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%I #5 May 31 2012 12:29:14
%S 2,5,3,2,17,2,7,5,7,23,41,2,11,5,3,59,29,71,2,17,11,3,43,41,37,7,31,
%T 17,13,7,5,47,59,47,151,2,23,17,79,5,3,59,2,113,2,29,71,23,17,83,5,67,
%U 61,131,53,47,43,41,31,17,13,11,7,67,239,53,227,47,2,107
%N a(n) is the prime p corresponding to the smallest integer k such that k^2 == p (mod prime(n)).
%C The corresponding values of k are {3, 4, 4, 6, 6, 5, 6, 6, 9, 8, 16, 7, 8, 8, 8, 27, …}
%H Michel Lagneau, <a href="/A212887/b212887.txt">Table of n, a(n) for n = 4..10000</a>
%e a(8) = 17 because 17 == 6^2 mod 19 where 19 = prime(8) and 6 is the smallest k.
%e Remark : 11 == 7^2 mod 19, but 7 > 6.
%p for n from 2 to 100 do:p:=ithprime(n):i:=0:for k from 0 to p-1 while(i=0) do: q:=irem(k^2,p):if type(q,prime)=true then i:=1:printf(`%d, `,q):else fi:od:od:
%Y Cf. A000224, A095972.
%K nonn
%O 4,1
%A _Michel Lagneau_, May 29 2012
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