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A129976
Numbers k such that the numerator of Sum_{j=0..k} k^j/j! is a prime number.
0
1, 2, 3, 4, 5, 6, 8, 10, 14, 21, 33, 36, 56, 68, 94, 378, 1943, 2389, 5455
OFFSET
1,2
COMMENTS
The corresponding primes are A120266(a(n)) = {2, 5, 13, 103, 1097, 1223, ...}
EXAMPLE
Sum_{j=0..4} 4^j/j! = 103/3. The numerator is a prime, hence 4 is in the sequence.
MATHEMATICA
Do[ f=Numerator[Sum[n^k/k!, {k, 0, n}]]; If[PrimeQ[f], Print[{n, f}]], {n, 1, 378}]
CROSSREFS
KEYWORD
hard,more,nonn
AUTHOR
Alexander Adamchuk, Jun 13 2007
EXTENSIONS
Edited by Stefan Steinerberger, Jul 22 2007
3 more terms from Ryan Propper, Jan 22 2008
STATUS
approved