corrected:

p = [TEX]e^{-lambda}[/TEX] for n=0 (because 0! = 1 by definition)

So solve for lambda in terms of p (or the constant 0.01), then substitute the definition of lambda and solve further.

[QUOTE=CRGreathouse;230984]I was looking for a function that, given a probability p (and N, t, L), would give an n such that

estimatePrimes(N,t,n,L)

is 1 - p.[/QUOTE]

what happened to this one ?

[QUOTE=science_man_88;231122]what happened to this one ?[/QUOTE]

That's what I was showing you in #634 etc.

lambda = ((t/log(N))*exp(Euler)*log(L))

p= [TEX]e^{-lambda}[/TEX]

p=[TEX]exp^{-((t/log(N))*exp(Euler)*log(L))}[/TEX]

OK, so take the (natural) log of both sides. (TeX note: e^{\gamma}, not e^{(Euler)}. "Euler" is what Pari calls it, "\gamma" is what TeX calls it.)

Remember, you're solving for t.

actually no I'm not stop assuming.

[QUOTE=science_man_88;231145]actually no I'm not stop assuming.[/QUOTE]

I don't know what that means. I didn't say you were assuming anything, nor that you were stopping any assumptions.

you say I'm solving for t without proof of that and my statement of the opposite yet you still claim it hence you assume it's true there's the assumption.

Dammit.. I accidentally clicked away at 1 day's worth of work.. Now I lost about 10 hours of work..

[QUOTE=3.14159;231152]Dammit.. I accidentally clicked away at 1 day's worth of work.. Now I lost about 10 hours of work..[/QUOTE]

that's better than 24 hours worth Pi !!!!!