Well, for starters, if P is your given prime number,

   2*}.i.&.(p:inv)P%2

would all be “almost prime” so that suggests one approach: calculate
i.&.(p:inv)P%2 then work your way through th low prime values while
trimming off excessive numbers from the remaining sequence...

—
Raul

On Monday, December 31, 2018, Bernie Eckhart <[email protected]>
wrote:

> 1, 3, 673, 2019
>
> 4 factors is almost a prime.
>
> Let's denote only 4 factor numbers like 2019, i.e. every factor is a unique
> integer that occurs only once, as "almost a prime" number.
>
> A problem is to compute efficiently all "almost a prime" numbers between 1
> and any given prime number.
>
> Almost certainly a waste of time unless some pattern might be discovered
> that allows one to predict and compute primes faster than other methods.
>
> I have no idea where to begin because I am not a mathematician and because
> I am very new to J.
>
> HAPPY 2019!
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm

Reply via email to