[QUOTE=science_man_88;230899]if I did the math correct that means 230 numbers near a googol seems to give the best chance of exactly 1 being prime at 36.7879 % I think[/QUOTE]

Exactly -- because log(1e100) is about 230.

Edit: Mini-Geek, your death will be swift.

I put L in the code but don't I have to check every numbers for factors below a given L ? so what do i do to the code i haven't figure it out yet lol.

that means for the 6n+1 and 6n-1 L would be at least 5

if I did the math right that would decrease 230 to about 80-81 cuts it down quite a bit.

[QUOTE=science_man_88;230905]I put L in the code but don't I have to check every numbers for factors below a given L ?[/QUOTE]

No!

You're not checking n numbers, you're predicting the results if a person checks n numbers. All you have to do is adjust the probability by the factor I gave.

[QUOTE=CRGreathouse;230915]No!

You're not checking n numbers, you're predicting the results if a person checks n numbers. All you have to do is adjust the probability by the factor I gave.[/QUOTE]

oh so multiply the final probability by that I see lol.

[CODE](12:53) gp > poisson2(N,t,n,L) = ((((t/log(N))^n)*exp(-(t/log(N))))/n!)*(exp(Euler)*log(L))
%212 = (N,t,n,L)->((((t/log(N))^n)*exp(-(t/log(N))))/n!)*(exp(Euler)*log(L))
(12:53) gp > poisson2(10^100,20,0,5)
%213 = 2.628049047414643169717703463
(12:53) gp >[/CODE]

if I was correct

[QUOTE=science_man_88;230917][CODE](12:53) gp > poisson2(N,t,n,L) = ((((t/log(N))^n)*exp(-(t/log(N))))/n!)*(exp(Euler)*log(L))
%212 = (N,t,n,L)->((((t/log(N))^n)*exp(-(t/log(N))))/n!)*(exp(Euler)*log(L))
(12:53) gp > poisson2(10^100,20,0,5)
%213 = 2.628049047414643169717703463
(12:53) gp >[/CODE]

if I was correct[/QUOTE]

Doesn't look right. Let's clean up the code and see if that makes it any easier:

[code]estimatePrimes(N,t,n,L)={
my(lambda=...);
(lambda^n*exp(-lambda))/n!
};
addhelp(estimatePrimes, "estimatePrimes(N,t,n,L): ...");[/code]

[CODE](13:23) gp > estimatePrimes(10^100,230,1,5)
%217 = 0.1634366358216543572411964273[/CODE]

with code:

[CODE]estimatePrimes(N,t,n,L)=my(lambda=(t/log(N))*(exp(Euler)*log(L)));(lambda^n*exp(-lambda))/n!;[/CODE]

I love how that change in code fits exactly the range of 80-81 I predicted.

The code looks good.

[QUOTE=science_man_88;230910]that means for the 6n+1 and 6n-1 L would be at least 5

if I did the math right that would decrease 230 to about 80-81 cuts it down quite a bit.[/QUOTE]

[QUOTE=science_man_88;230927]I love how that change in code fits exactly the range of 80-81 I predicted.[/QUOTE]

You expect about one prime per 230 for numbers near a googol, but one prime per 115 for odd numbers near a googol and one prime per 77 for odd numbers not divisible by 3 near a googol.

So 80-81 is close, but no cigar. :smile:

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.)