I am trying to check if a very big number ($>10^{10,000,000}$) is possibly prime. I have written a computer program to check if the number has any smallish (less than like $600,000,000$) factors... it doesn't. I know that the chance of a random number $p$ being prime is $\dfrac{1}{\ln p}$, but what if it doesn't have any factors less than $600,000,000$? Or more generally, what is the probability that $p$ is prime if it doesn't have any factors less than $x$?
I thought that it might be $\dfrac{\ln(x)}{\ln p}$ but since you only have to check up to the square root of the number to confirm it's prime that didn't make since. I would guess it might be $\dfrac{\ln(x)^2}{\ln p}$ but that's just a guess.
Any help is appreciated. Thanks in advance!
If a huge number has no prime factor below $n$ and $n$ itself is large (lets say $10^4$ or larger) , then the chance that the number is prime increases approximately by the factor $$e^{\gamma}\cdot \ln(n)$$ , so if the trial factor limit is $\ 10^k\ $ , the factor is about $\ 4.1\cdot k\ $.
So, the chance for the huge number $N$ to be prime is $$\frac{e^{\gamma}\cdot \ln(n)}{\ln(N)}$$
So, even if you check the prime factors upto $\ 10^{10}\ $ , the chance increases only by a factor $\ 41\ $. That means that a number with millions of digits still has a very low chance to be prime. This is the reason for the difficulty to find huge primes, trial factoring has a much smaller effect than one would expect.
A bit higher is the chance if you search for prime numbers of a special form, for example mersenne primes because of the form the prime factors must have, but I do not know how large this additional effect is.