A052274 Number of distinct 5th powers mod n.

1, 2, 3, 3, 5, 6, 7, 5, 7, 10, 3, 9, 13, 14, 15, 9, 17, 14, 19, 15, 21, 6, 23, 15, 5, 26, 19, 21, 29, 30, 7, 17, 9, 34, 35, 21, 37, 38, 39, 25, 9, 42, 43, 9, 35, 46, 47, 27, 43, 10, 51, 39, 53, 38, 15, 35, 57, 58, 59, 45, 13, 14, 49, 34, 65, 18, 67, 51, 69, 70

Offset: 1

Author

N. J. A. Sloane, Feb 05 2000

Keywords

• nonn
• mult

Comments

This sequence is multiplicative. - Leon P Smith, Apr 16 2005

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• S. Li, On the number of elements with maximal order in the multiplicative group modulo n, Acta Arithm. 86 (2) (1998) 113, see proof of theorem 2.1

Crossrefs

Cf. A000224 (squares), A046530 (cubic residues), A052273 (4th powers), A052275 (6th powers), A085310 (7th powers), A085311 (8th powers), A085312 (9th powers), A085313 (10th powers), A085314 (11th powers), A228849 (12th powers).

Programs

Maple

A052274 := proc(m)
    {seq( modp(b^5,m),b=0..m-1) };
    nops(%) ;
end proc:
seq(A052274(m),m=1..100) ; # R. J. Mathar, Sep 22 2017

Mathematica

With[{nn=100},Table[Length[Union[PowerMod[Range[nn],5,n]]],{n,nn}]] (* Harvey P. Dale, Mar 19 2016 *)

PARI

a(n)=my(f=factor(n)); prod(i=1, #f[, 1], my(k=f[i, 1]^f[i, 2]); #vecsort(vector(k, i, i^5%k), , 8)) \\ Charles R Greathouse IV, Sep 05 2013

Formula

Conjecture: a(5^e) = 1+floor[(5-1)*5^(e+3)/(5^5-1)] if e == {0,2,3,4} (mod 5). a(5^e) = 5+floor[(5-1)*5^(e+3)/(5^5-1)] if e==1 (mod 5). - R. J. Mathar, Oct 22 2017

Conjecture: a(p^e) = 1+floor[(p-1)*p^(e+4)/{gcd(p-1,5)*(p^5-1)}] for primes p<>5 - R. J. Mathar, Oct 22 2017