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%I #20 Jun 04 2019 10:37:35
%S 1,2,5,1,1,1,1,1,1,1,13,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%T 37,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%N a(n) = gcd(A(n), A(n+2))/gcd(d(n), d(n+2)) where A(n) = Sum_{k=0..n} n!/k! and d(n) = gcd(A(n), n!).
%C The next term > 1 is a(460) = 463. The primes 2, 5, 13, 37, 463 are the only terms > 1 up to n = 600000. If a(n) > 1 with n > 1, then a(n) = n+3 is prime. This uses A(n+2) = (n+2)(n+1)*A(n) + n+3. The terms > 1 are A064384 = primes p such that p divides 0!-1!+2!-3!+...+(-1)^{p-1}(p-1)!. The proof uses (n-1)!/(n-k-1)! = (n-1)(n-2)...(n-k) == (-1)^k k! (mod n). Cf. Cloitre's comment in A064383.
%C An integer p > 1 is in the sequence if and only if p is prime and p|A(p-1), where A(0) = 1 and A(n) = n*A(n-1)+1 for n > 0. - _Jonathan Sondow_, Dec 22 2006
%C Michael Mossinghoff has calculated that there are only five primes in the sequence up to 150 million. Heuristics suggest it contains infinitely many. - _Jonathan Sondow_, Jun 12 2007
%D R. K. Guy, Unsolved Problems in Number Theory, Springer-Verlag, 3rd edition, 2004, B43.
%H J. Sondow, <a href="https://www.jstor.org/stable/27642006">A geometric proof that e is irrational and a new measure of its irrationality</a>, Amer. Math. Monthly 113 (2006) 637-641.
%H J. Sondow, <a href="https://arxiv.org/abs/0704.1282">A geometric proof that e is irrational and a new measure of its irrationality</a>, arXiv:0704.1282 [math.HO], 2007-2010.
%H J. Sondow, <a href="http://home.earthlink.net/~jsondow/PrimesAndE.pdf">The Taylor series for e and the primes 2, 5, 13, 37, 463: a surprising connection</a>
%H J. Sondow and K. Schalm, <a href="http://arxiv.org/abs/0709.0671">Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II</a>, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AlternatingFactorial.html">Alternating Factorial</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/IntegerSequencePrimes.html">Integer Sequence Primes</a>
%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>
%F a(n) = A124780(n)/A124781(n) = A124782(n)/A123901(n).
%F a(n) = gcd(A(n), A(n+2))/gcd(A(n), A(n+2), n!) where A(n)=1+n+n(n-1)+...+n!. - _Jonathan Sondow_, Nov 10 2006
%F a(n) = gcd(N(n), N(n+2)), where N(n) = A061354(n) = numerator of Sum[1/k!,{k,0,n}]. - _Jonathan Sondow_, Jun 12 2007
%e a(2) = gcd(A(2), A(4))/gcd(d(2), d(4)) = gcd(5, 65)/gcd(1, 1) = 5/1 = 5.
%t (A[n_] := Sum[n!/k!, {k, 0, n}]; d[n_] := GCD[A[n],n! ]; Table[GCD[A[n],A[n+2]]/GCD[d[n],d[n+2]], {n,0,100}])
%o (PARI) A124779(n)={my(An=A000522(n),A2=A000522(n+2));gcd(An, A2)/gcd([An,n!,A2,(n+2)!])} \\ _M. F. Hasler_, Jun 04 2019
%Y A(n) = A000522, d(n) = A093101, gcd(A(n), A(n+2)) = A124780, gcd(d(n), d(n+2)) = A124781, (n+3)/gcd(A(n), A(n+2)) = A124782, (n+3)/gcd(d(n), d(n+2)) = A123901. Cf. A061354, A061355, A123899, A123900.
%Y Cf. A129924.
%K nonn
%O 0,2
%A _Jonathan Sondow_, Nov 07 2006
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