mersenneforum.org Sum of the reciprocals of all Mersenne prime exponents
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2021-08-19, 14:07   #12
Dobri

"Καλός"
May 2018

17×19 Posts

Quote:
 Originally Posted by jnml Coincidence? I don't think so /s
Well, the comment ends in /s (which means being sarcastic) so one should take it with a pinch of salt.

2021-08-19, 15:07   #13
Dobri

"Καλός"
May 2018

32310 Posts

Quote:
 Originally Posted by Dobri An attempt to apply the Wynn's Epsilon Method https://mathworld.wolfram.com/WynnsEpsilonMethod.html for convergence acceleration gives 2.813839133... which is greater than 2.813838890...
Assuming that SumEPSILON is close to the unknown SumTOTAL, one could entertain the idea of estimating a wide range for M52 as follows:

π(M52) (1 + π(M52)/π(M53) + π(M52)/π(M54) + ...) / (SumEPSILON - Sum51)

for some π(M52)/π(M53) < 1, π(M52)/π(M54) < 1,...

2021-08-19, 15:50   #14
Dr Sardonicus

Feb 2017
Nowhere

22×1,459 Posts

Quote:
 Originally Posted by charybdis I don't think it's even known that there are infinitely many composite 2^p-1 - which makes proving the infinitude of Mersenne primes seem pretty unrealistic.
The things we don't know about 2p - 1, p prime, are legion. We don't know whether infinitely many are prime. [If there are only finitely many Mersenne primes, the sum of the reciprocals of the exponents is a finite sum of rational numbers, hence is rational.) We don't know whether infinitely many are composite. (If 2p - 1 is prime for all but finitely many p, the sum diverges.) Just about everybody and his uncle believes there are infinitely many primes p = 4n+3 for which 8n+ 7 is also prime (though AFAIK nobody knows how to prove it), so "prime for all but finitely many p" seems unlikely. If 2p - 1 is prime for "enough" p (I think a positive proportion is enough but am not certain) the sum diverges.

We don't even know whether 2p - 1 is square-free for infinitely many primes p, though AFAIK not a single prime p is known for which 2p - 1 is divisible by the square of any prime.

 2021-08-20, 08:35 #15 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 346310 Posts The sequence 2^n-1 can be generalized to any exponential sequence (a*b^n+c)/gcd(a+c,b-1) (with a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, |c|>=1, gcd(a,c) = 1, gcd(b,c) = 1), if this sequence does not have full numerical covering set (e.g. 78557*2^n+1, (11047*3^n+1)/2, (419*4^n+1)/3, 509203*2^n-1, (334*10^n-1)/9, 14*8^n-1, etc.) or full algebraic covering set (e.g. 1*8^n+1, 8*27^n+1, (1*8^n-1)/7, (1*9^n-1)/8, 4*9^n-1, 2500*16^n+1, etc.) or partial algebraic/partial numerical covering set (e.g. 4*24^n-1, 4*39^n-1, (4*19^n-1)/3, (343*10^n-1)/9, 1369*30^n-1, 2500*55^n+1, etc.), we can find the sum of the reciprocals of the positive integers n such that (a*b^n+c)/gcd(a+c,b-1) is prime. Conjecture: If sequence (a*b^n+c)/gcd(a+c,b-1) does not have covering set (full numerical covering set, full algebraic covering set, or partial algebraic/partial numerical covering set), then the sum of the reciprocals of the positive integers n such that (a*b^n+c)/gcd(a+c,b-1) is prime is converge (i.e. not infinity) and transcendental number. (of course, this conjecture will imply that there are infinitely many such n) Last fiddled with by sweety439 on 2021-08-20 at 08:41
2021-08-20, 12:15   #16
charybdis

Apr 2020

751 Posts

Quote:
 Originally Posted by sweety439 Conjecture: If sequence (a*b^n+c)/gcd(a+c,b-1) does not have covering set (full numerical covering set, full algebraic covering set, or partial algebraic/partial numerical covering set), then the sum of the reciprocals of the positive integers n such that (a*b^n+c)/gcd(a+c,b-1) is prime is converge (i.e. not infinity) and transcendental number. (of course, this conjecture will imply that there are infinitely many such n)
This is almost certainly false. We expect that for any b there are finitely many primes of the form b^n+1, since these can only be prime when n is a power of 2 (generalized Fermat numbers). The same holds for (b^n+1)/2 when b is odd.

2021-08-20, 14:57   #17
RomanM

Jun 2021

2·52 Posts

Quote:
 Originally Posted by Dr Sardonicus It seems "obvious" that the Mersenne prime exponents (primes p such that 2p - 1 is prime) are sufficiently rare that the sum of their reciprocals converges. However, I am not aware of any proof of this.
Much better this way, this series converges absolutely) sum = sum + (-1)^n/Mexponent[[n]]

Last fiddled with by RomanM on 2021-08-20 at 15:38 Reason: Insert number, -0.2697096036

2021-08-20, 16:31   #18
Dobri

"Καλός"
May 2018

14316 Posts

Quote:
 Originally Posted by RomanM Much better this way, this series converges absolutely) sum = sum + (-1)^n/Mexponent[[n]]
Perhaps the multiplier (-1)^(n-1) could be used instead for the sum of the alternating sequence to be positive, 0.2697096036...

Also, the sum of the alternating sequence of the reciprocals of the number of primes π(p) less than or equal to the known Mersenne prime exponents tends to converge to 0.6628596647...

 2021-08-20, 21:54 #19 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 1000110110102 Posts I have no way of proving any of the following, so I state them as virtual (as opposed to literal) wagers: * There are infinite Mersenne primes * The sum of the reciprocals of Mersenne-Prime-Exponents does not converge My reasoning for positing The 2nd statement is that for the known Mersenne-Primes the tendency is for the next exponent of each exponent to be less than double. This would make it less convergent than the infinite series: (1/2)^1 + (1/2)^2 + (1/2)^3 + ..... Just my 2 cents.
2021-08-20, 22:11   #20
charybdis

Apr 2020

751 Posts

Quote:
 Originally Posted by a1call My reasoning for positing The 2nd statement is that for the known Mersenne-Primes the tendency is for the next exponent of each exponent to be less than double. This would make it less convergent than the infinite series: (1/2)^1 + (1/2)^2 + (1/2)^3 + .....
What about the sequence 1.1, 1.1^2, 1.1^3, ...? Every term is much less than double the previous term, but the geometric series (1/1.1) + (1/1.1)^2 + (1/1.1)^3 + ... still converges. This is basic high-school stuff.

Or what about square numbers? The ratio between successive squares tends to 1, but the sum of the reciprocals converges.

Last fiddled with by charybdis on 2021-08-20 at 22:12

2021-08-20, 22:45   #21
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

8DA16 Posts

Quote:
 Originally Posted by charybdis What about the sequence 1.1, 1.1^2, 1.1^3, ...? Every term is much less than double the previous term, but the geometric series (1/1.1) + (1/1.1)^2 + (1/1.1)^3 + ... still converges. This is basic high-school stuff. Or what about square numbers? The ratio between successive squares tends to 1, but the sum of the reciprocals converges.
sum_(n=1)^∞ 1.1^n diverges to ∞

https://www.wolframalpha.com/input/?...%5En+converges

2021-08-20, 22:47   #22
charybdis

Apr 2020

75110 Posts

Quote:
 Originally Posted by a1call sum_(n=1)^∞ 1.1^n diverges to ∞
Well yes, obviously it does. But if you read my post properly you'll see that I was talking about the sum of the reciprocals of that sequence.

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