On the sum of least prime factors in short intervals

Let $p(n)$ denote the least prime factor of $n$ and $L_C(x)=C{x}^{1\mathrm{/}2}(\log x{)}^2$ with $C>0$. The sum of $p(n)\mathrm{/}n$ over composite $n$ lying in the short interval $[x,x+L_C(x)]$, a question raised by Erdős and Graham, is studied. (i) The constant $c=8$ in the mean asymptotic is estimated $$S(x)=\sum _{n<x,n\text{ composite}}\frac{p(n)}{n}=\frac{c{x}^{1\mathrm{/}2}}{(\log x{)}^2}(1+O(\frac{1}{\log x}\left)\right)+O\left(\frac{x^{1\mathrm{/}3}}{\log x}\right)\mathrm{.}$$ (ii)For every fixed $C>0$, the window sums ${\mu }_C(x):={\sum }_{x\le n\le x+L_C(x)}p(n)\mathrm{/}n$ over composites have mean $4C$: $\frac{1}{X}{\sum }_{x\le X}{\mu }_C(x)=4C+O_C(1\mathrm{/}\log X)$, and second moment $\frac{1}{X}{\sum }_{x\le X}({\mu }_C(x)-4C{)}^2=O_C((\log X{)}^{-2})$. In particular ${\mu }_C(x)=4C+o(1)$ for almost all $x$. (iii)Under a weak Cramér-type hypothesis on primes in intervals of length $(\log y{)}^{2+o(1)}$, the estimate ${\mu }_C(x)=4C+O_C(1\mathrm{/}\log x)$ holds uniformly in $x$, giving an affirmative answer to the Erdős–Graham question. Unconditionally, the uniform statement remains open; proving the uniform statement unconditionally would require resolving short-interval prime estimates at scale $(\log y{)}^2$.