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A242746
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Least k > 0 such that (k!+n!)/(k+n) is an integer.
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1
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1, 1, 3, 1, 4, 1, 5, 3, 6, 1, 7, 1, 7, 6, 5, 1, 7, 1, 5, 4, 7, 1, 7, 6, 4, 9, 8, 1, 7, 1, 9, 8, 7, 6, 5, 1, 8, 7, 6, 1, 7, 1, 11, 10, 9, 1, 9, 8, 7, 10, 9, 1, 7, 6, 5, 7, 13, 1, 11, 1, 9, 8, 7, 8, 7, 1, 13, 12, 11, 1, 9, 1, 7, 6, 8, 8, 7, 1, 11, 10, 9, 1, 7, 6, 11, 10, 9, 1, 11
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OFFSET
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1,3
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COMMENTS
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a(n) = 1 iff n+1 is prime.
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LINKS
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EXAMPLE
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(1!+3!)/(1+3) = 7/4 is not an integer. (2!+3!)/(2+3) = 8/5 is not an integer. (3!+3!)/(3+3) = 12/6 = 2 is an integer. Thus a(3) = 3.
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MATHEMATICA
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lnk[n_]:=Module[{k=1, nf=n!}, While[!IntegerQ[(k!+nf)/(k+n)], k++]; k]; Array[ lnk, 90] (* Harvey P. Dale, Sep 26 2014 *)
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PROG
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(PARI) a(n) = for(k=1, oo, if((k!+n!)%(k+n)==0, return(k))); \\ Modified by Jinyuan Wang, Mar 13 2020
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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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