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%I M5212 N2266 #32 Jan 01 2023 12:36:11
%S 1,30,625,11515,203889,3602088,64720340,1194928020,22800117076,
%T 450996059800,9262414989464,197632289814960,4381123888865424,
%U 100869322905986496,2410630110159777216,59757230054773959552,1535299458203884231296,40848249256425236795904
%N Generalized Stirling numbers.
%C The asymptotic expansion of the higher order exponential integral E(x,m=5,n=4) ~ exp(-x)/x^5*(1 - 30/x + 625/x^2 - 11515/x^3 + 203889/x^4 - ... ) leads to the sequence given above. See A163931 for E(x,m,n) information and A163932 for a Maple procedure for the asymptotic expansion. - _Johannes W. Meijer_, Oct 20 2009
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A001719/b001719.txt">Table of n, a(n) for n = 0..100</a>
%H D. S. Mitrinovic, M. S. Mitrinovic, <a href="http://pefmath2.etf.rs/files/47/77.pdf">Tableaux d'une classe de nombres reliés aux nombres de Stirling</a>, Univ. Beograd. Pubi. Elektrotehn. Fak. Ser. Mat. Fiz. 77 (1962).
%F E.g.f.: (log(1-x)/(1-x))^4/24. - _Vladeta Jovovic_, May 05 2003
%F a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(k+4, 4)*4^k*Stirling1(n+4, k+4). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
%F If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then a(n-4) = |f(n,4,4)|, for n>=4. - _Milan Janjic_, Dec 21 2008
%t nn = 24; t = Range[0, nn]! CoefficientList[Series[(Log[1 - x]/(1 - x))^4/24, {x, 0, nn}], x]; Drop[t, 4] (* _T. D. Noe_, Aug 09 2012 *)
%o (PARI) a(n) = sum(k=0, n, (-1)^(n+k)*binomial(k+4, 4)*4^k*stirling(n+4, k+4, 1)); \\ _Michel Marcus_, Jan 20 2016
%Y Cf. A000254, A001706, A001713.
%K nonn
%O 0,2
%A _N. J. A. Sloane_
%E More terms from _Vladeta Jovovic_, May 05 2003
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