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A210693
Smallest number k (different from n) such that n^k (mod 2k+1) = k^n (mod 2n+1).
0
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, 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, 9, 1, 3, 8, 1, 9, 5, 1, 2003, 99, 1, 6, 5, 1, 557, 3, 1, 16
OFFSET
1,1
COMMENTS
A majority of numbers k satisfies the equation n^k (mod 2k+1) = k^n (mod 2n+1) = r = 1.
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, …
EXAMPLE
a(5) = 9 because 5^9 (mod 19) = 9^5 (mod 11) = 1;
a(17) = 5 because 17^5 (mod 11) = 5^17 (mod 35) = 10;
a(62) = 15 because 62^15 (mod 31) = 15^62 (mod 125) = 0.
MAPLE
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:
CROSSREFS
Sequence in context: A222284 A330683 A020949 * A168212 A014449 A281387
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 30 2012
STATUS
approved