Mersenne Prime Sequence
 

Could someone please check the attached theorem for errors and post a reply for the location of any errors?[ATTACH]7579[/ATTACH]

[QUOTE=Stan;286809]Could someone please check the attached theorem for errors and post a reply for the location of any errors?[ATTACH]7579[/ATTACH][/QUOTE]
See [URL="http://mersenneforum.org/showpost.php?p=286808&postcount=26"]here[/URL] and [URL="http://mersenneforum.org/showpost.php?p=286812&postcount=28"]here[/URL].

[QUOTE=ccorn;286814]See [URL="http://mersenneforum.org/showpost.php?p=286808&postcount=26"]here[/URL] and [URL="http://mersenneforum.org/showpost.php?p=286812&postcount=28"]here[/URL].[/QUOTE]
ccorn quotes "5*phi(5) | phi(33) but 25 does not divide 33", but my statement includes that the prime 5 [COLOR=black][U]has[/U][/COLOR] to divide (33 - 1) and it does not.

[QUOTE=Stan;286819]ccorn quotes "5*phi(5) | phi(33) but 25 does not divide 33", but my statement includes that the prime 5 [COLOR=black][U]has[/U][/COLOR] to divide (33 - 1) and it does not.[/QUOTE]
The second "[tex]\Rightarrow[/tex]" in the fifth line from the bottom and the following "[tex]\Rightarrow[/tex]" aren't. ([I]non sequitur[/I], so to say.)
I expect that you will notice the gaps (if you have not done so already) when you try to improve the written reasoning at those points.

Stan,

Could you please explain how your proof path would differ for a similar sequence:
n[SUB]0[/SUB] = 2^5-1 (prime)
n[SUB]1[/SUB] = 2^n[SUB]0[/SUB]-1 (prime)
n[SUB]2[/SUB] = 2^n[SUB]1[/SUB]-1 (composite, has known factors)

What is the specific reason that we wouldn't be able to plug it in the same proof and demonstrate that n[SUB]2[/SUB] is actually prime?

[QUOTE=Batalov;286841]Stan,

Could you please explain how your proof path would differ for a similar sequence:
n[SUB]0[/SUB] = 2^5-1 (prime)
n[SUB]1[/SUB] = 2^n[SUB]0[/SUB]-1 (prime)
n[SUB]2[/SUB] = 2^n[SUB]1[/SUB]-1 (composite, has known factors)

What is the specific reason that we wouldn't be able to plug it in the same proof and demonstrate that n[SUB]2[/SUB] is actually prime?[/QUOTE]
2^5-1 = 1+2.3.5 and phi(2^5-1) = 2.3.5
phi(2^5-1) does not divide phi(2^n[SUB]0[/SUB]-1), therefore no sequence.
My proof relies on the chain:
phi(2^n[SUB]0[/SUB]-1) | phi(2^n[SUB]1[/SUB]-1) | phi(2^n[SUB]2[/SUB]-1) etc.

Mersenne Primes
 

I believe the proof of my theorem to be now complete but I still need it checking. Any comments would be appreciated.[ATTACH]7581[/ATTACH]
The attached PDF file has been updated.

n[SUB]0[/SUB] = 19 (prime)
n[SUB]1[/SUB] = 2^n[SUB]0[/SUB]-1 (prime)
n[SUB]2[/SUB] = 2^n[SUB]1[/SUB]-1 (?composite?)

[QUOTE=Stan;286889]I believe the proof of my theorem to be now complete but I still need it checking. Any comments would be appreciated.[ATTACH]7581[/ATTACH]
The attached PDF file has been updated.[/QUOTE]
There is still no attempt to thoroughly derive the two alleged [I]non sequitur[/I]s. Try to actually prove those points. This will turn out to be insightful, whatever the outcome.

[QUOTE=Stan;286819]ccorn quotes "5*phi(5) | phi(33) but 25 does not divide 33", but my statement includes that the prime 5 [COLOR=black][U]has[/U][/COLOR] to divide (33 - 1) and it does not.[/QUOTE]
Well then: 5*phi(5) | phi(66) and 5 | (66-1), but 25 does not divide 66.

[QUOTE=Stan;286859]2^5-1 = 1+2.3.5 and phi(2^5-1) = 2.3.5
phi(2^5-1) does not divide phi(2^n[SUB]0[/SUB]-1), therefore no sequence.
My proof relies on the chain:
phi(2^n[SUB]0[/SUB]-1) | phi(2^n[SUB]1[/SUB]-1) | phi(2^n[SUB]2[/SUB]-1) etc.[/QUOTE].

[QUOTE=ccorn;286947]Well then: 5*phi(5) | phi(66) and 5 | (66-1), but 25 does not divide 66.[/QUOTE]
Once again you are correct, but I have not been sufficiently explicit since all the primes in the sequence (2^n[SUB]i[/SUB])-1; 0 < i < 5, are congruent to 3 mod.4
So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1.

[QUOTE=Stan;287058]Once again you are correct, but I have not been sufficiently explicit since all the primes in the sequence (2^n[SUB]i[/SUB])-1; 0 < i < 5, are congruent to 3 mod.4
So I require p*phi(p) | phi(m) and p | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1.[/QUOTE]
If you require p | (m - 1), you cannot conclude p^2 | m anyway...

P.S.: You have presented no reasoning why from such phi(a) | phi(b) you could possibly conclude a | b. In fact, many coprime numbers share the same totient function value, so you cannot easily carry divisibiliy properties across applications of the totient function. With very strong conditions on a or b this might be possible (e. g. if a and b are increasing powers of 2 or 6), but you would still need to provide a proof why such a rather singular case should happen.

P.P.S.: This is the second of the two alleged [I]non sequitur[/I]s, the first one deserves attention too.

[QUOTE=ccorn;287060]If you require p | (m - 1), you cannot conclude p^2 | m anyway...

P.S.: You have presented no reasoning why from such phi(a) | phi(b) you could possibly conclude a | b. In fact, many coprime numbers share the same totient function value, so you cannot easily carry divisibiliy properties across applications of the totient function. With very strong conditions on a or b this might be possible (e. g. if a and b are increasing powers of 2 or 6), but you would still need to provide a proof why such a rather singular case should happen.[/QUOTE]
I take it that you have read the PDF I asked to be checked for errors.
I do not assume phi(a) | phi(b) implies a | b but I have discovered a situation where p*phi(p) | phi(m) and p*phi(p) | (m-1) from which I propose that m is a prime.

[QUOTE=Stan;287064]I take it that you have read the PDF I asked to be checked for errors.
I do not assume phi(a) | phi(b) implies a | b but I have discovered a situation where p*phi(p) | phi(m) and p*phi(p) | (m-1) from which I propose that m is a prime.[/QUOTE]
Aha. For what reason? For the time being, take p = 5, m=341 as a counterexample. There may well be a large gcd(phi(m),m-1).
Also, consider Batalov's second counterexample.

[QUOTE=ccorn;287068]Aha. For what reason? For the time being, take p = 5, m=341 as a counterexample. There may well be a large gcd(phi(m),m-1).
Also, consider Batalov's second counterexample.[/QUOTE]
So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1. In fact, p = 2^(2^q - 1) - 1
where q is a prime of the same form.
In Batalov's second counterexample, 19 is not 2^(a prime) - 1.

[QUOTE=Stan;287075]19 is not 2^(a prime) - 1.[/QUOTE]
Is your starting element 2 a 2^(a prime) - 1?

[QUOTE=Stan;287075]So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1. In fact, p = 2^(2^q - 1) - 1
where q is a prime of the same form.
In Batalov's second counterexample, 19 is not 2^(a prime) - 1.[/QUOTE]
Where is the logical link (the reasoning) between all those requirements and your desired conclusion?

The current state, as I perceive it, is that you list a number of true properties (mostly laid down in propositions 1 and 2) and then claim what you like.

For example, you essentially claim that gcd(phi(m),m-1) = m-1 (where m=n[SUB]5[/SUB]) but you provide only a divisor of the gcd. (And only a divisor of that divisor is actually granted, due to the first [I]non sequitur[/I].) You will agree that this is not a proof.

[QUOTE=Stan;287075]So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1. In fact, p = 2^(2^q - 1) - 1
where q is a prime of the same form.
In Batalov's second counterexample, 19 is not 2^(a prime) - 1.[/QUOTE]

2^(2^q-1)-1 means p is a double Mersenne if m=2^p-1 then you are talking about triple Mersennes ?

2^3-1 = 2^(2^2-1)-1
2^7-1=2^(2^3-1)-1 = 2^(2^(2^2-1)-1)-1

so we can assume that you speak of all the terms > M7=MM3=MMM2 correct ?

of course M127=MM7=MMM3=MMMM2

[QUOTE=science_man_88;287114]2^(2^q-1)-1 means p is a double Mersenne if m=2^p-1 then you are talking about triple Mersennes ?

2^3-1 = 2^(2^2-1)-1
2^7-1=2^(2^3-1)-1 = 2^(2^(2^2-1)-1)-1

so we can assume that you speak of all the terms > M7=MM3=MMM2 correct ?

of course M127=MM7=MMM3=MMMM2[/QUOTE]

oh and how could I forget , say p= 3 mod 4 2*p+1 = (3*2+1) mod 4 = 7 mod 4 = 3 mod 4 so every mersenne >=3 works out as a possible p.

Carmichael Numbers
 

Is it true that all Carmichael numbers are congruent to 1 modulo 4?

[QUOTE=Stan;287808]Is it true that all Carmichael numbers are congruent to 1 modulo 4?[/QUOTE]

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=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.

[QUOTE=Webber;350657]After a cursory glance, I can find no logical errors but will examine further and inform you should I find an error.[/QUOTE]

Dear "Stan Webber": Dude, if you're going to try to upvote your own crank-drivel by registering under a second userID, you should at least make the effort to disguise the fact. Buh-bye.

What was his original ID?

Edit: Never mind. I found it on the previous page (Stan).