A195637 Number of distinct residues of k^n (mod n), k=0..n-1.
1, 2, 3, 2, 5, 4, 7, 2, 3, 6, 11, 4, 13, 8, 15, 2, 17, 4, 19, 4, 9, 12, 23, 4, 5, 14, 3, 8, 29, 12, 31, 2, 33, 18, 35, 4, 37, 20, 15, 4, 41, 8, 43, 12, 15, 24, 47, 4, 7, 6, 51, 8, 53, 4, 15, 8, 21, 30, 59, 8, 61, 32, 9, 2, 65, 24, 67, 10, 69, 24, 71, 4, 73
Offset: 1
Examples
a(18) = 4 because k^18 == 0, 1, 9, 10 (mod 18) => 4 distinct residues. From _R. J. Mathar_, Aug 27 2013: (Start) The triangle of k^n (mod n) starts in row n=1 with columns k>=0 as: 0; 0, 1; 0, 1, 2; 0, 1, 0, 1; 0, 1, 2, 3, 4; 0, 1, 4, 3, 4, 1; 0, 1, 2, 3, 4, 5, 6; 0, 1, 0, 1, 0, 1, 0, 1; 0, 1, 8, 0, 1, 8, 0, 1, 8; 0, 1, 4, 9, 6, 5, 6, 9, 4, 1; 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; 0, 1, 4, 9, 4, 1, 0, 1, 4, 9, 4, 1; 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; Its row sums are 0, 1, 3, 2, 10, 13, 21, 4, 27, 45, 55, 38, 78, 77, 105, 8,.... (End)
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
Programs
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Maple
a:= n-> nops({seq(k&^n mod n, k=0..n-1)}): seq(a(n), n=1..100);
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Mathematica
Table[Length[Union[PowerMod[Range[0, n - 1], n, n]]], {n, 100}] (* T. D. Noe, Sep 21 2011 *)
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PARI
a(n)=if(isprime(n), n, #Set(vector(n,i,lift(Mod(i-1,n)^n)))) \\ Charles R Greathouse IV, Jul 31 2016
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Python
def A195637(n): return len({pow(x,n,n) for x in range(n)}) # Chai Wah Wu, Aug 22 2023
Comments