A123365 Values of k such that A046530(k) = (k+2)/3, where A046530(k) is the number of distinct residues of cubes mod k.

1, 7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 157, 163, 181, 193, 199, 211, 223, 229, 241, 271, 277, 283, 307, 313, 331, 337, 349, 367, 373, 379, 397, 409, 421, 433, 439, 457, 463, 487, 499, 523, 541, 547, 571, 577, 601, 607, 613, 619

Offset: 1

Author

John W. Layman, Oct 12 2006

Keywords

• nonn

Comments

Conjecture: With the exception of the first term a(1)=1, this is exactly the sequence of primes of the form 6k+1 (A002476). This has been verified up to a(n)=2000.

Links

• R. J. Mathar, Table of n, a(n) for n = 1..10000
• Jon Maiga, Computer-generated formulas for A123365, Sequence Machine.

Crossrefs

Cf. A002476, A046530, A123722, A123723, A177965.

Programs

Maple

n := 1 :
a := 1 :
while n <= 10000 do
    printf("%d %d\n",n,a) ;
    a := a+1 ;
    while A046530(a) <> (a+2)/3 do
        a := a+1 ;
    end do:
    n := n+1 ;
end do: # creates b-file, R. J. Mathar, Sep 21 2017