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1, 1, 1, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 26, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 13, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 26, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 65, 2, 1, 2, 1, 10, 1, 2, 1, 74, 5, 2, 1, 26, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 13, 2, 1, 2, 5, 2, 1, 2, 1
(list;
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refs;
listen;
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internal format)
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OFFSET
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0,4
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COMMENTS
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a(n) divides n+3 because 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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a(n) = gcd(A(n), A(n+2), n!) where A(n)=1+n+n(n-1)+...+n!. - Jonathan Sondow, Nov 13 2006
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EXAMPLE
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a(3) = gcd(d(3),d(5)) = gcd(gcd(3!,16), gcd(5!,326)) = gcd(2,2) = 2.
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MATHEMATICA
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(A[n_] := Sum[n!/k!, {k, 0, n}]; d[n_] := GCD[n!, A[n]]; Table[GCD[d[n], d[n+2]], {n, 0, 100}])
(* Second program, faster: *)
Table[GCD @@ Map[GCD[#!, Floor[E*#!] - Boole[# == 0]] &, n + {0, 2}], {n, 0, 96}] (* Michael De Vlieger, Jul 12 2017 *)
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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
m1=m2=1; for(n=0, 4096, m=m1; m1=m2; m2 = A093101(n+2); m124781 = gcd(m, m2); write("b093101.txt", n, " ", m); write("b124781.txt", n, " ", m124781); write("b123901.txt", n, " ", (n+3)/m124781)); \\ Antti Karttunen, Jul 12 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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