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A210642
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a(n) = least integer m > 1 such that k! == n! (mod m) for no 0 < k < n.
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1
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2, 2, 3, 4, 5, 9, 7, 13, 17, 17, 11, 13, 13, 19, 23, 17, 17, 29, 19, 23, 31, 31, 23, 41, 31, 29, 31, 37, 29, 31, 31, 37, 41, 41, 59, 37, 37, 59, 43, 41, 41, 59, 43, 67, 53, 53, 47, 53, 67, 59, 61, 53, 53, 79, 59, 59, 67, 73, 59, 67
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OFFSET
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1,1
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COMMENTS
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Conjecture: a(n) is a prime not exceeding 2n with the only exceptions a(4)=4 and a(6)=9.
Note that a(n) is at least n and there is at least a prime in the interval [n,2n] by the Bertrand Postulate first confirmed by Chebyshev.
Compare this sequence with A208494.
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LINKS
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EXAMPLE
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We have a(4)=4, because 4 divides none of 4!-1!=23, 4!-2!=22, 4!-3!=18, and both 2 and 3 divide 4!-3!=18.
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MATHEMATICA
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R[n_, m_]:=If[n==1, 1, Product[If[Mod[n!-k!, m]==0, 0, 1], {k, 1, n-1}]] Do[Do[If[R[n, m]==1, Print[n, " ", m]; Goto[aa]], {m, Max[2, n], 2n}]; Print[n]; Label[aa]; Continue, {n, 1, 2500}]
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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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