# Even Perfect numbers $n$ with $n+1$ prime

The set $S$ of even perfect numbers $n$ such that $n+1$ is a prime number contains $$6,28,33550336,137438691328$$

Latter number found by Joerg Arndt, corresponds to $M_{19}$ (mersenne)

Question: Is $S$ reduced to these $4$ numbers.

New: Joerg Arndt checked up to exponent $110503$ that the corresponding number $n+1$ is composite. (Improved $19$ to $110503$).

Which function of $x$ migh describe well the size of the set of elements in $S$ less than $x$

divided

by the size of the set of all even perfect numbers less than $x$; mainly with big $x.$

So, I am asking for relative size not absolute size. E.g., if I were asking for relative density of the prime numbers congruent to $3$ modulo $4$: I do not want to use the big machinery of the prime number theorem, or Dirichlet's Theorem to deduce how many should be there. I just want (in these case) to know how to describe in terms of $x$

number of primes congruent to $3$ modulo $4$ and less than $x$

divided by

number of primes less than $x$

How many such numbers $n$ we may expect inside the known 47 perfect numbers ?

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One reason not to consider the- 1 version is that almost all such numbers are multiples of 3. It might be prudent to check a few more congruences and see what that says about the Mersenne exponent n before going much farther. Gerhard "Ask Me About System Design" Paseman, 2011.04.23 – Gerhard Paseman Apr 24 '11 at 3:43
If the Mersenne exponent is 7 mod 10; your number is composite. This takes out about a third of the known perfect numbers. 2 mod 3 takes out about another third. There are other congruences on the exponent that show some of your numbers aren't prime; you can use them to get an upper bound or even a density estimate. Gerhard "Ask Me About System Design" Paseman, 2011.04.25 – Gerhard Paseman Apr 25 '11 at 7:20
Simulation suggests the next candidate corresponds to Mersenne exponent 61, with at most five more candidates among the known perfect numbers after that. Gerhard "Ask Me About System Design" Paseman, 2011.04.25 – Gerhard Paseman Apr 25 '11 at 8:04
Based on Tapio Rajala's statements to a parallel (not quite duplicate) question, I now guess that there are four prime candidates among the 47. Gerhard "12 Guesses For 10 Cents" Paseman, 2011.05.07 – Gerhard Paseman May 7 '11 at 17:32

There's a conjecture (for which I can't find a source now) that the number of Mersenne primes $2^n-1$ with $n < x$ is $c \log x$ for some constant $c$. Differentiating this, the "probability" that $2^n-1$ is prime is about $c/n$. (This is unconditional; that is, I'm not assuming $n$ is prime.)

The even perfect numbers are exactly of the form $2^{n-1}(2^n-1)$ with $2^n-1$ prime.

So the "probability" that $2^n-1$ and $2^{n-1} (2^n-1) + 1$ is prime, assuming independence, is $c/n$ times the probability that $2^{n-1} (2^n-1) + 1$ is prime. $2^{n-1} (2^n-1) + 1$ is roughly $2^{2n}$, so by the prime number theorem its "probability" of being prime is about $1/log(2^{2n})$, or again a constant divided by $n$. That is, the "probability" that $2^n-1$ is prime and the corresponding number is one less than a prime is $c/n^2$; since $\sum_{n \ge 1} cn^{-2}$ is finite this leads us to suspect that there are finitely many solutions.

Of course none of this is anywhere near being a proof...

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Based on these heuristics. How many such numbers $n$ we may expect inside the known 47 perfect numbers ? – Luis H Gallardo Apr 25 '11 at 2:23

For dealing with large potential primes a good choice is openpfgw

Using openpfgw I finished the list to $1 3466 917$ in about 20 minutes without finding new primes.

[added] The only prime perfect + 1 candidate from the known Mersenne primes is for $M_{20996011}$ - I am running ECM factoring on it.

[later] François Brunault found that $M_{20996011}$ is divisible by $1552147$ which settles the question for the known perfect numbers.

Here is the log:

./pfgw64 -f10 -lmer1log.txt /tmp/mer.txt

2^0*(2^1-1)+1 is trivially prime!: 2
2^1*(2^2-1)+1 is trivially prime!: 7
2^2*(2^3-1)+1 is trivially prime!: 29
2^4*(2^5-1)+1 trivially factors as: 7*71
2^6*(2^7-1)+1 trivially factors as: 11*739
2^12*(2^13-1)+1 is trivially prime!: 33550337
2^16*(2^17-1)+1 has factors: 7
2^18*(2^19-1)+1 is 3-PRP! (0.0000s+0.0009s)
2^30*(2^31-1)+1 has factors: 29
2^60*(2^61-1)+1 is composite: RES64: [36E090A8C361AD6C] (0.0000s+0.0003s)
2^88*(2^89-1)+1 has factors: 7
2^106*(2^107-1)+1 has factors: 7
2^126*(2^127-1)+1 has factors: 11
2^520*(2^521-1)+1 has factors: 7
2^606*(2^607-1)+1 has factors: 11
2^1278*(2^1279-1)+1 is composite: RES64: [570A6B3FD91E6339] (0.8700s+0.0011s)
2^2202*(2^2203-1)+1 is composite: RES64: [ECB4FE924C674723] (4.6906s+0.0010s)
2^2280*(2^2281-1)+1 has factors: 197
2^3216*(2^3217-1)+1 has factors: 11
2^4252*(2^4253-1)+1 has factors: 7
2^4422*(2^4423-1)+1 is composite: RES64: [F3603EEF4BD4F197] (17.0237s+0.0031s)
2^9688*(2^9689-1)+1 has factors: 7
2^9940*(2^9941-1)+1 has factors: 7
2^11212*(2^11213-1)+1 has factors: 7
2^19936*(2^19937-1)+1 has factors: 7
2^21700*(2^21701-1)+1 has factors: 7
2^23208*(2^23209-1)+1 has factors: 35603
2^44496*(2^44497-1)+1 has factors: 11
2^86242*(2^86243-1)+1 has factors: 7
2^110502*(2^110503-1)+1 has factors: 491
2^132048*(2^132049-1)+1 is composite: RES64: [1B3B60AEC3578817] (744.2790s+111.7145s)
2^216090*(2^216091-1)+1 has factors: 4673
2^756838*(2^756839-1)+1 has factors: 7
2^859432*(2^859433-1)+1 has factors: 7
2^1257786*(2^1257787-1)+1 has factors: 11
2^1398268*(2^1398269-1)+1 has factors: 7
2^2976220*(2^2976221-1)+1 has factors: 7
2^3021376*(2^3021377-1)+1 has factors: 7
2^6972592*(2^6972593-1)+1 has factors: 7
2^13466916*(2^13466917-1)+1 has factors: 11

The rest Mersenne primes lead to small factors:

Format ($p$,factor) (24036583,149),(25964951,7),( 30402457,11),( 32582657,7),( 37156667,7),( 42643801,3593),( 43112609,7)

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The only remaining candidate seems to be divisible by p=1552147 (computed used fast exponentiation mod p). – François Brunault Nov 9 '11 at 16:08
@François Indeed, will update the answer. – joro Nov 9 '11 at 16:17
It's good to confirm Tapio Rajala's findings. Gerhard "Ask Me About System Design" Paseman, 2011.11.09 – Gerhard Paseman Nov 9 '11 at 16:37

The numbers involved are pretty huge - have you tried all the Mersenne primes' perfect numbers yet?

The other answer might be referring to Wagstaff's conjecture about the number of these primes being less than $e^{\gamma}/\log(2) *\log(\log(x))$; see e.g. here, here, or here for some references (some better than others).

I would imagine that this would be helpful in solving this, but gives a sense of just how hard it would be to prove anything.

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Based on these heuristics. How many such numbers $n$ we may expect inside the known 47 perfect numbers ? – Luis H Gallardo Apr 25 '11 at 2:24

I used an awk program to generate congruences on n such that, if the Mersenne exponent n satisfied such a congruence, then the corresponding candidate had a small prime factor, which usually was smaller than the candidate prime. Using moduli up to 4800, I found that the candidate corresponding to 216091 exponent was a multiple of 4673, most of the remaining candidates were divisible by smaller primes. At this writing, 61, 1279, 23209, and 20996011, are the exponents whose corresponding candidates may be prime, if I didn't foul up the coding. So my guess is: at most eight.

Well, the coding might have been good, but the filtering wasn't: some composite moduli slipped in. The latest version (morally just trial factorization) gives pairs $(m,f)$ with $f$ a factor of candidate using exponent $m$: (20996011,1552147), (110503,491), (216091,4673), (23209,35603), (42643801,3593),(4423,2163571) (2203,60449). Most of the rest of the candidates have small factors or are already seen to be prime. Still to crack are 61, 1279, and 132049. Gerhard "Ask Me About System Design" Paseman, 2011.05.03 – Gerhard Paseman May 3 '11 at 18:42