20190507, 09:39  #1 
Mar 2018
17·31 Posts 
Pg primes minus 7 divisible by 1063
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 Now it turns out that: (pg(k)7)/1063+1 is prime only for k=7 and k=8 up to k=100.000 Because k=1272 is the next k after k=8 for which (pg(k)7) is divisible by 1063, do you think that could be the reason why there are no more primes of the form : (pg(k)7)/1063+1? 
20190507, 15:45  #2 
Aug 2006
2·29·103 Posts 
You have a form which is around 4^k and which is an integer around 1/1063 of the time (computing the exact probability is a pain), so you might expect the k to work with 'probability' 1/(1063k log 4) or so. This is small over this interval (but diverges over k large enough). That seems like enough of an explanation to me.

20190507, 15:49  #3  
Mar 2018
1000001111_{2} Posts 
Quote:
Last fiddled with by enzocreti on 20190507 at 15:51 

20190507, 16:37  #4 
Aug 2006
1011101010110_{2} Posts 
What do you mean by that?
If you merely mean that the function is deterministic, I know  that's why I put shock quotes around "probability" above. If you mean that there is some special significance to the form, please explain what it is. If you mean that the proportion of primes is different than what you'd expect by standard heuristics, then please explain why and what you should expect instead. 
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