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%I #8 Jun 08 2021 02:24:36
%S 1,2,3,4,5,6,8,10,14,21,33,36,56,68,94,378,1943,2389,5455
%N Numbers k such that the numerator of Sum_{j=0..k} k^j/j! is a prime number.
%C The corresponding primes are A120266(a(n)) = {2, 5, 13, 103, 1097, 1223, ...}
%e Sum_{j=0..4} 4^j/j! = 103/3. The numerator is a prime, hence 4 is in the sequence.
%t Do[ f=Numerator[Sum[n^k/k!,{k,0,n}]]; If[PrimeQ[f], Print[{n,f}]], {n, 1, 378}]
%Y Cf. A120266, A119029, A120267.
%K hard,more,nonn
%O 1,2
%A _Alexander Adamchuk_, Jun 13 2007
%E Edited by _Stefan Steinerberger_, Jul 22 2007
%E 3 more terms from _Ryan Propper_, Jan 22 2008