%I #21 Apr 18 2013 02:45:58
%S 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,
%T 29,31,37,29,31,31,37,41,41,59,37,37,59,43,41,41,59,43,67,53,53,47,53,
%U 67,59,61,53,53,79,59,59,67,73,59,67
%N a(n) = least integer m > 1 such that k! == n! (mod m) for no 0 < k < n.
%C Conjecture: a(n) is a prime not exceeding 2n with the only exceptions a(4)=4 and a(6)=9.
%C 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.
%C Compare this sequence with A208494.
%H Zhi-Wei Sun, <a href="/A210642/b210642.txt">Table of n, a(n) for n = 1..2500</a>
%H Oliver Gerard, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;125e9edf.1203">Re: A new conjecture on primes</a>, a message to Number Theory List, March 23, 2012.
%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;79d1a9b5.1203">A new conjecture on primes</a>, a message to Number Theory List, March 20, 2012.
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2013.02.003">On functions taking only prime values</a>, J. Number Theory 133(2013), no.8, 2794-2812.
%e 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.
%t 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}]
%Y Cf. A000040, A208494, A210640, A210393, A210394, A210186, A210144, A208643, A207982.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Mar 26 2012
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