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%I #95 Mar 18 2022 13:02:22
%S 2,2,10,2,26,2,42,8,82,2,122,16,170,2,226,2,290,12,362,2,170,50,530,2,
%T 626,2,90,80,842,70,962,36,130,92,1226,2,1370,138,1522,2,1626,178,
%U 1554,152,2026,152,2210,232,2402,12,2602,272,2810,2,306,2,1010,338,3482
%N Number of distinct residues of k^(n^2) (mod n^2+1), k=0..n^2.
%C a(A005574(n)) = 2.
%C a(n) = n for n = 2, 8, 128, ...
%C a(n) = n^2+1 (subsequence of A134406) for n = 1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25, 29, 31, ...
%C a(n) > 2 and a(n) <= n for n = 8, 18, 50, 60, 64, 72, 98, 112, 128, 132, 162, ... .
%C For n odd, gcd(a(n),n) = 1 except for n = 7, 27, 63, 75, 93, 105, 111, 125, 135, 153, 177, 207, 213, ...
%C For n even, gcd(a(n),n) = 2 for n in {A005574} union {22, 34, 38, 42, 46, 50, 58, 62, 78, 82, 86, 98, 102, 106, 114, 118, 122, 138, ...}
%C gcd(a(n),n) > 2 for n = 7, 8, 12, 18, 27, 28, 30, 32, 44, 48, 52, 60, 63, 64, 68, ...
%e a(2) = 2 because k^(2^2) == 0, 1 (mod 5) implies 2 distinct residues.
%e The table of k^(n^2) (mod n^2+1) of residues starts in row n=1 with columns k>=2 as:
%e 0,1;
%e 0,1,1,1,1;
%e 0,1,2,3,4,5,6,7,8,9;
%e 0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
%e 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25;
%e Its row sums are 1, 4, 45, 16, 325, ...
%p a:= n-> nops ({seq (k&^(n^2) mod (n^2+1), k=0..n^2)}):
%p seq (a(n), n=1..100);
%t Table[Length[Union[PowerMod[Range[0,n^2],n^2,n^2+1]]],{n,100}]
%o (PARI) a(n) = #Set(vector(n^2+1, k, k--; Mod(k, n^2+1)^n^2)); \\ _Michel Marcus_, Mar 18 2022
%Y Cf. A005574, A134406, A195637.
%K nonn
%O 1,1
%A _Michel Lagneau_, Mar 18 2022
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