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

%I A123365 #18 Feb 07 2022 20:15:51
%S A123365 1,7,13,19,31,37,43,61,67,73,79,97,103,109,127,139,151,157,163,181,
%T A123365 193,199,211,223,229,241,271,277,283,307,313,331,337,349,367,373,379,
%U A123365 397,409,421,433,439,457,463,487,499,523,541,547,571,577,601,607,613,619
%N A123365 Values of k such that A046530(k) = (k+2)/3, where A046530(k) is the number of distinct residues of cubes mod k.
%C A123365 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 A123365 R. J. Mathar, <a href="/A123365/b123365.txt">Table of n, a(n) for n = 1..10000</a>
%H A123365 Jon Maiga, <a href="http://sequencedb.net/s/A123365">Computer-generated formulas for A123365</a>, Sequence Machine.
%p A123365 n := 1 :
%p A123365 a := 1 :
%p A123365 while n <= 10000 do
%p A123365     printf("%d %d\n",n,a) ;
%p A123365     a := a+1 ;
%p A123365     while A046530(a) <> (a+2)/3 do
%p A123365         a := a+1 ;
%p A123365     end do:
%p A123365     n := n+1 ;
%p A123365 end do: # creates b-file, _R. J. Mathar_, Sep 21 2017
%Y A123365 Cf. A002476, A046530, A123722, A123723, A177965.
%K A123365 nonn
%O A123365 1,2
%A A123365 _John W. Layman_, Oct 12 2006