20181115, 08:15  #1 
Mar 2018
2^{4}·31 Posts 
A question about primes of a particular form
The numbers pg(k) are so defined:
pg(k)=(2^k1)*10^m+2^(k1)1. where m is the number of decimal digits of 2^(k1)1. So for example pg(2)=(3*10)+21=31 pg(3)=7*10+3=73. pg(4)=157 In other words pg(k) is obteined by concatenating the decimal digits of 2^k  1 and 2^(k1)  1. With Pfgw i found that up to k=21.000, the k's such that (pg(k)+11)/42 is a prime are these: 3 6 12 36 105 156 336 2286 4272 4427 11979 20076 so for example (pg(3)+11)/42 is a prime. I wonder why there is only one case out of 12, thqat is pg(4427) in which k=4427 is NOT a multiple of 3. In all other cases k=3,6,12,36,105,156,336,2286,4272,11979,20076 k is a multiple of 3. 
20181115, 10:37  #2 
Mar 2018
2^{4}·31 Posts 
...continues...
can a k of the form 3s+1 exist such that (pg(k)+11)/42 is prime? where s is some positive integer.
Last fiddled with by enzocreti on 20181115 at 10:58 
20181116, 08:04  #3 
Mar 2018
1F0_{16} Posts 
other k's found
I found other k's leading to a prime (yet multiples of 3)
29343, 29988, 30405 
20181116, 08:52  #4 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
2^{2}×37×61 Posts 

20181116, 09:24  #5 
Jun 2003
4593_{10} Posts 

20181116, 10:51  #6 
Mar 2018
111110000_{2} Posts 
k=100

20181116, 15:32  #7 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
9028_{10} Posts 
When a person claims that that they found "with Pfgw", they imply integer division.
pfgw divides. Examples: pfgw N k q"8/3" 8/3 is trivially prime!: 2 pfgw N k q"(1267650600228229401496703205375633825300114114700748351602687+11)/42" (1267650600228229401496703205375633825300114114700748351602687+11)/42 is 3PRP! cat > help 71419 1108537 782395429 131693398271 pfgw N k hhelp t q"((2^1001)*10^30+2^991+11)/42" Primality testing ((2^1001)*10^30+2^991+11)/42 [N1, BrillhartLehmerSelfridge] Reading factors from helper file help Running N1 test using base 2 ((2^1001)*10^30+2^991+11)/42 is prime! (0.0144s+0.0005s) 
20181116, 20:54  #8 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
2^{2}·37·61 Posts 
If one uses "division only when it divides", then the answer is very trivial.
It has nothing to do with primality, only with divisibility of pg(k)+11 by 42. Define: m = 2^{k1}  1, d = ceil(log_{10}m), and then pg(k) = (2*m+1)*10^d+m. Now, consider k and d separately. When does 2  pg(k)+11? Answer: for all k and d. When does 3  pg(k)+11? Answer: for all k and d. When does 7  pg(k)+11? Answer: if(d%6 == 1) { for all k } else {only for k = multiple of 3}. Do the homework. See that the above three statements are true. Now, d%6 == 1 is true only for ~1/6 of all k and 1/3 of these k's are a multiple of 3. So, the k :: 3k have a ~ 6:1 times better odds to produce a prime than the other k (considering the primality is an essentially random function wrt k, d) For k = 4427, d = 1333 which is 1 (mod 6), indeed. 
20181117, 01:00  #9 
Feb 2017
Nowhere
3251_{10} Posts 
A slightly different approach: As Batalov showed, 6(pg(k) + 11) always. For 7, we have:
If 3k then Mod(2^k  1, 7) = 0 and Mod(2^(k1)  1, 7) = 3, so Mod(pg(k),7) = 3 for any m. If 3 does not divide k, the fact that 10 is a primitive root (mod 7) comes into play. In each case, there will only be one residue class of Mod(m, 6) that makes Mod(pg(k),7) = 3. If k == 1 (mod 3) then Mod(2^(k1)  1, 7) = 0 and Mod(2^k  1, 7) = 1, so Mod(pg(k), 7) = Mod(10^m, 7). This is 3 when Mod(m, 6) = 1. If k == 2 (mod 3) then we have Mod(pg(k), 7) = Mod(3*10^m + 1, 7), which is again 3 when Mod(m,6) = 1. Of the 18 possible pairs (Mod(k,3), Mod(m, 6)) then, the 6 pairs with Mod(k,3) = 0 , the pair (Mod(k,3) = 1, Mod(m,6) = 1), and the pair (Mod(k,3) = 2, Mod(m,6) = 1) allow Mod(pw(k), 7) = 3. For the remaining ten pairs, Mod(pg(k), 7) is not 3. So, pg(k) + 4 is divisible by 7 about 6 times as often for 3k as for k == 1 (mod 3), and about 6 times as often as for k == 2 (mod 3). A numerical check up to k = 10000 confirms this: ? c0=0;c1=0;c2=0;a=log(2)/log(10);f=a;m=1;for(i=1,10000,e1=i%3;e2=m%6;if(e1==0,c0++);if(e1==1&&e2==1,c1++);if(e1==2&&e2==1,c2++);f+=a;if(f>1,f;m++));print(c0" "c1" "c2) 3333 556 557 
20181117, 13:21  #10 
Feb 2017
Nowhere
3,251 Posts 
I fouled up. I want m to be the number of decimal digits of 2^(i1)  1, not 2^i  1. I should have initialized i at 2, not 1. (Or left i initialized at 1, initialized f at 0, and had a value pg(1) = 10.)
? c0=0;c1=0;c2=0;a=log(2)/log(10);f=a;m=1;for(i=2,10000,e1=i%3;e2=m%6;if(e1==0,c0++);if(e1==1&&e2==1,c1++);if(e1==2&&e2==1,c2++);f+=a;if(f>1,f;m++));print(c0" "c1" "c2) 3333 556 556 While I was at it, I ran the numbers out farther, to 100,000 an 300,000: ? c0=0;c1=0;c2=0;a=log(2)/log(10);f=a;m=1;for(i=2,100000,e1=i%3;e2=m%6;if(e1==0,c0++);if(e1==1&&e2==1,c1++);if(e1==2&&e2==1,c2++);f+=a;if(f>1,f;m++));print(c0" "c1" "c2) 33333 5556 5556 ? c0=0;c1=0;c2=0;a=log(2)/log(10);f=a;m=1;for(i=2,300000,e1=i%3;e2=m%6;if(e1==0,c0++);if(e1==1&&e2==1,c1++);if(e1==2&&e2==1,c2++);f+=a;if(f>1,f;m++));print(c0" "c1" "c2) 100000 16666 16668 (EDIT: I'm sure e1 and e2 can be computed more expeditiously.) Last fiddled with by Dr Sardonicus on 20181117 at 13:29 
20181119, 09:25  #11 
Mar 2018
2^{4}×31 Posts 
A question about pg primes whose exponent is congruent to 7 mod 10
pg(k) numbers are so defined:
pg(k)=(2^k1)*10^d+2^(k1)1, where d is the number of decimal digits of 2^(k1)1. In other words they are numbers obtained by the concatenation of two consecutive Mersenne numbers. Examples are 12763 and 157. Now I consider when pg(k) is prime and k is congruent to 7 mod 10. The values I found are k=7,67,360787. It seems that if pg(k) is prime and k is congruent to 7 mod 10, then k is also congruent to 1 (mod 6). Is there any Mathematical reason? Could exist a pg(k) prime with k congruent to 7 mod 10 and k NOT congruent to 1 mod 6? And all the pg(k) which are primes with k congruent to 7 mod 10, are all of the form 900s+163? 
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