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A263026 a(n) = Sum_{k=1..n} stirling2(n,k)*((k+1)!)^3/8. 1

%I #39 Dec 13 2023 08:40:12

%S 1,28,1810,226558,48859606,16717044358,8536211225830,6206816010688678,

%T 6191950081736354086,8223501207813329312038,

%U 14182148054223247947725350,31102596462109513014876988198,85207893723061275473574262742566,287156553366174285430392015701185318,1174632657911183483067648902342293048870

%N a(n) = Sum_{k=1..n} stirling2(n,k)*((k+1)!)^3/8.

%F Representation as a sum of infinite series of special values of Meijer G functions, a(n) = (1/8) Sum_{k>=0} MeijerG([[1-k],[]],[[2,2,2],[]],1)) k^n/k!. The Meijer G functions in the above formula cannot be represented through any other special function.

%p # This program is intended for quick evaluation of a(n)

%p with(combinat):

%p a:= n-> add(stirling2(n, k)*((k+1)!)^3, k=1..n)/8:

%p seq(a(n), n=1..15);

%p # Maple program for the evaluation and verification of the infinite series representation:

%p a:= n-> evalf(sum(k^n*evalf(MeijerG([[1-k],[]],[[2,2,2],[]],1))/k!,k=0..infinity)/8); # n=1,2,... .

%p # This infinite series is slowly converging and the use of above formula will presumably not give the result in a reasonable time. Instead it is practical to replace the upper summation limit k = infinity by some kmax, say kmax = 6000. For example this yields for a(4) = 226558 the approximation 226557.9980714 in about 100 sec. Increasing kmax improves this approximation.

%t Table[Sum[StirlingS2[n, k] ((k + 1)!)^3/8, {k, n}], {n, 15}] (* _Michael De Vlieger_, Oct 09 2015 *)

%Y Cf. A261833, A262960.

%K nonn

%O 1,2

%A _Karol A. Penson_ and Katarzyna Gorska, Oct 08 2015

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Last modified April 26 19:25 EDT 2024. Contains 372004 sequences. (Running on oeis4.)