[QUOTE=CRGreathouse;230933]So now can you write a different function that takes a limit L, base b, an exponent e, and a k-range kmin,kmax and determines
1. The expected number of candidates remaining after sieving to L?
2. The number of primes in the range? (Use your estimatePrimes function.)[/QUOTE]

I'll try but I'm not promising anything yet lol.

well k*b^e +1 ? I can make a limit variable that's not too hard

[QUOTE=science_man_88;230939]I'll try but I'm not promising anything yet lol.[/QUOTE]

Of course. But you've already built a useful tool; I'm just trying to get you to make another.

well if you look at it the rough estimate of candidates in the range is #number_in _range *probability or 1/average gap.

as for the candidates left after L unless i know the upper bound I'm not so sure how to conquer it I could assume that if i can find a upper bound at this many per gap etc. I can expect this many that might be simple if I can do the math with them in reasonable time.

[QUOTE=science_man_88;230942]as for the candidates left after L unless i know the upper bound I'm not so sure how to conquer it I could assume that if i can find a upper bound at this many per gap etc. I can expect this many that might be simple if I can do the math with them in reasonable time.[/QUOTE]

See post #593. (1 - 1/f) * 100% of candidates are sieved out, leaving (1/f) * 100% of candidates that may be prime with 'probability' f times greater than usual.

in the range yes outside the range unless I know what percentage of candidates are in each range or something I still see no clue.

I don't understand what you wrote or how it relates to what I wrote.

what I'm saying is

up to L is definable range
up to unknown value is quite defined in my eyes enough to help me find what percentage is left of all candidates.

Yep, not getting it. You use L to calculate a new lambda; you don't do anything else with it. You're not working in ranges defined by it.

then i don't know of a way to do it unless I loop forever changing lambda and never stop until I can prove no more candidates exist I have no way of doing either really.

I might get it if I followed the:

[B]K[/B]eep
[B]i[/B]t
[B]s[/B]imple
[B]s[/B]tupid

method

[QUOTE=science_man_88;230959]then i don't know of a way to do it unless I loop forever changing lambda and never stop until I can prove no more candidates exist I have no way of doing either really.[/QUOTE]

I have no idea why when I asked for the expected number of remaining candidates, you thought this would require an infinite loop.