[QUOTE=R.D. Silverman;287811]Don't be lazy.

A simple Google search will reveal the answer. (hint: Wikipedia)

One might ask: what motivates this question? do you have some reason to
suspect that it is true?[/QUOTE]
I have tried Wikipedia and several others but the only facts found are:
All Carmichael numbers C are square free,
All Carmichael numbers C have at least three factors,
If p is a prime dividing C then p - 1 divides C - 1.
The smallest Carmichael number is 561.
If (6k + 1), (12k + 1) and (18k + 1) are each prime, then there product gives a Carmichael Number.

Wiki gives you a link to OEIS sequence (among other tidbits).
You can take just the first terms and check
[SPOILER]and you will easily find 8911
[/SPOILER]

[QUOTE=Stan;287815]I have tried Wikipedia and several others but the only facts found are:

<snip>

.[/QUOTE]

You didn't look very hard.

[QUOTE=Batalov;287818]Wiki gives you a link to OEIS sequence (among other tidbits).
You can take just the first terms and check
[SPOILER]and you will easily find 8911
[/SPOILER][/QUOTE]

I should mention [url=https://oeis.org/A185321]A185321[/url] which José María Grau Ribas recently added to the OEIS.

[QUOTE]n5 has 51,217,600,457,105,052,828,189,829,037,592,592,347 digits.[/QUOTE]

How was this calculated?!

[QUOTE="Stan"]
[LEFT]...n[SUB]5[/SUB] has 51,217,600,457,105,052,828,189,829,037,592,592,347 digits.[/LEFT]
This also proves the sequence: n[SUB]0[/SUB] = 2, n[SUB]i+1[/SUB]= 2^n[SUB]i[/SUB]- 1, i E N defines a set of Mersenne primes.
[/QUOTE]

Quite a few digits less actually. :-)
Set precision in gp/pari ([SPOILER]\p 60[/SPOILER]) and enter [SPOILER]log(2)/log(10)*(2^127-1)[/SPOILER].

[QUOTE=siegert81;288875]How was this calculated?![/QUOTE]
n[SIZE=1]5 [SIZE=2]= 2^n[SIZE=1]4[SIZE=2] - 1 , log[SIZE=1]10[SIZE=2](n[SIZE=1]5[SIZE=2]) is approximately [/SIZE][/SIZE][/SIZE][/SIZE](n[SIZE=1]4[SIZE=2]*(log2) base 10[/SIZE][/SIZE][/SIZE][/SIZE][/SIZE][/SIZE]) +1.
So n[SIZE=1]5[SIZE=2] is approximately 10^([/SIZE][/SIZE][SIZE=1][SIZE=2][SIZE=1][SIZE=2]n[SIZE=1]4[SIZE=2]*(log2) base 10[/SIZE][/SIZE][/SIZE][/SIZE][/SIZE][/SIZE]).

Mersenne Primes
 

Could anyone verify or find a fault in the reasoning in the attached theorem?
Any replies, good or bad with reasons, would be appreciated.

[QUOTE=Stan;350636]Could anyone verify or find a fault in the reasoning in the attached theorem?
Any replies, good or bad with reasons, would be appreciated.[/QUOTE]
There is not only one flaw, but many. Easier than pointing them out is to ask you where did you use the fact that the sequence starts with 2? You can start the sequence with 5, your proof will flow exactly the same, but it generates the sequence 2^5-1, 2^31-1, etc, from which the third term is a proved composite.

What is the point of the new thread?

I am merging it to the old one.

Mersenne Primes
 

Could anyone please check the attached theory for errors?
I would be grateful for any replies, good or bad.
Thanks.