...when 2*3^n + 1 is prime
 

[CODE]2*3^n + 1 is prime for n = 0, 1, 2, 4, 5, 6, 9, 16, 17, 30, 54, 57, 60, 65, 132, 180, 320, 696, 782, 822, 897, 1252, 1454, 4217, 5480, 6225, 7842, 12096, 13782, 17720, 43956, 64822, 82780, 105106, 152529, 165896, 191814, 529680, 1074726, 1086112, 1175232 ...[/CODE]

[url]https://primes.utm.edu/bios/code.php?code=p199[/url]

Is there a collaborative search for primes of this format in progress?
Is it known, what is the largest exponent tested so far?
Thank you for your time.:smile:

I don't know the answers myself, I do have ideas. 2*3^n+1 is only prime iff 3^n is not of form: (2*j+1)*k+j for some natural numbers j,k .

Thank you for route reply SM.
At least now I know I'm not the last man on Earth. It is still an open question of there is anyone else.
I am interested in the format because it is probably the simplest format after Mersenne & Fermat Primes. I am surprised there is not that much of an effort spent at finding primes of this format as a matter of principle.

A bit OT and 9 levels more complicated to find, but interesting in its pressing the primality software to their limits from [B]Batalov[/B].

[url]https://primes.utm.edu/primes/page.php?id=118946[/url]


I think once you have the N-1 proof, others are unnecessary.
It is the simplest to Program.

[QUOTE=a1call;493755]I am interested in the format because it is probably the simplest format after Mersenne & Fermat Primes. I am surprised there is not that much of an effort spent at finding primes of this format as a matter of principle.[/QUOTE]

3*2^n-1 is more natural than 2*3^n-1, and the 321 project (well, now subproject of primegrid) searches those fairly extensively.

b^n-1 is also a natural extension of Mersennes.

[QUOTE=VBCurtis;493767]3*2^n-1 is more natural than 2*3^n-1, and the 321 project (well, now subproject of primegrid) searches those fairly extensively.[/QUOTE]
It occurred to me to wonder, which primes could be divisors of 3*2^n - 1 for positive integers n. Clearly, they are the primes for which the multiplicative order of 3 (mod p) divides the multiplicative order of 2 (mod p). Running a mindless check out to the limit ten million indicates that about .576 of primes qualify.

My first look at the proportion of primes which may be divisors was out to the limit 1000, and was remarkably close to the constant gamma, .577+. This prompted me to check further, and the proportion got closer to .576. I have no idea what the limiting value of the proportion may be, or even if there is a limit.

Up to 200, the possible prime divisors are

[5, 11, 13, 19, 23, 29, 37, 47, 53, 59, 61, 67, 71, 83, 97, 101, 107, 131, 139, 149, 163, 167, 173, 179, 181, 191, 193, 197].

[QUOTE=Dr Sardonicus;493954]It occurred to me to wonder, which primes could be divisors of 3*2^n - 1 for positive integers n.[/QUOTE]

See [url=https://oeis.org/A001915]A001915[/url] (which includes 2 but otherwise matches). Note that Pólya's proved infinitude.

[QUOTE=VBCurtis;493767]3*2^n-1 is more natural than 2*3^n-1, and the 321 project (well, now subproject of primegrid) searches those fairly extensively.

b^n-1 is also a natural extension of Mersennes.[/QUOTE]
I would argue that 2*3^n+1 is more basic.
If I'm not mistaking (for n>=1), it will have at least 1 prime factor of the form 2k*3^n+1 where k may or may not be even. All its factors are of the form k'3^n'+1.
And it can be equivalently to Fermat Little T. be primality checked to the powers less than p-1.
I will let you figure out the last part on your own.

[QUOTE=a1call;493975]I would argue that 2*3^n+1 is more basic.
[/QUOTE]

I quite like 3^m-2. If we consider prime exponents only: n=3^p-2. Then n-1 is divisible by p. So 1==3^(n-1)==2^((n-1)/p) mod n. And p==3 mod 4 implies n is divisible by 5. For example prime n=3^41-2=36472996377170786401. Then (n-1)/p=889585277491970400. And 2^((n-1)/p)==1 mod n

[QUOTE=paulunderwood;493979]I quite like 3^m-2. If we consider prime exponents only: n=3^p-2 ... [/QUOTE]
Why prime exponents only? This form has no inclination towards m needing to be prime.
[CODE]2: 2
4: 2 2
5: 5
6: 2 3
9: 3 3
22: 2 11
37: 37
41: 41
90: 2 3 3 5
102: 2 3 17
105: 3 5 7
317: 317
520: 2 2 2 5 13
541: 541
561: 3 11 17
648: 2 2 2 3 3 3 3
780: 2 2 3 5 13
786: 2 3 131
957: 3 11 29
1353: 3 11 41
2224: 2 2 2 2 139
2521: 2521
6184: 2 2 2 773
7989: 3 2663
8890: 2 5 7 127
19217: 11 1747
20746: 2 11 23 41
31722: 2 3 17 311
37056: 2 2 2 2 2 2 3 193
69581: 17 4093
195430: 2 5 19543
225922: 2 37 43 71
506233: 7 13 5563
761457: 3 253819
[/CODE]
[url]http://oeis.org/A014224[/url]
They just cannot be 3 (mod 4), obviously, but that is a simple result of sieving.

[QUOTE=a1call;493975]I would argue that 2*3^n+1 is more basic.
If I'm not mistaking (for n>=1), it will have at least 1 prime factor of the form 2k*3^n+1 where k may or may not be even.[/QUOTE]
I'm not getting any younger.:smile:
Let's try that again.
It has at least one prime factor of the form
2k*3^n'+1 where k may or may not be even.:smile::pcl::paul: