# $\displaystyle\sum@\slimits@@@_{n\leq x}y^{\Omega(n)}=O\left(\frac{x(\mathop{% log}\nolimits x)^{y-1}}{2-y}\right)$ for $1\leq y<2$

Within this entry, $\Omega$ 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 $1\leq y<2$, $\displaystyle\sum_{n\leq x}y^{\Omega(n)}=O\left(\frac{x(\log x)^{y-1}}{2-y}\right)$.

###### Proof.

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

$\displaystyle\sum_{n\leq x}y^{\Omega(n)}$ $\displaystyle=\sum_{d\leq x}\sum_{m\leq\frac{x}{d}}g(d)$ by the convolution method $\displaystyle=O\left(\sum_{d\leq x}g(d)\cdot\frac{x}{d}\right)$ $\displaystyle=O\left(x\prod_{p\leq x}\left(1+\sum_{k}\frac{g(p^{k})}{p^{k}}% \right)\right)$ $\displaystyle=O\left(x\prod_{p\leq x}\left(1+\sum_{k}\frac{\mu(1)y^{k}+\mu(p)y% ^{k-1}}{p^{k}}\right)\right)$ $\displaystyle=O\left(x\prod_{p\leq x}\left(1+\frac{y-1}{p}\sum_{k}\left(\frac{% y}{p}\right)^{k-1}\right)\right)$ $\displaystyle=O\left(x\prod_{p\leq x}\left(1+\frac{y-1}{p}\cdot\frac{1}{1-% \frac{y}{p}}\right)\right)$ $\displaystyle=O\left(x\prod_{p\leq x}\left(1+\frac{y-1}{p-y}\right)\right)$ $\displaystyle=O\left(x\left(1+\frac{y-1}{2-y}\right)\prod_{3\leq p\leq x}\left% (1+\frac{y-1}{p-y}\right)\right)$ $\displaystyle=O\left(x\left(\frac{2-y+y-1}{2-y}\right)\prod_{3\leq p\leq x}% \exp\left(\frac{y-1}{p-y}\right)\right)$ $\displaystyle=O\left(\frac{x}{2-y}\left(\exp\left(\sum_{3\leq p\leq x}\frac{y-% 1}{p-y}\right)\right)\right)$ $\displaystyle=O\left(\frac{x}{2-y}\left(\exp\left(\sum_{3\leq p\leq x}\frac{1}% {p-y}\right)\right)^{y-1}\right)$ $\displaystyle=O\left(\frac{x}{2-y}\left(\exp\left(\log\log x+O(1)\right)\right% )^{y-1}\right)$ $\displaystyle=O\left(\frac{x}{2-y}\left(ce^{\log\log x}\right)^{y-1}\right)$ for some $c>0$ $\displaystyle=O\left(\frac{x}{2-y}\left(\max\{1,c\}\log x\right)^{y-1}\right)$ $\displaystyle=O\left(\frac{x}{2-y}\left(\max\{1,c\}\right)^{2-1}\left(\log x% \right)^{y-1}\right)$ $\displaystyle=O\left(\frac{x\left(\log x\right)^{y-1}}{2-y}\right)$.

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

Title $\displaystyle\sum@\slimits@@@_{n\leq x}y^{\Omega(n)}=O\left(\frac{x(\mathop{% log}\nolimits x)^{y-1}}{2-y}\right)$ for $1\leq y<2$ displaystylesumnleXYOmeganOleftfracxlogXy12yrightFor1leY2 2013-03-22 16:09:15 2013-03-22 16:09:15 Wkbj79 (1863) Wkbj79 (1863) 18 Wkbj79 (1863) Theorem msc 11N37 AsymptoticEstimate ConvolutionMethod DisplaystyleXlog2xOleftsum_nLeX2OmeganRight DisplaystyleSum_nLeXYomeganO_yxlogXy1ForYGe0 DisplaystyleSum_nLeXTaunaO_axlogX2a1ForAGe0 2omeganLeTaunLe2Omegan