 mersenneforum.org A question about primes of a particular form
 Register FAQ Search Today's Posts Mark Forums Read  2018-11-15, 08:15 #1 enzocreti   Mar 2018 24·31 Posts A question about primes of a particular form The numbers pg(k) are so defined: pg(k)=(2^k-1)*10^m+2^(k-1)-1. where m is the number of decimal digits of 2^(k-1)-1. So for example pg(2)=(3*10)+2-1=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^(k-1) - 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.   2018-11-15, 10:37 #2 enzocreti   Mar 2018 24·31 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 2018-11-15 at 10:58   2018-11-16, 08:04 #3 enzocreti   Mar 2018 1F016 Posts other k's found I found other k's leading to a prime (yet multiples of 3) 29343, 29988, 30405   2018-11-16, 08:52   #4
Batalov

"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

22×37×61 Posts Quote:
 Originally Posted by enzocreti can a k of the form 3s+1 exist such that (pg(k)+11)/42 is prime? where s is some positive integer.
You are simply doing it wrong.   2018-11-16, 09:24   #5
axn

Jun 2003

459310 Posts Quote:
 Originally Posted by Batalov How about k=100? You are simply doing it wrong.
pg(100)+11 is not divisible by 42. So...    2018-11-16, 10:51   #6
enzocreti

Mar 2018

1111100002 Posts k=100

Quote:
 Originally Posted by Batalov How about k=100? You are simply doing it wrong.
for k=100, (pg(100)+11)/42 is NOT prime   2018-11-16, 15:32   #7
Batalov

"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

902810 Posts Quote:
 Originally Posted by enzocreti With Pfgw i found that
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 3-PRP!

cat > help
71419
1108537
782395429
131693398271

pfgw -N -k -hhelp -t -q"((2^100-1)*10^30+2^99-1+11)/42"
Primality testing ((2^100-1)*10^30+2^99-1+11)/42 [N-1, Brillhart-Lehmer-Selfridge]
Reading factors from helper file help
Running N-1 test using base 2
((2^100-1)*10^30+2^99-1+11)/42 is prime! (0.0144s+0.0005s)   2018-11-16, 20:54 #8 Batalov   "Serge" Mar 2008 Phi(3,3^1118781+1)/3 22·37·61 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 = 2k-1 - 1, d = ceil(log10m), 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 :: 3|k 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.   2018-11-17, 01:00 #9 Dr Sardonicus   Feb 2017 Nowhere 325110 Posts A slightly different approach: As Batalov showed, 6|(pg(k) + 11) always. For 7, we have: If 3|k then Mod(2^k - 1, 7) = 0 and Mod(2^(k-1) - 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^(k-1) - 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 3|k 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   2018-11-17, 13:21 #10 Dr Sardonicus   Feb 2017 Nowhere 3,251 Posts I fouled up. I want m to be the number of decimal digits of 2^(i-1) - 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 2018-11-17 at 13:29   2018-11-19, 09:25 #11 enzocreti   Mar 2018 24×31 Posts A question about pg primes whose exponent is congruent to 7 mod 10 pg(k) numbers are so defined: pg(k)=(2^k-1)*10^d+2^(k-1)-1, where d is the number of decimal digits of 2^(k-1)-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?   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post sweety439 sweety439 162 2020-05-15 18:33 JeppeSN And now for something completely different 27 2018-04-12 14:20 carpetpool carpetpool 3 2017-01-26 01:29 YuL Math 21 2012-10-23 11:06 Dougy Math 8 2009-09-03 02:44

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