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If p is in the sequence then p divides 0!-1!+2!-3!+...+(-1)^N N! for all sufficiently large N. Naive heuristics suggest that the sequence should be infinite but very sparse. I have searched up to 600000 and have not found another prime with this property.
Same as the terms > 1 in A124779. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Nov 09 2006
A prime p is in the sequence if and only if p|A(p-1), where A(0) = 1 and A(n) = n*A(n-1)+1 = A000522(n). - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Dec 22 2006
It appears that also a(n) = primes p such that p divides A061354(p-1), where A061354(n) = Numerator of Sum_{k=0..n} 1/k!. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jun 14 2007
Michael Mossinghoff has calculated that 2, 5, 13, 37, 463 are the only terms up to 150 million. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Jun 12 2007
For proofs of Adamchuk's and my Comments, see the link "The Taylor series for e ...". - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Jun 18 2007
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