# A VERY interesting Fermat-like sequence: A_n=4^3^n+2^3^n+1

1. May 7, 2005

### T.Rex

Hi,
In 1995, Yannick Saouter produced the study of a family of numbers close to the Fermat numbers: $$A_n=4^{3^n}+2^{3^n}+1$$ .
(See: http://www.inria.fr/rrrt/rr-2728.html)

Saouter proved that this A_n serie shares many properties with the Fermat numbers:

3.4 A_n numbers are pairwise relatively primes

3.3 $$A_n \text{ is prime iff } 5^{(A_n-1)/2} \equiv -1 \pmod{A_n}$$

3.5 $$p | A_n ==> p = 1 \pmod{2.3^{n+1}}$$

I've also discovered (and checked with PARI) that the following property is true:
$$A_n = 3+2(2^{3^{\scriptstyle n}-1}+1)\prod_{i=0}^{n-1}A_i$$

(I've summarized the properties at: http://tony.reix.free.fr/Mersenne/PropertiesOfFermatLikeTNumbers.pdf)

A VERY interesting thing is that the primality of these numbers can be checked with the Pépin's test, with 5 instead of 3, like Fermat numbers.

Saouter provides the divisors of several of these numbers (n up to 39).
It appears:
- that 10 of them have no divisors known
- only the 3 first Saouter numbes are prime (same as Fermat numbers).

Is someone interested in studying these numbers in more details ?

Is it possible to adapt some Pépin's test code to this kind of numbers ?

Regards,
Tony

2. May 7, 2005

### Zurtex

Erm interesting, but due to their size it's a bit computationally awkward to calculate isn't it? I mean I'm sure there are quite a few sequences that behave in similar ways but of such great size that they can't be calculated in any short amount of time.

Not criticizing you at all, it's something I couldn't do, just wondering.

3. May 8, 2005

### T.Rex

You're right. These numbers grow awfully fast, much faster than Fermat numbers.
What I think really interesting is that they share several properties with Fermat numbers. Do they also seem to have a finite number of primes ? What about the equivalent of Sierpinski's problem Saouter has studied ? Maybe a proof for Saouter numbers could help for Fermat numbers ?
Tony

4. May 8, 2005

### T.Rex

Hi Zurtex,
I've updated the paper with some proof and with the number of digits of A_n.
For n=16, A_n has ~26 millions of digits. Very big !
Tony