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A387595
Numbers k that divide both x^2 + 1 and 2^x + 1 for some number x.
2
1, 5, 13, 17, 29, 37, 41, 53, 61, 65, 97, 101, 109, 113, 137, 145, 149, 157, 173, 181, 185, 193, 197, 229, 241, 257, 265, 269, 277, 281, 293, 313, 317, 349, 353, 373, 377, 389, 397, 401, 409, 421, 433, 449, 457, 461, 481, 509, 521, 533, 541, 545, 557, 565, 569, 577, 593, 613, 617, 641, 653, 661
OFFSET
1,2
COMMENTS
Numbers that divide some term of A066787.
Every divisor of a term is a term.
The first term that is a square > 1 is 1093^2, with 49351666^2 + 1 and 2^49351666 + 1 both divisible by 1093^2.
All prime divisors of a term that is a square must be Wieferich primes == 1 (mod 4). The only known Wieferich prime == 1 (mod 4) is 1093.
EXAMPLE
a(3) = 13 is a term because 13 divides both 18^2 + 1 = 325 = 13 * 25 and 2^18 + 1 = 262145 = 13 * 20165.
MAPLE
# Note: due to a bug in Maple, this program fails for 1093^2 (at least up to Maple 2025)
filter:= proc(n) local a, b, b0, b1, x, t, tt;
uses NumberTheory;
if n::even then return false fi;
a:=[msolve(x^2 + 1 = 0, n)];
if a = [] then return false fi;
a:= map(t -> rhs(op(t)), a);
b:=msolve(2^x + 1 = 0, t, n);
if b = {} or b = NULL then return false fi;
b:= rhs(op(b));
tt:= indets(b)[1];
b0:= subs(tt=0, b); b1:= coeff(b, tt);
for x in a do
if ChineseRemainder([x, b0], [n, b1]) <> FAIL then return true fi
od;
false
end proc:
filter(1):= true:
select(filter, [seq(seq(i+6*j, i=[1, 5]), j=0..1000)]);
CROSSREFS
Subset of A004613. Cf. A001220, A066787.
Sequence in context: A279857 A077426 A231754 * A175768 A351535 A002144
KEYWORD
nonn
AUTHOR
Robert Israel, Sep 02 2025
STATUS
approved