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2022-08-17, 12:01   #56
Dr Sardonicus

Feb 2017
Nowhere

61·97 Posts

Quote:
 Originally Posted by Dobri On the basis of quite limited observations for k up to 11 (see posts #29 and #30), the values of k for another factor are obtained as follows: - the sum of a prime number (or 1) and 2n such as 5 + 22 = 32 and 1 + 23 = 32; - the sum of 1 and a prime number giving 2n such as 1 + 3 = 22 and 1 + 7 = 23; - the sum of two prime numbers giving 2n such as 3 + 5 = 23; or - the sum of 2n and a prime number giving a prime number such as 22 + 3 = 7 and 23 + 3 = 11.
Quote:
 Originally Posted by Dobri For k = 13, only cases with Δk = 1, 4, 9, and 12 for a given Mersenne number are found, where Δk = k2 - k1 with k2 = 13 and k2 > k1. Here 1 + 12 = 13 and 4 + 9 = 13.
In the sum 4 + 9 = 13 neither 4 nor 9 is prime, and neither 4 nor 9 is equal to 1, so the sum does not fit any of the categories you list.

2022-08-17, 14:22   #57
Dobri

"Καλός"
May 2018

25·11 Posts

Quote:
 Originally Posted by Dr Sardonicus In the sum 4 + 9 = 13 neither 4 nor 9 is prime, and neither 4 nor 9 is equal to 1, so the sum does not fit any of the categories you list.
Affirmative, this is a new category, although 4 = 22 and 9 was already included as 5 + 22 = 9 in the expanding scheme of values of k.
A distant loose analogy with a growing Pascal's triangle comes in mind...
It is to be seen to what extent 2n, n = 2, 3,..., would appear in the additive scheme with the increase of the values of k.

 2022-08-18, 00:37 #58 Dobri   "Καλός" May 2018 1011000002 Posts k = 14: None; k = 15: 3 + 12 = 7 + 23 = 11 + 22 = 15.
 2022-08-18, 00:57 #59 Dobri   "Καλός" May 2018 25×11 Posts k = 16: 1 + 15 = 3 + 13 = 4 + 12 = 7 + 9 = 16.
 2022-08-18, 01:06 #60 Dobri   "Καλός" May 2018 25·11 Posts k = 17: 5 + 12 = 23 + 9 = 17; k = 18: None.
2022-08-18, 10:31   #61
User140242

Jul 2022

29 Posts

Quote:
 Originally Posted by Dobri No cases have been forgotten. In the quoted example, 237-1 = 137438953471, then 13743895347 / 223 = 616318177, and k2 = (616318177-1)/(2×37) = 8328624, thus Δk = 8328624 - 3 = 8328621 = 3×7×396601, which is unrelated to k = 13.

I didn't quite understand what you were looking for but the only possible values ​​of k%12 are these:

if p%12 == 1 then k%12 = (0, 3, 8, 11)
if p%12 == 5 then k%12 = (0, 3, 4, 7)
if p%12 == 7 then k%12 = (0, 5, 8, 9)
if p%12 == 11 then k%12 = (0, 1, 4, 9).

for example if k=13 then k%12=1 then the possible cases of have when p%12=11
and possible k%12 are (0, 1, 4, 9) with k%12=0 corresponding to 12
so you find that only cases 0+1=0+12+1=13 and 4+9=13

if k=14 then k%12=2 there are no cases

if k=15 then k%12=3 the possible cases are
p%12=1 then k%12=(0, 3, 8, 11)
and p%12=5 then k%12=(0, 3, 4, 7)
from which you find that only cases 3+0 or 3+0+12=15 and 4+11=15 and 7+8=15

if k=16 then k%12=4 the possible cases are
p%12=11 then k%12=(0, 1, 4, 9)
and p%12=5 then k%12=(0, 3, 4, 7)
from which you find that only cases 3+1=3+12+1=15+1=16 and 7+9=16 and 4+0=4+0+12=16 and 3+1=3+1+12=3+13=16

Last fiddled with by User140242 on 2022-08-18 at 10:39

 2022-08-18, 12:00 #62 Dobri   "Καλός" May 2018 25·11 Posts ... k = 19: 3 + 24 = 22 + 15 = 7 + 12 = 19; k = 20: 3 + 17 = 5 + 15 = 23 + 12 = 9 + 11 = 20; k = 21: 1 + 20 = 22 + 17 = 5 + 24 = 23 + 13 = 9 + 12 = 21; k = 22: None; k = 23: 3 + 20 = 23 + 15 = 11 + 12 = 23; k = 24: 1 + 23 = 3 + 21 = 22 + 20 = 5 + 19 = 7 + 17 = 23 + 24 = 9 + 15 = 2×12 = 24; k = 25: 1 + 24 = 22 + 21 = 9 + 24 = 12 + 13 = 25; k = 26: None; k = 27: 3 + 24 = 22 + 23 = 7 + 20 = 23 +19 = 11 + 24 = 12 + 15 = 27; k = 28: 1 + 27 = 3 + 25 = 22 + 24 = 7 + 21 = 9 + 19 = 12 + 24 = 13 + 15 = 28; k = 29: 5 + 24 = 23 + 21 = 9 + 20 = 12 + 17 = 29; k = 30: None; k = 31: 3 + 28 = 22 + 27 = 7 + 24 = 12 + 19 = 15 + 24 = 31; k = 32: 3 + 29 = 5 + 27 = 23 + 24 = 9 + 23 = 11 + 21 = 12 + 20 = 15 + 17 = 32; ... It is known that there are no factors 2pk + 1 of Mersenne numbers for k = 2(2m + 1), m = 0, 1, 2,... The exhaustive computations indicate that if there is a factor 2pk2 + 1 of a Mersenne number 2p - 1 for a given pair of k2 and p, then there is no factor 2pk1 + 1 for said p for the positive values of k1 = k2 - 2(2m + 1), m = 0, 1, 2,....
2022-08-18, 12:40   #63
Dr Sardonicus

Feb 2017
Nowhere

591710 Posts

Quote:
 Originally Posted by Dobri The exhaustive computations indicate that if there is a factor 2pk2 + 1 of a Mersenne number 2p - 1 for a given pair of k2 and p, then there is no factor 2pk1 + 1 for said p for the positive values of k1 = k2 - 2(2m + 1), m = 0, 1, 2,....
What a waste of computing power.

2022-08-19, 03:01   #64
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

272B16 Posts

Quote:
 Originally Posted by Dr Sardonicus What a waste of computing power.
Agree. But let them discharge the energy. I went the same way for a while, 20 years ago (probably few other now-famous forumites too, hehe).

2022-08-19, 09:57   #65
Dobri

"Καλός"
May 2018

25×11 Posts

Quote:
 Originally Posted by Dr Sardonicus What a waste of computing power.
On the contrary, a waste of computing power was avoided.
This made it possible to obtain the results in post #62 much faster than the ones posted earlier.
After stumbling upon this result during the evaluation of pairs of factors, I was able to modify my custom source code and deal with Δk instead of 2kp+1 == 1 or 7 (mod 8) using a single subtraction instead of addition and two multiplications which reduced the computation time.
For this reason, perhaps the statement in post #62 deserves the status of a lemma despite its simple proof.

2022-08-19, 10:36   #66
User140242

Jul 2022

29 Posts

Quote:
 Originally Posted by Dobri On the contrary, a waste of computing power was avoided. This made it possible to obtain the results in post #62 much faster than the ones posted earlier. ...

In my opinion the result highlighted post #62 does not allow to exclude factors that are not possible in fact for example if k=16 and p=5 mod 12 it is also necessary to exclude the factors k2%12=1=16-15 and k2%12=9=16-7

with this code if you take (RW[i] -1)/2/P you can easily find the possible values ​​of k%PrimesBaseProd with PrimesBaseProd=12 if you set n_PB=2

Code:
#include <iostream>
#include <vector>
#include <cstdlib>
#include <stdint.h>

int main()
{

int64_t P=86000063;
int64_t PrimesBase[5]={2,3,5,7,11};
int n_PB = 5;

int64_t  bW;
std::vector<int64_t> RW;
if (n_PB>1)
{
int64_t  PrimesBaseProd=2;
for (int i=0; i<n_PB;i++)
PrimesBaseProd*=PrimesBase[i];
bW=2*P*PrimesBaseProd;
int64_t r_t1,r_t7;
int j;
r_t1=1;
r_t7=1+2*P;
if (P%4==1)
r_t7+=4*P;
for (int64_t k=1; k<=PrimesBaseProd/4;k++)
{
for (j=1;j<n_PB;j++)
if (r_t1%PrimesBase[j]==0)
break;
if (j==n_PB)
RW.push_back(r_t1);
for (j=1;j<n_PB;j++)
if (r_t7%PrimesBase[j]==0)
break;
if (j==n_PB)
RW.push_back(r_t7);
r_t1+=8*P;
r_t7+=8*P;
}
}
else
{
bW=8*P;
RW.push_back(1);
RW.push_back(1+2*P);
if (P%4==1)
RW[1]+=4*P;
}
int64_t  nR=RW.size();

return 0;

}

Last fiddled with by User140242 on 2022-08-19 at 11:00

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