Thread for posting tiny primes

[CRGreathouse]

Originally Posted by science_man_88

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

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

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

[science_man_88]

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.

[science_man_88]

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.

[CRGreathouse]

Originally Posted by science_man_88

I put L in the code but don't I have to check every numbers for factors below a given L ?

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.

[science_man_88]

Originally Posted by CRGreathouse

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.

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

[science_man_88]

(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 >

if I was correct

[CRGreathouse]

Originally Posted by science_man_88

(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 >

if I was correct

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

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

[science_man_88]

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

with code:

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

[science_man_88]

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

[CRGreathouse]

The code looks good.

Originally Posted by science_man_88

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.

Originally Posted by science_man_88

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

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.

[CRGreathouse]

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.)