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

${\displaystyle \sum _{n\le x}}{y}^{\mathrm{\Omega }(n)}$	$={\displaystyle \sum _{d\le x}}{\displaystyle \sum _{m\le \frac{x}{d}}}g(d)$ by the convolution method
	$=O\left({\displaystyle \sum _{d\le x}}g(d)\cdot {\displaystyle \frac{x}{d}}\right)$
	$=O\left(x{\displaystyle \prod _{p\le x}}\left(1+{\displaystyle \sum _k}{\displaystyle \frac{g(p^k)}{p^k}}\right)\right)$
	$=O\left(x{\displaystyle \prod _{p\le x}}\left(1+{\displaystyle \sum _k}{\displaystyle \frac{\mu (1)y^k+\mu (p)y^{k-1}}{p^k}}\right)\right)$
	$=O\left(x{\displaystyle \prod _{p\le x}}\left(1+{\displaystyle \frac{y-1}{p}}{\displaystyle \sum _k}{\left({\displaystyle \frac{y}{p}}\right)}^{k-1}\right)\right)$
	$=O\left(x{\displaystyle \prod _{p\le x}}\left(1+{\displaystyle \frac{y-1}{p}}\cdot {\displaystyle \frac{1}{1-\frac{y}{p}}}\right)\right)$
	$=O\left(x{\displaystyle \prod _{p\le x}}\left(1+{\displaystyle \frac{y-1}{p-y}}\right)\right)$
	$=O\left(x\left(1+{\displaystyle \frac{y-1}{2-y}}\right){\displaystyle \prod _{3\le p\le x}}\left(1+{\displaystyle \frac{y-1}{p-y}}\right)\right)$
	$=O\left(x\left({\displaystyle \frac{2-y+y-1}{2-y}}\right){\displaystyle \prod _{3\le p\le x}}\exp \left({\displaystyle \frac{y-1}{p-y}}\right)\right)$
	$=O\left({\displaystyle \frac{x}{2-y}}\left(\exp \left({\displaystyle \sum _{3\le p\le x}}{\displaystyle \frac{y-1}{p-y}}\right)\right)\right)$
	$=O\left({\displaystyle \frac{x}{2-y}}{\left(\exp \left({\displaystyle \sum _{3\le p\le x}}{\displaystyle \frac{1}{p-y}}\right)\right)}^{y-1}\right)$
	$=O\left({\displaystyle \frac{x}{2-y}}{\left(\exp \left(\log \log x+O(1)\right)\right)}^{y-1}\right)$
	$=O\left({\displaystyle \frac{x}{2-y}}{\left(c{e}^{\log \log x}\right)}^{y-1}\right)$ for some $c>0$
	$=O\left({\displaystyle \frac{x}{2-y}}{\left(\max \{1,c\}\log x\right)}^{y-1}\right)$
	$=O\left({\displaystyle \frac{x}{2-y}}{\left(\max \{1,c\}\right)}^{2-1}{\left(\log x\right)}^{y-1}\right)$
	$=O\left({\displaystyle \frac{x{\left(\log x\right)}^{y-1}}{2-y}}\right)$.

∎

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.

Title	${\displaystyle \sum \mathrm{@}\mathrm{@}\mathrm{@}{\mathrm{@}}_{n\le x}{y}^{\mathrm{\Omega }(n)}}=O\left({\displaystyle \frac{x{(logx)}^{y-1}}{2-y}}\right)$ for
Canonical name	displaystylesumnleXYOmeganOleftfracxlogXy12yrightFor1leY2
Date of creation	2013-03-22 16:09:15
Last modified on	2013-03-22 16:09:15
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	18
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11N37
Related topic	AsymptoticEstimate
Related topic	ConvolutionMethod
Related topic	DisplaystyleXlog2xOleftsum_nLeX2OmeganRight
Related topic	DisplaystyleSum_nLeXYomeganO_yxlogXy1ForYGe0
Related topic	DisplaystyleSum_nLeXTaunaO_axlogX2a1ForAGe0
Related topic	2omeganLeTaunLe2Omegan