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
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
KEYWORD
nonn,easy
AUTHOR
Lear Young, May 22 2013
STATUS
approved