

A124779


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!).


9



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
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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)^{p1}(p1)!. The proof uses (n1)!/(nk1)! = (n1)(n2)...(nk) == (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 pA(p1), where A(0) = 1 and A(n) = n*A(n1)+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, SpringerVerlag, 3rd edition, 2004, B43.


LINKS

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.


FORMULA

a(n) = gcd(A(n), A(n+2))/gcd(A(n), A(n+2), n!) where A(n)=1+n+n(n1)+...+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}])


PROG



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.


KEYWORD

nonn


AUTHOR



STATUS

approved



