431</a>
%H A028245 Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>
%H A028245 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (15,-85,225,-274,120).
%F A028245 a(n) = 24*S(n, 5) = 24*A000481(n). - _Emeric Deutsch_, May 02 2004
%F A028245 G.f.: -24*x^5/((x-1)*(4*x-1)*(3*x-1)*(2*x-1)*(5*x-1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009; checked and corrected by _R. J. Mathar_, Sep 16 2009
%F A028245 E.g.f.: (Sum_{k=0..5} (-1)^(5-k)*binomial(5,k)*exp(k*x))/5. with a(0) = 0. - _Wolfdieter Lang_, May 03 2017
%t A028245 24StirlingS2[Range[30],5] (* _Harvey P. Dale_, Jun 18 2013 *)
%t A028245 Table[5^(n - 1) - 4*4^(n - 1) + 6*3^(n - 1) - 4*2^(n - 1) + 1, {n, 21}] (* or *)
%t A028245 Rest@ CoefficientList[Series[-24 x^5/((x - 1) (4 x - 1) (3 x - 1) (2 x - 1) (5 x - 1)), {x, 0, 21}], x] (* _Michael De Vlieger_, Sep 24 2016 *)
%o A028245 (PARI) for(n=1,30, print1(24*stirling(n,5,2), ", ")) \\ _G. C. Greubel_, Nov 19 2017
%o A028245 (Magma) [5^(n-1) - 4*4^(n-1) + 6*3^(n-1) - 4*2^(n-1) + 1: n in [1..30]]; // _G. C. Greubel_, Nov 19 2017
%Y A028245 Cf. A000481, A008277, A163626, A000225, A028243, A028244.
%K A028245 nonn,easy
%O A028245 1,5
%A A028245 _N. J. A. Sloane_, Doug McKenzie mckfam4(AT)aol.com