Estimation of the magnitude of the second largest prime factor

Question

Very special cases ignored, a large number is difficult to be completely factored if the second largest prime factor is large.

Can we estimate the magnitude of the second largest prime factor of a random number $N$ , lets say , in the range $[10^{99},10^{100}]$ ?

For clarification : We assume that $N$ is not a power of a prime and only look at the distinct prime factors and choose the second largest of them.

I am only aware of an estimation of the number of distinct prime factors (roughly $\ln(\ln(N))$) , but this does not help to estimate how large the second largest prime factor will be in average.

Answer 1

We can estimate the number of numbers in $[10^{99},10^{100}]$ that have two prime factors greater than $10^{20}$ using the prime number theorem. If we pick two numbers $a,b$ greater than $10^{20}$ the chance they are both prime is $\frac 1{\log a \log b}$ Letting $a \lt b$ we get the number of numbers with these two as prime factors is $$\int_{10^{20}}^{10^{49}}\int_{a}^{\frac{10^{100}}a}\frac {da\ db}{\log a \log b}=\int_{10^{20}}^{10^{49}}\operatorname{Li}\left(\frac{10^{100}}a\right)-\operatorname{Li}\left(a\right)da$$ Alpha gives an evaluation of $4.53287\cdot 10^{99}$ which seems amazingly high. It would say $45\%$ of the numbers have two factors greater than $10^{20}$. We have overcounted numbers with three or four factors greater than $10^{20}$.

Answer 2

The random model for the primes says that when taking $n$ uniformly in $[1,x] $, for $p \le x^r$ we can consider that $n \bmod p$ is uniformly distributed and independent from one $p$ to the other.

Thus with $2ndLpf$ the second largest prime factor we'd have $$Pr[2ndLpf(n) = p] \approx \frac1p\sum_{p<q <x}\frac1q\prod_{q<Q <x} (1-\frac1Q))\approx \sum_{p< q<x} \frac1{pq}\frac{\ln q}{\ln x}\sim \sum_{p<m <x} \frac1{pm \ln x}$$

$$\Bbb{E}[2ndLpf(n)] = \sum_{p \le x^r}p \ Pr(2ndLpf(n) = p)\approx \sum_{p \le x}\sum_{p<m <x} \frac1{m \ln x}\\ \approx \sum_{k \le x}\frac{1}{\ln k}\sum_{k<m <x} \frac1{ m \ln x}\sim \sum_{k \le x} \frac{\ln x-\ln k}{\ln k \ln x}$$