OFFSET
1,1
COMMENTS
Associated primes q = 31, 257, 89, 151, 73, ...
If p is in the sequence, then 2^(q-p) == 1 (mod pq).
If p is in the sequence, then 2^gcd(p-1,q-1) == 1 (mod pq). - Robert Israel, Aug 03 2016
From Thomas Ordowski, Aug 03 2016: (Start)
Theorem: if p == 1 (mod 4) and q = 2p-1 are primes, then 2^(q-p) == 1 (mod pq); such p = 37, 97, 157, 229, ...
Thus q divides 2^(p-1)-1 and p divides 2^(q-1)-1.
Problem: are there infinitely many pairs of primes p < q such that 2^(q-p) == 1 (mod pq)?
Lemma: let p < q are primes, then 2^(pq-1) == 1 (mod pq) if and only if 2^(q-p) == 1 (mod pq). (End)
EXAMPLE
For p=11 and q=31, 2^(p-1)-1 = 3*p*q and 2^(q-1)-1 = 3*3*7*p*q*151*331.
MAPLE
filter:= proc(p)
local Q, q, t;
if not isprime(p) then return false fi;
Q:= select(type, map(t -> t[1], ifactors(2^(p-1)-1, easy)[2]), integer);
q:= min(select(`>`, Q, p));
if not q::integer then
q:= min(select(`>`, numtheory:-factorset(2^(p-1)-1), p));
if not q::integer then return false fi;
fi;
if 2 &^(q-1) mod p <> 1 then return false fi;
for t from p+2 to q-2 by 2 do
if isprime(t) and 2 &^(q-1) mod t = 1 then return false fi
od;
true
end proc;
select(filter, [seq(p, p=3..700, 2)]); # Robert Israel, Aug 03 2016
PROG
(PARI) is(n)=if(!isprime(n), return(0)); my(f=factor(2^(n-1)-1)[, 1], q); f=select(k->k>n, f); if(#f==0, return(0)); q=f[1]; forprime(p=n+1, q-1, if(Mod(2, p)^(q-1)==1, return(0))); Mod(2, n)^(q-1)==1 \\ Charles R Greathouse IV, Aug 03 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Ordowski, Aug 03 2016
EXTENSIONS
a(2) inserted by Charles R Greathouse IV, Aug 03 2016
a(6)-a(21) from Charles R Greathouse IV, Aug 03 2016
a(22)-a(30) from Robert Israel, Aug 03 2016
a(31)-a(50) from Charles R Greathouse IV, Aug 03 2016
STATUS
approved