${\displaystyle \sum \mathrm{@}\mathrm{@}\mathrm{@}{\mathrm{@}}_{n\le x}{y}^{\mathrm{\Omega }(n)}}=O\left({\displaystyle \frac{x{(logx)}^{y-1}}{2-y}}\right)$ for

Within this entry, $\mathrm{\Omega }$ refers to the number of (nondistinct) prime factors function (http://planetmath.org/NumberOfNondistinctPrimeFactorsFunction), $\mu$ refers to the Möbius function, $\log$ refers to the natural logarithm, $p$ refers to a prime, and $d$, $k$, $m$, and $n$ refer to positive integers.

Theorem.

For , ${\displaystyle \mathrm{\sum }_{n\mathrm{\le }x}}{y}^{\mathrm{\Omega }\mathit{}\mathrm{(}n\mathrm{)}}\mathrm{=}O\mathit{}\mathrm{\left(}{\displaystyle \frac{x\mathit{}{\mathrm{(}\log \mathit{}x\mathrm{)}}^{y\mathrm{-}\mathrm{1}}}{\mathrm{2}\mathrm{-}y}}\mathrm{\right)}$.

Proof.

Let $g$ be a function such that $y^{\mathrm{\Omega }}=1*g$. Then $g$ is multiplicative and $g=\mu *y^{\mathrm{\Omega }}$. Thus:

∎

Note that a result for $y=2$ (and therefore for $y\ge 2$), such as ${\displaystyle \sum _{n\le x}}{2}^{\mathrm{\Omega }(n)}=O(x\log x)$, is unobtainable, as evidenced by this theorem (http://planetmath.org/DisplaystyleXlog2xOleftsum_nLeX2OmeganRight). On the other hand, the asymptotic estimates ${\displaystyle \sum _{n\le x}}{2}^{\omega (n)}=O(x\log x)$ and ${\displaystyle \sum _{n\le x}}\tau (n)=O(x\log x)$ are true.