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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!). +0
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 (list; graph; listen)
OFFSET

0,2

COMMENT

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 (jsondow(AT)alumni.princeton.edu), 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 (jsondow(AT)alumni.princeton.edu), 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

Index entries for sequences related to factorial numbers

J. Sondow, The Taylor series for e and the primes 2, 5, 13, 37, 463: a surprising connection

J. Sondow, Which Partial Sums of the Taylor Series for e Are Convergents to e? (and a Link to the Primes $2, 5, 13, 37, 463, ...$) with an Appendix "Periodic Behaviour of Some Recurrence Sequences Related to $e$, Modulo Powers of 2" by Kyle Schalm

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 (jsondow(AT)alumni.princeton.edu), 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 (jsondow(AT)alumni.princeton.edu), 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: A062627 A011217 A078506 this_sequence A092134 A024548 A091772

Adjacent sequences: A124776 A124777 A124778 this_sequence A124780 A124781 A124782

KEYWORD

nonn

AUTHOR

Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Nov 07 2006

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