20220817, 12:01  #56  
Feb 2017
Nowhere
2^{2}·3·499 Posts 
Quote:


20220817, 14:22  #57  
"Καλός"
May 2018
3×11^{2} Posts 
Quote:
A distant loose analogy with a growing Pascal's triangle comes in mind... It is to be seen to what extent 2^{n}, n = 2, 3,..., would appear in the additive scheme with the increase of the values of k. 

20220818, 00:37  #58 
"Καλός"
May 2018
553_{8} Posts 
k = 14: None;
k = 15: 3 + 12 = 7 + 2^{3} = 11 + 2^{2} = 15. 
20220818, 00:57  #59 
"Καλός"
May 2018
3·11^{2} Posts 
k = 16: 1 + 15 = 3 + 13 = 4 + 12 = 7 + 9 = 16.

20220818, 01:06  #60 
"Καλός"
May 2018
3×11^{2} Posts 
k = 17: 5 + 12 = 2^{3} + 9 = 17;
k = 18: None. 
20220818, 10:31  #61  
Jul 2022
64_{8} Posts 
Quote:
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 20220818 at 10:39 

20220818, 12:00  #62 
"Καλός"
May 2018
3·11^{2} Posts 
...
k = 19: 3 + 2^{4} = 2^{2} + 15 = 7 + 12 = 19; k = 20: 3 + 17 = 5 + 15 = 2^{3} + 12 = 9 + 11 = 20; k = 21: 1 + 20 = 2^{2} + 17 = 5 + 2^{4} = 2^{3} + 13 = 9 + 12 = 21; k = 22: None; k = 23: 3 + 20 = 2^{3 }+ 15 = 11 + 12 = 23; k = 24: 1 + 23 = 3 + 21 = 2^{2} + 20 = 5 + 19 = 7 + 17 = 2^{3} + 2^{4} = 9 + 15 = 2×12 = 24; k = 25: 1 + 24 = 2^{2} + 21 = 9 + 2^{4} = 12 + 13 = 25; k = 26: None; k = 27: 3 + 24 = 2^{2} + 23 = 7 + 20 = 2^{3} +19 = 11 + 2^{4} = 12 + 15 = 27; k = 28: 1 + 27 = 3 + 25 = 2^{2 }+ 24 = 7 + 21 = 9 + 19 = 12 + 2^{4} = 13 + 15 = 28; k = 29: 5 + 24 = 2^{3} + 21 = 9 + 20 = 12 + 17 = 29; k = 30: None; k = 31: 3 + 28 = 2^{2} + 27 = 7 + 24 = 12 + 19 = 15 + 2^{4} = 31; k = 32: 3 + 29 = 5 + 27 = 2^{3} + 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 2pk_{2} + 1 of a Mersenne number 2^{p}  1 for a given pair of k_{2} and p, then there is no factor 2pk_{1} + 1 for said p for the positive values of k_{1} = k_{2}  2(2m + 1), m = 0, 1, 2,.... 
20220818, 12:40  #63 
Feb 2017
Nowhere
2^{2}·3·499 Posts 
What a waste of computing power.

20220819, 03:01  #64 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10011110000110_{2} Posts 

20220819, 09:57  #65 
"Καλός"
May 2018
553_{8} Posts 
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. 
20220819, 10:36  #66  
Jul 2022
2^{2}·13 Posts 
Quote:
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=1615 and k2%12=9=167 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 20220819 at 11:00 

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