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%I #28 Apr 02 2020 10:36:55
%S 1,1,2,3,2,30,24,21,90,126,210,660,462,8580,6006,1980,4410,157080,
%T 39270,2106720,510510,5087250,1963500,91861770,29099070,1806420,
%U 17117100,48498450,135795660,340510170,562582020,5642366730,1539968430,47683165530,17440042620
%N Least k such that Sum_{i=0..n} k^i / i! is a positive integer.
%C Note that Sum_{i=0..n-1} k^i / i! has a denominator that divides (n-1)! for n > 0. Therefore, for the expression to be an integer, k^n / n! must have a denominator that divides (n-1)!. Thus, k^n is divisible by n, a(n) = k is divisible by A007947(n).
%C a(n) is the smallest integer k such that Gamma(n+1,k)*e^k/n! is a positive integer, where Gamma is the upper incomplete gamma function. - _Chai Wah Wu_, Apr 02 2020
%H Bert Dobbelaere, <a href="/A332734/b332734.txt">Table of n, a(n) for n = 0..100</a>
%F a(n) <= A034386(n).
%e For n = 4, k > 0 if Sum_{i=0..4} k^i / i! is positive. a(4) = 2 since 1 + 1/1 + 1/2 + 1/6 + 1/24 = 65/24 is not an integer and 1 + 2/1 + 4/2 + 8/6 + 16/24 = 7 is an integer.
%o (PARI) a(n) = for(k=1, oo, if((s=sum(i=2, n, k^i/i!))==floor(s), return(k)));
%o (PARI) a(n) = {if (n==0, return (1)); my(m = factorback(factorint(n)[, 1]), k = m); while (denominator(sum(i=0, n, k^i/i!)) != 1, k += m); k;} \\ _Michel Marcus_, Mar 06 2020
%Y Cf. A000142, A007947, A034386.
%K nonn
%O 0,3
%A _Jinyuan Wang_, Mar 06 2020
%E a(24)-a(30) from _Michel Marcus_, Mar 06 2020
%E More terms from _Bert Dobbelaere_, Mar 09 2020
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