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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 (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.
LINKS
J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly 113 (2006) 637-641.
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
Eric Weisstein's World of Mathematics, Integer Sequence Primes
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}])
PROG
(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
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
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Nov 07 2006
STATUS
approved

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Last modified April 23 20:27 EDT 2024. Contains 371916 sequences. (Running on oeis4.)