Conjecture regarding the sum of prime factors of $x!$ and $(x-1)!$? I think using dodgy means I can show the following:
Let $\lambda(x)$ be the sum of  primes in $x$. For example:
$$ \lambda(2) = 2$$
$$ \lambda(4) = 2 + 2 = 4$$
$$ \lambda(6) = 3 + 2 = 5$$
Then for large $n$:
$$ \frac{\lambda (n!)}{\lambda ((n-1)!)} \sim \frac{\ln (n!)}{\ln (n-1)!}$$
Or more precisely substracting $1$:
$$ \frac{\lambda (n)}{\lambda ((n-1)!)} \sim \frac{\ln (n)}{\ln (n-1)!}$$
Or even better:
$$ \frac{\lambda (n)}{\lambda ((n-1)!)} - \frac{\ln (n)}{\ln (n-1)!} = o(1)$$
Can someone prove or disprove this conjecture?
For example:
$$ \frac{\lambda(8!)}{\lambda(7!)} = \frac{31}{26} \approx 1.23$$
And
$$ \frac{\ln(8!)}{\ln(7!)} \approx 1.244 $$
Substracting $1$:
$$ .23  \approx .24$$
 A: The question does not make sense, why ?
First $ \frac{\lambda((n+1)!)}{\lambda(n!)} \sim \frac{\ln((n+1)!}{\ln n!} \implies \frac{\lambda((n+1)!)}{\lambda(n!)}\sim 1 $ but this is not that impressive, it only means that for every $\epsilon>0$ we have that $ (1-\epsilon)^n < \lambda(n!) < (1+\epsilon)^n$ for large enough $n$ (so basically $ \lambda(n!)$ grows sub-exponential), for instance i could easily said that :$$ \frac{\lambda((n+1)!)}{\lambda(n!)} \sim \frac{(n+1)^3}{n^3}$$
Second $ \frac{\lambda(n+1)}{\lambda(n!)} \sim \frac{\ln(n+1)}{\ln n!}$ is simply wrong take $n$ such that $ n+1$ is prime number and so $ \lambda(n+1) = n+1$  and $ \lambda (n!) \sim \frac{\pi^2 n^2}{12 \ln n}$ and $\ln n! \sim n \ln n$ and so what you are saying is that $ \frac{12 \ln n}{\pi^2 n} \sim \frac{1}{n}$ which obviously is not correct.
Third This part tells you nothing simply because $ \frac{\lambda(n+1)}{\lambda(n!)} = o(1)$ so i could easily said that : $$ \frac{\lambda(n+1)}{\lambda(n!)} - \frac{1}{\ln \ln n} = o(1)$$
Also the function $\lambda(n)$ is very chaotic as most of the arithmetic function in number theory because they are related to the prime numbers and those guys are the definition of chaotic in small intervals( for instance when $n+1$ is highly composite number $\frac{\lambda(n+1)}{\lambda(n!)}$ will approach $0$ waaay faster than if $n+1$ was prime number).
In these cases we find the Average\Sum of such function to have beautiful results such as $ \lambda (n!) \sim \frac{\pi^2 n^2}{12 \ln n}$.
Hope its helpfull.
