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A226014 Primes p such that A179382((p+1)/2) = (p-1)/(2^x) for some x>0.

%I A226014 #39 Sep 26 2015 13:20:15
%S A226014 3,7,11,13,17,19,29,31,37,41,53,59,61,67,83,97,101,107,113,127,131,
%T A226014 137,139,149,163,173,179,181,193,197,211,227,257,269,281,293,313,317,
%U A226014 347,349,353,373,379,389,401,409,419,421,443,449,461,467,491,509,521,523,541,547,557,563,569,577,587,593,613,617,619,653,659,661,677,701,709,757,761,769,773,787,797,809,821,827,829,853,857,859,877,883,907,929,941,947,977
%N A226014 Primes p such that A179382((p+1)/2) = (p-1)/(2^x) for some x>0.
%C A226014 It is conjectured that:
%C A226014 Let n be an odd number and the period of 1/n is n-1 or a divisor of n-1. Call c=A179382((n+1)/2) the "cycle length of n". If c divides n-1 or n+1 = 2^x for some x>0, then n is prime. For details see link and Cf. - _Lear Young_, with contributions from Peter Košinár, _Giovanni Resta_, _Charles R Greathouse IV_, May 22 2013
%C A226014 The numbers in the sequence are the values of n in the above conjecture.
%H A226014 Hagen von Eitzen, <a href="http://math.stackexchange.com/questions/394408/how-to-prove-these-two-ways-give-the-same-numbers">Details of the "cycle length of n"</a>
%e A226014 929 : (929-1)/(2^2)=232=A179382((929+1)/2) and znorder(Mod(10,929))=464=(929-1)/2
%o A226014 (PARI) oddres(n)=n>>valuation(n, 2)
%o A226014 cyc(d)=my(k=1, t=1); while((t=oddres(t+d))>1, k++); k
%o A226014 forstep(n=3, 1e3, [4, 2, 2, 2], x=cyc(n);z=znorder(Mod(10, n));if((x==1 || (n%x==1 && oddres((n-1)/x)==1)) && (n%z==1 || n%z==0), print1(n", ")))
%o A226014 \\ _Charles R Greathouse IV_, May 22 2013
%Y A226014 Cf. A179382, A136042 (both sequences related to the way to get the "cycle length of n").
%K A226014 nonn,easy
%O A226014 1,1
%A A226014 _Lear Young_, May 22 2013