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

%I #18 Feb 07 2022 20:15:51

%S 1,7,13,19,31,37,43,61,67,73,79,97,103,109,127,139,151,157,163,181,

%T 193,199,211,223,229,241,271,277,283,307,313,331,337,349,367,373,379,

%U 397,409,421,433,439,457,463,487,499,523,541,547,571,577,601,607,613,619

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

%C 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.

%H R. J. Mathar, <a href="/A123365/b123365.txt">Table of n, a(n) for n = 1..10000</a>

%H Jon Maiga, <a href="http://sequencedb.net/s/A123365">Computer-generated formulas for A123365</a>, Sequence Machine.

%p n := 1 :

%p a := 1 :

%p while n <= 10000 do

%p printf("%d %d\n",n,a) ;

%p a := a+1 ;

%p while A046530(a) <> (a+2)/3 do

%p a := a+1 ;

%p end do:

%p n := n+1 ;

%p end do: # creates b-file, _R. J. Mathar_, Sep 21 2017

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

%K nonn

%O 1,2

%A _John W. Layman_, Oct 12 2006