%I #7 Sep 03 2012 14:30:09

%S 1,3,5,9,11,15,19,25,29,35,39,45,47,49,51,59,61,65,69,71,79,85,95,101,

%T 121,131,139,141,145,159,165,169,171,175,181,195,199,201,205,209,219,

%U 221,231,245,261,271,275,279,289,299,309,315,321,325,329,335,345

%N Numbers n such that k == k^(q-1) mod q for k = 1, 2, ..., q-1, where q = n^2+1.

%C It is interesting to note that this sequence is identical to A002731 except for the numbers 1 and 47. For instance, a(13) = 47 but (47^2+1)/2 = 1105 is not prime, but 47^2+1 = 2210 => k^2209 == {1, 2, 3, ..., 2208, 2209} mod 2210 for k = {1, 2, ..., 2210}.

%C Conclusion: the two numbers of this sequence 1, 47 are not in A002731. Are there other numbers?

%e 3 is in the sequence because, for q = 3^2 + 1 = 10 we obtain the congruences:

%e 1^9 = 1 == 1 mod 10;

%e 2^9 = 512 == 2 mod 10;

%e 3^9 = 19683 == 3 mod 10;

%e 4^9 = 262144 == 4 mod 10;

%e 5^9 = 1953125 == 5 mod 10;

%e 6^9 = 10077696 == 6 mod 10,

%e 7^9 = 40353607 == 7 mod 10;

%e 8^9 = 134217728 == 8 mod 10;

%e 9^9 = 387420489 == 9 mod 10.

%p with(numtheory):for n from 1 by 2 to 500 do:q:=n^2+1:if type(x,prime)=false then j:=0:for i from 1 to q do: if irem(i^(q-1),q)=i then j:=j+1:else fi:od:if j=q-1 then printf(`%d, `, n):else fi:fi:od:

%t f[n_] := Module[{q = n^2 + 1}, And @@ Table[PowerMod[k, q - 1, q] == k, {k, q - 1}]]; Select[Range[345], f] (* _T. D. Noe_, Sep 03 2012 *)

%Y Cf. A002731.

%K nonn

%O 1,2

%A _Michel Lagneau_, Sep 01 2012