$M_{M_{M_{p}}} = {2^2}^{2^{p - 1}} - 1, p \geq 1$ and either a prime or a non-prime

$p = 1, {2^2}^{2^{1 - 1}} - 1 = 4 - 1 = 3$ is a prime

$p = 2, {2^2}^{2^{2 - 1}} - 1 = 16 - 1 = 15 \neq$ a prime

$p = 3, {2^2}^{2^{3 - 1}} - 1 = 65536 - 1 = 65535 \neq$ a prime

$p = 4, {2^2}^{2^{4 - 1}} - 1 = 1.15792089×10^{77} - 1 = 1.15792089×10^{77} \neq$ a prime

$p = 5, {2^2}^{2^{5 - 1}} - 1 = 2^{65536} - 1 \neq$ a prime:

Therefore, there is no triple Mersenne prime where $p$ is a prime, however, when $p = 1$, a non-prime, it produces a prime. I have also discovered Triple Mersenne Numbers!

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## Comments

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TopNewest@Yajat Shamji, why is it undefined? Its value is still $2 ^ {2 ^ {16}} - 1$

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Trying to find that out now, I have further simplified it to $2^{65536} - 1$. @Aryan Sanghi

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It is very big, but that is defined, isn't it? Also, maybe at p = 11, it is a prime, but so big we can't evaluate, isn't it?

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@Aryan Sanghi.

What program you got? You sound very confident,Log in to reply

$a ^ 2 - b ^2$ which can be factored as (a + b)(a - b). I hope I explained well. :)

Actually above number is defined, codes can't evaluate that much big numbers accurately(I do code and that's why I am saying). Above number is composite for every positive integer p as it can be written of formLog in to reply

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$p = 4$ is a prime - let me check first.

I will check but I believe thatLog in to reply

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You just checked for numbers 2,3,4,5 then how do you know that there can't be any prime after that?

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As it can be written of form a² - b² which can be factored into (a + b)(a - b)

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That is OK, but the proof in the note is not valid in my opinion.

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Just checking for some numbers is not a proof, I think it should be done algebraically.

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You can read the long discussion we had in previous comment.

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@Yajat Shamji, good job!! I never thought of p = 1.

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@Aryan Sanghi, what was the proof you were telling of there? I wish to know it!

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Simply that 2^2^2^(p-1) can be written as a² and 1 can be written as 1² which can be factored to (a + 1)(a -1).

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I am making a note on Quadruple Mersenne Numbers and Primes. What do you think, @Aryan Sanghi, @Vinayak Srivastava?

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I guess you should rather post a question on it.

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What should the question be? @Aryan Sanghi

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What should the question be? @Vinayak Srivastava

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In mathematics it is not sufficient to check only some cases to prove something, either you check for all primes (which are infinite), or either you give a mathematical proof so that the statement can mathematically be accepted.

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@Zakir Husain, thank you so much for reminding me but I think you should ask @Aryan Sanghi. He has got the proof.

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@Mahdi Raza, @Alak Bhattacharya, @Gandoff Tan, @Hamza Anushath, @Aryan Sanghi

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Will nobody comment on the fact that I have found Triple Mersenne numbers? @Vinayak Srivastava, @Aryan Sanghi

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Can you tell any such prime you have discovered? :)

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I haven't discovered any prime as of yet.

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$2^{131072} - 1$ is undefined according to all calculators or is above their amount of numbers limit.

I am working on it - currentlyLog in to reply

@Aryan Sanghi said, $2^{131072}$ is a square and so it will be a difference of two squares. Also, $a-b\neq 1$, so it

It is not prime. Aswillhave at least 2 factors other than itself and 1.Log in to reply

stop talking about Triple Mersenne primes?! I have discoveredTriple Mersenne numbers!Log in to reply

20 +factors!Log in to reply

$n=1$, then $2^{2^{{2^{0}}}}-1=2^{{2^1}}-1=4-1=3$, which is prime. I don't know of any other case.

There won't be any prime of such type, except whenLog in to reply

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is a Triple Mersenne PrimeI am editing the note right now!Log in to reply

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@Yajat Shamji to see.)

But what is the flaw in my comment? (I know it, you know it, I wantLog in to reply

$1$ is not a prime - look at the note now!

I knowLog in to reply

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stop talking about Triple Mersenne primes?! I have discoveredTriple Mersenne numbers!Log in to reply

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