A question about primes of a particular form - Page 2

enzocreti · 2018-11-19, 12:01

Have you an explanation why up to k=541000 there are 9 pg primes congruent to 5 mod 7 and none congruent to 6 mod 7?

enzocreti · 2018-11-19, 19:55

pg(k)=(2^k-1)*10^d+2^(k-1)-1, where d is the number of decimal digits of 2^(k-1)-1.
pg(3) for example is 73.

pg(4)=157.

pg(7)=12763 (prime) divides pg(7717), pg(14259), pg(15906),...
so I wonder if there is a proof that 12763 divides an infinite number of pg(k)'s.
The second question is: is 12763 the largest pg prime that divides at least two pg(k)'s?

science_man_88 · 2018-11-19, 20:13

Quote:

Originally Posted by enzocreti

pg(k)=(2^k-1)*10^d+2^(k-1)-1, where d is the number of decimal digits of 2^(k-1)-1.

aka $pg(k)= (2^k-1)\cdot10^{\lfloor{\log_{10}(2^{k-1}-1)}\rfloor+1}+2^{k-1}-1$

all in 1 variable now.

enzocreti · 2018-11-20, 07:16

I found that 255127 divides pg(k) for these values of k.:284274 1129738 1189846 1214317 1301821 1362842 1445186 1795733 1853089 2203032 2
237654 2267753 3055770 3080516 3532082 3624320 3842054 4653541 4839828 5220495 5
436726 5444103 5828733 5956001 6144125 6432347 6821804 7135640 7173850 7458223 7
513523 7690720 7979828 8006289 8010227 8162195 8195920 8255472 8412247 8449267 8
590602 8936597 9571824 9625677 9853929

I yet haven't find a pg(k) divisible by the next pg prime 40952047.

Batalov · 2018-11-20, 07:17

"Except for the last page, the previous thread became a bit long."

enzocreti · 2018-11-20, 07:29

so for example 255127 divides (2^284274-1)*10^85575+2^284273-1=pg(284274)

enzocreti · 2018-11-20, 07:31

Must pg(k) have a particular form to be divisible by 255127?

enzocreti · 2018-11-20, 09:47

pg(k)=(2^k-1)*10^d+2^(k-1)-1, where d is the number of decimal digits of 2^(k-1)-1.
pg(3) for example is 73.
With Julia software I found that pg(k)'s which are divisible by 212605 are the following:
pg(3333)
pg(683733)
pg(887433)
pg(1091133)

3333,683733,887433,1091133 are all ending with digits 33 and are all congruent to 183 mod 210.
So I wonder if for pg(k) to be divisible by 212605 it is necessary that k is congruent to 183 mod 210 and k congruent to 33 (mod 10).

enzocreti · 2018-11-20, 10:54

found another example 212605 divides ec(1294833)

enzocreti · 2018-11-20, 11:07

another example found 212605 divides pg(1498533)

Dr Sardonicus · 2018-11-20, 14:47

p = 40952047 divides pg(k) for k=118327344; for this k, m=35620080.

2018-11-19, 12:01	#12
enzocreti Mar 2018 2·5·53 Posts	pg primes congruent to 5 mod 7 Have you an explanation why up to k=541000 there are 9 pg primes congruent to 5 mod 7 and none congruent to 6 mod 7?

2018-11-19, 19:55	#13
enzocreti Mar 2018 530₁₀ Posts	Pg primes dividing pg numbers pg(k)=(2^k-1)*10^d+2^(k-1)-1, where d is the number of decimal digits of 2^(k-1)-1. pg(3) for example is 73. pg(4)=157. pg(7)=12763 (prime) divides pg(7717), pg(14259), pg(15906),... so I wonder if there is a proof that 12763 divides an infinite number of pg(k)'s. The second question is: is 12763 the largest pg prime that divides at least two pg(k)'s?

2018-11-20, 07:16	#15
enzocreti Mar 2018 1000010010₂ Posts	I found that 255127 divides pg(k) for these values of k.:284274 1129738 1189846 1214317 1301821 1362842 1445186 1795733 1853089 2203032 2 237654 2267753 3055770 3080516 3532082 3624320 3842054 4653541 4839828 5220495 5 436726 5444103 5828733 5956001 6144125 6432347 6821804 7135640 7173850 7458223 7 513523 7690720 7979828 8006289 8010227 8162195 8195920 8255472 8412247 8449267 8 590602 8936597 9571824 9625677 9853929 I yet haven't find a pg(k) divisible by the next pg prime 40952047. Last fiddled with by enzocreti on 2018-11-20 at 07:16

2018-11-20, 07:29	#17
enzocreti Mar 2018 1000010010₂ Posts	ec primes so for example 255127 divides (2^284274-1)*10^85575+2^284273-1=pg(284274)

2018-11-20, 07:31	#18
enzocreti Mar 2018 2×5×53 Posts	continue... Must pg(k) have a particular form to be divisible by 255127?

2018-11-20, 07:17	#16
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 3⁶×13 Posts	"Except for the last page, the previous thread became a bit long."

2018-11-20, 09:47	#19
enzocreti Mar 2018 2·5·53 Posts	PG numbers which are divisible by 212605 pg(k)=(2^k-1)10^d+2^(k-1)-1, where d is the number of decimal digits of 2^(k-1)-1. pg(3) for example is 73. With Julia software I found that pg(k)'s which are divisible by 212605 are the following: pg(3333) pg(683733) pg(887433) pg(1091133) 3333,683733,887433,1091133 are all ending with digits 33 and are all congruent to 183 mod 210. So I wonder if for pg(k) to be divisible by 212605 it is necessary that k is congruent to 183 mod 210 and k congruent to 33 (mod 10). Last fiddled with by enzocreti on 2018-11-20 at 10:43*

2018-11-20, 10:54	#20
enzocreti Mar 2018 1022₈ Posts	...continue... found another example 212605 divides ec(1294833)

2018-11-20, 11:07	#21
enzocreti Mar 2018 2×5×53 Posts	...continues... another example found 212605 divides pg(1498533)

2018-11-20, 14:47	#22
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	p = 40952047 divides pg(k) for k=118327344; for this k, m=35620080.

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