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A124782
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a(n) = (n+3)/gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!.
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4
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3, 2, 1, 3, 7, 4, 9, 1, 11, 6, 1, 7, 3, 8, 17, 9, 19, 2, 21, 11, 23, 12, 5, 1, 27, 14, 29, 3, 31, 16, 33, 17, 7, 18, 1, 19, 3, 4, 41, 21, 43, 22, 9, 23, 47, 24, 49, 5, 51, 2, 53, 27, 11, 28, 57, 29, 59, 6, 61, 31, 63, 32, 1, 33, 67, 34, 69, 7, 71, 36, 73, 1, 15, 38, 77, 3, 79, 8, 81, 41
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OFFSET
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0,1
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COMMENTS
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a(n) is an integer since A(n+2) = (n+2)(n+1)*A(n) + n+3.
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LINKS
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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.
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FORMULA
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EXAMPLE
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a(3) = (3+3)/gcd(A(3), A(5)) = 6/gcd(16, 326) = 6/2 = 3.
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MATHEMATICA
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(A[n_] := Sum[n!/k!, {k, 0, n}]; Table[(n+3)/GCD[A[n], A[n+2]], {n, 0, 80}])
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PROG
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(PARI)
A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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