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%I #16 Oct 19 2017 10:38:28
%S 11,67,24077,29041891,33699452071
%N Smallest prime p such that 1/2, 2/3, 3/4, ..., (m-1)/m are n-th power non-residues modulo p for maximum possible m (=A000236(n)).
%C A000236(n) is the maximum length of a run of consecutive residues modulo prime p, starting with 1, where no two adjacent elements belong to the same n-th power residue class (in other words, there is no n-th power residue modulo p in the sequence of ratios 1/2, 2/3, ..., (A000236(n)-1)/A000236(n)). a(n) equals the smallest p admitting a run of maximum length A000236(n).
%e a(2)=11 since A000236(2)=3 and 1/2=6, 2/3=8 are nonsquares modulo 11, and there is no smaller prime modulo which 1/2 and 2/3 are nonsquares.
%Y Cf. A000236, A000445.
%K hard,nonn,more
%O 2,1
%A _Max Alekseyev_, Aug 21 2005
%E a(6) from _Don Reble_, added by _Max Alekseyev_, Sep 03 2017
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