gcd(n,2^(2n^2)-2) is usually 1 or 2

[Joshua Zucker]

Hi folks,
Someone on another mailing list (I don't know why) asked about a proof
that gcd(n, 2^(2n^2)-2) = 1 for n odd, 2 for n even.

But of course it's not true, though the exceptions are pretty rare,
and at least the small exceptions all involve a factor of either 47 or
127, which seems interesting.

But I don't know enough number theory to say whether this sequence (of
exceptions to the 1 or 2 pattern) is interesting or not.

Can someone help me decide whether this is worth submitting or not?

n such that gcd(n, 2^(2n^2 - 2)) is greater than 2:
254 423 635 658 1143 1504 1524 1739 2032 2413 2585 2820 2921 3302 3666
3810 3901 4191 4699 4747 4982 5080 5588 5828 5969 6063 6477 6858 6909
7024 7144 7366 7747 7990 8225 8255 8636 8843 9071 9144 9306 9525

and the corresponding gcds:
254 47 127 94 127 94 254 47 254 127 47 94 127 254 94 254 47 127 127 47
94 254 254 94 127 47 127 254 47 878 94 254 127 94 47 127 254 239 47
254 94 127

Hm, that 878 and 239 seem to be interlopers in a crowd full of 47s and 127s.

What's special about the primes that appear in these GCDs?  Up to n =
100000, all we get are the obvious 2, and then a lot of  47 and 127
(about 410 of them), and also a few others (about 35) built from these
primes:
439 239 1289 383 479 2351 3463 1487 11447 863 4513 2687

and most of those small ones occur several times -- it sure doesn't
look "random".

Maybe there is something interesting here after all?

--Joshua Zucker