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 A124779 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!). 8
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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. 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 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 REFERENCES R. K. Guy, Unsolved Problems in Number Theory, Springer-Verlag, 3rd edition, 2004, B43. J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly 113 (2006) 637-641. LINKS Eric Weisstein's World of Mathematics, Integer Sequence Primes J. Sondow and K. Schalm, 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, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010. Eric Weisstein's World of Mathematics, Alternating Factorial FORMULA a(n) = A124780(n)/A124781(n) = A124782(n)/A123901(n) 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 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 EXAMPLE a(2) = GCD(A(2), A(4))/GCD(d(2), d(4)) = GCD(5, 65)/GCD(1, 1) = 5/1 = 5 MATHEMATICA (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}]) CROSSREFS 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. Cf. A129924. Sequence in context: A011217 A078506 A323909 * A284750 A092134 A181779 Adjacent sequences:  A124776 A124777 A124778 * A124780 A124781 A124782 KEYWORD nonn AUTHOR Jonathan Sondow, Nov 07 2006 STATUS approved

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Last modified April 26 12:07 EDT 2019. Contains 322472 sequences. (Running on oeis4.)