100$ prize for the solution of a conjecture

[enzocreti]

[QUOTE=Dr Sardonicus;490416]If I have gotten the data right and have done my sums correctly, the values of n known so far yielding prps are

[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019

I mean sum the reciprocals of this vector

[Dr Sardonicus]

Originally Posted by enzocreti

curiously if you sum the reciprocals of the exponents leading to a prime (1/2+1/3+1/4+...+1/285019) you get a value 1.64...which is very close to pi^2/6

I note that there are between 4 and 8 values of k with d decimal digits yielding PRP's for d = 1 to 5, and there are 4 known with 6 decimal digits.

The sum of the 1/k so far is less than pi^2/6 by about .001. Assuming the number of k's with a given number of digits yielding PRP's stays less than (say) 20, the sum of the reciprocals isn't going to get much closer.

[enzocreti]

Originally Posted by Dr Sardonicus

I note that there are between 4 and 8 values of k with d decimal digits yielding PRP's for d = 1 to 5, and there are 4 known with 6 decimal digits.

The sum of the 1/k so far is less than pi^2/6 by about .001. Assuming the number of k's with a given number of digits yielding PRP's stays less than (say) 20, the sum of the reciprocals isn't going to get much closer.

Yes just a curiosity...i dont know how much closer it could get..anyway pi^2/6 seems to be a bound

[enzocreti]

Originally Posted by ATH

Counter example: k=19179 k%7=6:

10^5774*(2^19179-1)+2^(19179-1)-1 is 2-PRP! (1.6956s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 3-PRP! (1.6811s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 5-PRP! (1.6777s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 7-PRP! (1.6963s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 11-PRP! (1.6753s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 13-PRP! (1.6836s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 17-PRP! (1.6772s+0.0005s)

EDIT: Nevermind it is n%7 that has to be 6, not k%7.

Now
Try to concatenate 2^k-1 and 2^(k-2)-1. Examples 71, 153,...can you find a prime of this form congruent to 3 or 4 mod 7?

[enzocreti]

https://solutionsti360.ca/MATH/pfgw/stats.html
if you want you could join in!

[pinhodecarlos]

Originally Posted by enzocreti

https://solutionsti360.ca/MATH/pfgw/stats.html
if you want you could join in!

The server doesn't detect unique clients therefore when you fire more than one it will process the same work in all instances, on the same machine.
Also there is a lot of idle time between server connection, would be wise to download more than one task.

[pinhodecarlos]

Also I think there's a new version of PFGW based on gwnum v28.9. Maybe on this version you can set the number of threads to use...

[enzocreti]

Originally Posted by Dr Sardonicus

I note that there are between 4 and 8 values of k with d decimal digits yielding PRP's for d = 1 to 5, and there are 4 known with 6 decimal digits.

The sum of the 1/k so far is less than pi^2/6 by about .001. Assuming the number of k's with a given number of digits yielding PRP's stays less than (say) 20, the sum of the reciprocals isn't going to get much closer.

I found another probable prime with residue 5. This is incredible. Residue 5 occurs with frequency 1/9 as the residue 6. But we found eight probable primes with residue 5 and none with residue 6. Would you admit that this is very very very stange? I wonder if these numbers have some hidden structure.

[enzocreti]

The exponents leading to a prime are:
[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]
These exponents seem to be NOT random at all...
for example 19179=2131*9
And if an exponent is divisible by 42 then it is also divisible by 13
If an exponent is divisible by 43 then it is also divisible by 5

[enzocreti]

look infact at the exponents divisible by 43 :215, 69660 and 92020. They are are multiples of 5.

The exponents multiples of 42 are 75894 and 56238. Both are multiples of 13.

[enzocreti]

Despite 3k views, nobody found a counter-example to the conjecture. So I replace the offer of 100$ for finding a prime factor of 2^2^2^2+3^3^3^3.