OFFSET
1,1
COMMENTS
It is possible to find an odd positive integer k and a set S = {p(1), ..., p(s)} containing only primes which appeared in the sequence such that for any nonnegative integer n, k*2^n + 1 == 0 (mod p(i)) and k*2^n - 1 == 0 (mod p(j)) for some p(i) and some p(j) from the set S.
MATHEMATICA
Select[Divisors[2^720-1], PrimeQ]
PROG
(Magma) PrimeDivisors(2^720-1);
(PARI) forprime(p=1, , if(Mod(2, p)^720==1, print1(p, ", "))) \\ Felix Fröhlich, Aug 23 2017
CROSSREFS
KEYWORD
nonn,easy,fini,full
AUTHOR
Arkadiusz Wesolowski, Aug 23 2017
STATUS
approved