After arguably too many New Year’s Eve drinks, and misreading the requirement
as four prime divisors, my effort on the iPad led to iOS J701 crashing, and so
to bed.
It’s a lot easier if you only want two prime factors. Unlike Skip, I’ve assumed
you only want squarefree factors, so here’s a function in the spirit of Raul’s
comments:
ppp =: 3 : 0
n =. y
p =. p: i. >: _1 p: -:n
l =. ''
while. #p do.
NB. multiply first by rest, and remove too high elements
l =. l, b =. n (>: # ]) ({. * }.) p
NB. trim inputs to too high elements
p =. }. p{.~ # b
end.
/:~ l NB. sorted list
)
ppp 30
6 10 14 15 21 22 26
_5{. ppp 2030
1994 2005 2018 2019 2026 NB. Even evens allowed!
Happy Nearly Prime 2019 Sent from my iPad Air running J701 under iOS 12.1.1,
screen not too bad!
Mike
> On 1 Jan 2019, at 02:23, 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!
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