%I #83 Mar 13 2016 10:33:14
%S 2,3,17,27,293,791,10583,25685,448303,251411,4503535,6331107,
%T 4436875097,3335427631,19619696071,75379875277,3019260651391,
%U 16773385986619,3047463007411973,2732480436961811,398377271835431771,173581842021095897,1716900426430701553,35001773773285490879,6684326532123939298051
%N Numerator of (1/e)*Sum_{k>=0} (1/k!)*(Sum_{j=0..k} j^n).
%C Denominators are given by A130190, which in that entry are associated with the denominators of the z-sequence for a certain Sheffer matrix (triangle), but also apply here. See also the formula given there for the denominator of a certain sum involving the Stirling2 numbers.
%C Note that a(0) = 2 uses 0^0 := 1. For n >= 1 use triangle A079618 and the formula (1/e)*Sum_{k>=0} ((k+1)^n)/k! = Bell(n+1) = A000110(n+1). - _Wolfdieter Lang_, Feb 03 2015
%C If instead of Sum_{j=0..k} j^n one uses the sum with falling factorials, namely F(k, n) := Sum_{j=0..k} A008279(j, n) = A008279(k+1, n+1)/(n+1) the result for the rationals R(n) = (1/e)*Sum_{k>=0} (1/k!)*F(k, n) becomes very simple, namely R(n) = (n+2)/(n+1), n >= 0. - _Wolfdieter Lang_, Feb 03 2015
%H G. C. Greubel, <a href="/A248716/b248716.txt">Table of n, a(n) for n = 0..251</a>
%F a(n) = numerator(r(n)) with the rationals r(n) = (1/e)*Sum_{k>=0} (1/k!)*(Sum_{j=0..k} j^n), n >= 0, where 0^0 := 1.
%F a(n) = numerator(r(n)), with r(n) = (1/A064538(n))*Sum_{k=0..n} T(n+1,k+1)*Bell(k+1), with T(n,k) = A079618(n,k) and Bell(n) = A000110(n), for n >= 1. a(0) = 2 using 0^0 := 1. See comments above. - _Wolfdieter Lang_, Feb 03 2015
%e Terms up to n = 10, with denominators, are 2/1, 3/2, 17/6, 27/4, 293/15, 791/12, 10583/42, 25685/24, 448303/90, 251411/10, 4503535/33, ... .
%e From _Wolfdieter Lang_, Feb 03 2015: (Start)
%e With triangle A079618, A064538 and the Bell numbers A000110 the rationals r(n) are:
%e n=4: (1/30)*(-1*1 + 0*2 + 10*5 + 15*15 + 6*52) = 293/15.
%e n=9: (1/20)*(0*1 + (-3)*2 + 0*5 + 10*15 + 0*52 + (-14)*203 + 0*877 + 15*4140 + 10*21147 + 2*115975) = 251411/10.
%e (End)
%t Numerator[Table[(1/Exp[1])*Sum[Sum[j^n/k!, {j, 0, k}], {k, 0, Infinity}],
%t {n, 0, 100}]]
%Y Cf. A000110, A064538, A079618, A130190.
%K nonn,easy
%O 0,1
%A _Richard R. Forberg_, Dec 28 2014
%E Edited. Comment and formula rewritten. Cross references added. - _Wolfdieter Lang_, Feb 03 2015