# Sequence A049096 with Counterexample or New Definition

Gottfried Helms Annette.Warlich at t-online.de
Tue Nov 1 12:47:00 CET 2005

```Seqfans, hallo Rainer  -

Am 01.11.2005 11:22 schrieb Rainer Rosenthal:
> The number k=182 is missing in
> http://www.research.att.com/projects/OEIS?Anum=A049096
> because 2^k+1 = 0 (mod 1093^2) and the definition is
> "**2^n + 1 is divisible by a square"
>
> Possibility #1: stick to the definition, enhancing
> it slightly to "2^a(n) + 1 is divisible by a square"
> and inserting the number 182 between 177 and 183.
>
> Possibility #2: change the definition according
> to the author's (Vladeta Jovovic) conjecture:
> "2^n+1 is square-free iff gcd(n,2^n+1)=1"
> which is proven wrong by n=182.
> This will give the longer but correct definition:
> "2^a(n) + 1 is divisible by a square and coprime to a(n)"
>
> My favorite is possibility #1 with a comment regarding
> the exceptional value a(43) = 182.
>
>    ....,,;;---''^^^''---;;,,,....
>
since this a problem expressing a property which is in direct
connection to a type of primes known as wieferich-primes,
and it is not known, how many wieferich primes exist (even
not whether there are inifnitely many) I think an additional
remark should be added, saying, that we don't know whether
there are more exceptions above, say 10^???, the current
limit to where we checked the existence of such wieferich
primes.

WIth p=1093 it is, that its group order is (1093-1) div 3 =364,
and 2^364-1 is thus divisible by 1093 (and even by 1093^2,
p being a wieferich prime), 364 is even and thus one of the
factors
(2^182-1)*(2*182+1)
must be divisible by 1093^2 (cannot be both; their difference
is only 2)

So for each wieferich prime p with an even group order k we
have to check, whether 2^(k/2)-1 or 2^(k/2)+1 is divisible
by p^2 - but we cannot exclude definitely the existence of
infinitely many of such primes.

Regards-
and Grüßle

Gottfried

```