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%I #9 Sep 22 2013 14:02:57
%S 4,11,11,1,9,9,1,4,5,1,2,18,1,5,5,1,5,3,1,5,23,1,3,48,1,4,221,1,3,3,1,
%T 8,14,1,4,2,1,30,2,1,10,2,1,2,9,1,13,6,1,20,2,1,3,9,1,2,10,1,7,3,1,15,
%U 9,1,3,8,1,9,5,1,2003,99,1,6,5,1,557,3,1,16
%N Smallest number k (different from n) such that n^k (mod 2k+1) = k^n (mod 2n+1).
%C A majority of numbers k satisfies the equation n^k (mod 2k+1) = k^n (mod 2n+1) = r = 1.
%C The values of n such that r <> 1 are given by n = 17, 38, 42, 47, 57, 59, …including the values with r = 0 given by n = 62, 84, 171, …
%e a(5) = 9 because 5^9 (mod 19) = 9^5 (mod 11) = 1;
%e a(17) = 5 because 17^5 (mod 11) = 5^17 (mod 35) = 10;
%e a(62) = 15 because 62^15 (mod 31) = 15^62 (mod 125) = 0.
%p with(numtheory): for n from 1 to 100 do:ii:=0:for k from 1 to 10000 while(ii=0) do:if n<>k and irem(n^k,2*k+1) = irem(k^n,2*n+1) then ii:=1:printf(`%d, `, k):else fi:od:od:
%K nonn
%O 1,1
%A _Michel Lagneau_, Mar 30 2012
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