A question about primes of a particular form

[enzocreti]

Originally Posted by Dr Sardonicus

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

Great! but I think that you could not find a pg(k) divisible by the next p=524287262143

[Dr Sardonicus]

Originally Posted by enzocreti

Great! but I think that you could not find a pg(k) divisible by the next p=524287262143

So? It would be a stupid waste of time to look. With p = 40952047, I reckoned I had a decent chance of finding another solution in a reasonable length of time, so I went ahead and ran the numbers.

But with p = 524287262143, I don't foresee finding another solution in a reasonable length of time. I don't even want to run the calculations to tell how far to search k-values in order to have a reasonable chance of success. I'm pretty sure, though, that it would probably be at least 2*p, maybe 4*p or more. Possibly much more.

I am, however, quite confident that p = 524287262143 divides pg(k) not only for infinitely many k, but for a positive proportion of positive integers k. The reason is quite simple. The remainder of pg(k) (mod p) is determined by the remainders of k (mod 87381210357) and m (mod 74898180306), respectively. We already know that p|pg(k) when k == 19 (mod 87381210357) and m == 6 (mod 74898180306).

So, all that is required, is to prove that the pairs of integer remainders (k%87381210357, m%74898180306) all occur equally often in the sequence. This would insure that at least 1/(87381210357*74898180306) of the terms are divisible by p. It would be more, depending on how many pairs (k%87381210357, m%74898180306) make pg(k) divisible by p.

I think the equal-distribution result is provable, and depends only on the fact that log(2)/log(10) is irrational. That would definitively answer the question about pg(k) primes (or any other prime other than 2 or 5) dividing terms of your sequence.

To me, the equal-distribution result is of more interest than your sequence, but so far I've been too lazy to look it up or try to prove it myself.

[CRGreathouse]

I find:

3333, 683733, 887433, 1091133, 1294833, 1498533, 2178933, 2382633, 2586333, 2790033, 2993733, 3674133, 3877833, 4081533, 4285233, 4488933, 5169333, 5373033, 5576733, 5780433, 5984133, 6187833, 6868233, 7071933, 7275633, 7479333, 7683033, 8363433, 8567133, 8770833, 8974533, 9178233, 9858633, 10062333, 10266033, 10469733, 10673433, 11353833, 11557533, 11761233, 11964933, 12168633, 12372333, 13052733, 13256433, 13460133, 13663833, 13867533, 14547933, 14751633, 14955333, 15159033, 15362733, 16043133, 16246833, 16450533, 16654233, 16857933, 17061633, 17742033, 17945733, 18149433, 18353133, 18556833, 19237233, 19440933, 19644633, 19848333, 20052033, 20732433, 20936133, 21139833, 21343533, 21547233, 22227633, 22431333, 22635033, 22838733, 23042433, 23246133, 23926533, 24130233, 24333933, 24537633, 24741333, 25421733, 25625433, 25829133, 26032833, 26236533, 26916933, 27120633, 27324333, 27528033, 27731733, 28412133, 28615833, 28819533, 29023233, 29226933, 29430633, 30111033, 30314733, 30518433, 30722133, 30925833, 31606233, 31809933, 32013633, 32217333, 32421033, 33101433, 33305133, 33508833, 33712533, 33916233, 34596633, 34800333, 35004033, 35207733, 35411433, 35615133, 36295533, 36499233, 36702933, 36906633, 37110333, 37790733, 37994433, 38198133, 38401833, 38605533, 39285933, 39489633, 39693333, 39897033, 40100733, 40781133, 40984833, 41188533, 41392233, 41595933, 41799633, 42480033, 42683733, 42887433, 43091133, 43294833, 43975233, 44178933, 44382633, 44586333, 44790033, 45470433, 45674133, 45877833, 46081533, 46285233, 46965633, 47169333, 47373033, 47576733, 47780433, 47984133, 48664533, 48868233, 49071933, 49275633, 49479333, 50159733, 50363433, 50567133, 50770833, 50974533, 51654933, 51858633, 52062333, 52266033, 52469733, 52673433, 53353833, 53557533, 53761233, 53964933, 54168633, 54849033, 55052733, 55256433, 55460133, 55663833, 56344233, 56547933, 56751633, 56955333, 57159033, 57839433, 58043133, 58246833, 58450533, 58654233, 58857933, 59538333, 59742033, 59945733, 60149433, 60353133, 61033533, 61237233, 61440933, 61644633, 61848333, 62528733, 62732433, 62936133, 63139833, 63343533, 64023933, 64227633, 64431333, 64635033, 64838733, 65042433, 65722833, 65926533, 66130233, 66333933, 66537633, 67218033, 67421733, 67625433, 67829133, 68032833, 68713233, 68916933, 69120633, 69324333, 69528033, 70208433, 70412133, 70615833, 70819533, 71023233, 71226933, 71907333, 72111033, 72314733, 72518433, 72722133, 73402533, 73606233, 73809933, 74013633, 74217333, 74897733, 75101433, 75305133, 75508833, 75712533, 76392933, 76596633, 76800333, 77004033, 77207733, 77411433, 78091833, 78295533, 78499233, 78702933, 78906633, 79587033, 79790733, 79994433, 80198133, 80401833, 81082233, 81285933, 81489633, 81693333, 81897033, 82577433, 82781133, 82984833, 83188533, 83392233, 83595933, 84276333, 84480033, 84683733, 84887433, 85091133, 85771533, 85975233, 86178933, 86382633, 86586333, 87266733, 87470433, 87674133, 87877833, 88081533, 88285233, 88965633, 89169333, 89373033, 89576733, 89780433, 90460833, 90664533, 90868233, 91071933, 91275633, 91956033, 92159733, 92363433, 92567133, 92770833, 93451233, 93654933, 93858633, 94062333, 94266033, 94469733, 95150133, 95353833, 95557533, 95761233, 95964933, 96645333, 96849033, 97052733, 97256433, 97460133, 98140533, 98344233, 98547933, 98751633, 98955333, 99635733, 99839433

All are 1233 mod 2100, as predicted. There are no other modular relations I can see (the gcd of the first differences is 2100).

I used this code:

glue(x,y)=x*10^#Str(y)+y;
pg(k)=glue(2^k-1,2^(k-1)-1);
pgmod(k,m)=if(k<99,return(Mod(pg(k),m))); my(d=(k-1)*log(2)\log(10)+1,t=Mod(2,m)^(k-1)); (2*t-1)*Mod(10,m)^d+t-1;
for(k=1,1e8, if(pgmod(k,212605)==0, print1(k,if(k%2100==1233,", ","***, "))))

[enzocreti]

for i=2:1000
if mod(pg(i)^2-pg(i)+1,31)== 0 println(i)
end
end
19
69
119
169
219
269
319
369
419
469
519
569
why pg must be congruent to 19 mod 50?
I mean why for pg(k)^2-pg(k)+1 to be divisible by 31, it is necessary that k is 19 mod 50?

[science_man_88]

Originally Posted by enzocreti

for i=2:1000
if mod(pg(i)^2-pg(i)+1,31)== 0 println(i)
end
end
19
69
119
169
219
269
319
369
419
469
519
569
why pg must be congruent to 19 mod 50?
I mean why for pg(k)^2-pg(k)+1 to be divisible by 31, it is necessary that k is 19 mod 50?

pg(k)^2-pg(k) will always be even so pg(k)^2-pg(k)+1 will always be odd. this means it could be:
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41;43,45,47, or 49 mod 50 . Mersenne numbers are 0,1,3,7,and 15 mod 31.

This means pg(k) is any of 0*x+15, 1*x+0, 3*x+1, or 7*x+3 mod 31 . The first two need impossible scenarios to work, the second to last needs x equal to 10 mod 31, and the last needs x equal to 4 mod 31 . x is our power of ten, mod 31.
powers of ten mod 31 go 10,7,8,18,25,2,20,14,16,5,19,4,9,28,1,...
where ... means it repeats. this forces d to be 1 or 12 mod 15. Which can create 0 mod 31 in these digit lengths ? What is their mod 50 for k ?

those last two will be left to the question asker.

[Dr Sardonicus]

Originally Posted by enzocreti

<snip>
I mean why for pg(k)^2-pg(k)+1 to be divisible by 31, it is necessary that k is 19 mod 50?

It isn't. However, k does have to be congruent to 4 (mod 5). I ran it out to 2000. The very next k after 1000 is congruent to 14 (mod 50). This continues for a while. But (running it out farther) then it is is 9 (mod 50). Then, 4 (mod 50). Then, 49 (mod 50). Then, 44 (mod 50). Then, ...?

k m
19 6
69 21
119 36
169 51
219 66
269 81
319 96
369 111
419 126
469 141
519 156
569 171
1114 336
1164 351
1214 366
1264 381
1314 396
1364 411
1414 426
1464 441
1514 456
1564 471
1614 486
1664 501
1714 516
1764 531
1814 546
1864 561
1914 576
1964 591

[Uncwilly]

The 3 threads about the pg(X) format numbers have been merged. Until there is enough different discussion going on about these numbers that requires different threads keep it here.

[enzocreti]

Ok, is there some particular reason why there is no pg(k) prime 6 mod 7 ? I arrived to k=565.000 and no prime 6 mod 7 found. Is just a mere coincidence and how much further do you think i have to go for having some chance to find it?

[enzocreti]

Are there pg(k)'s which are divisible by the number of digits of 2^(k-1)?

[LaurV]

No, there is no reason why a prime of this form is not 6 (mod 7). There should be an infinite amount of them.

Yes, there are pg(k) divisible by m (with the former notation in the thread), most probably an infinite amount of them too. For example, every forth is divisible by 5 (because all the powers of 2 end in 2, 4, 8, 6 and repeat, so every forth of your numbers ends in 5) so the one ending in 65535 is divisible by 5. If you go high enough, you will find other such "curiosities".

What it the reason of interest in these numbers? (We think you are just trolling...)

[Dr Sardonicus]

Originally Posted by enzocreti

Are there pg(k)'s which are divisible by the number of digits of 2^(k-1)?

Yes. Up to the limit 10,000,000 here they are. (The only significance to me of m being prime is, the formula (mod m) can be simplified. The only significance of this whole topic for me is, it gives me practice with using basic results of elementary number theory, and in writing Pari scripts.)

k m (Asterisk indicates that m is prime)
2 1
3 1
4 1
17 5*
41 13*
76 23*
81 25
102 31*
201 61*
235 71*
336 101*
11623 3499*
15065 4535
24074 7247*
158831 47813
423602 127517
851413 256301*
3150441 948377*
6289421 1893305