Pg primes minus 7 divisible by 1063
 

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


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?

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.

[QUOTE=CRGreathouse;516038]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.[/QUOTE]

Yes anyway these numbers are not random at all...also exponents leading to a prime are not random...so for example the 7-th exponent leading to a prime is 19 and it is prime the 14-th is a multiple of 215 the 21-th is a prime the 28-th is 1 mod 215 and the 35-th is a prime i mean the primes 10^d*(2^k-1)+2^(k-1)-1

[QUOTE=enzocreti;516042]Yes anyway these numbers are not random at all[/QUOTE]

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.