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

3, 7, 11, 13, 17, 19, 29, 31, 37, 41, 53, 59, 61, 67, 83, 97, 101, 107, 113, 127, 131, 137, 139, 149, 163, 173, 179, 181, 193, 197, 211, 227, 257, 269, 281, 293, 313, 317, 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

Offset

1,1

Comments

It is conjectured that:

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

The numbers in the sequence are the values of n in the above conjecture.

Links

Table of n, a(n) for n=1..93.

Hagen von Eitzen, Details of the "cycle length of n"

Example

929 : (929-1)/(2^2)=232=A179382((929+1)/2) and znorder(Mod(10,929))=464=(929-1)/2

Prog

(PARI) oddres(n)=n>>valuation(n, 2)
cyc(d)=my(k=1, t=1); while((t=oddres(t+d))>1, k++); k
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", ")))
\\ Charles R Greathouse IV, May 22 2013

Crossrefs

Cf. A179382, A136042 (both sequences related to the way to get the "cycle length of n").

Sequence in context: A020600 A020598 A310193 * A023236 A038920 A361822

Adjacent sequences: A226011 A226012 A226013 * A226015 A226016 A226017

Keyword

nonn,easy

Author

Lear Young, May 22 2013

Status

approved