# exponents

• Jun 12th 2009, 05:42 PM
Chandru1
exponents
For $\displaystyle n>1$, define $\displaystyle E(n)$ oto be the highest exponent to which a prime divides it. For instance, $\displaystyle E(12)=E(36)=2$. Prove that $\displaystyle \lim_{N \to \infty} \frac{1}{N} \sum\limits_{n=2}^{N}E(n)$exists.
• Jun 13th 2009, 05:05 AM
PaulRS
Previous Result: The cardinal of: $\displaystyle \left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}p^j \not | x{\text{ }}\forall p{\text{ prime}}} \right\}$ is $\displaystyle \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n} {{k^j }}} \right\rfloor \cdot \mu \left( k \right)}$ where $\displaystyle \left\lfloor x \right\rfloor$ is the floor function and $\displaystyle \mu(n)$ is the Möbius function.

Proof

Consider: $\displaystyle A= \bigcup\limits_{p{\text{ prime}}} {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. {p^j } \right|x} \right\}}$ , the cardinal of this set is, by inclusion-exclusion: $\displaystyle \sum\limits_{p{\text{ prime}}} {\left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. {p^j } \right|x} \right\}} \right|} - \sum\limits_{p > q{\text{ primes}}} {\left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. {\left( {pq} \right)^j } \right|x} \right\}} \right|} \pm ...$

But: $\displaystyle \left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}\left. m \right|x} \right\}} \right| = \left\lfloor {\tfrac{n} {m}} \right\rfloor$ thus note that we can write: $\displaystyle A = - \sum\limits_{k = 2}^\infty {\left\lfloor {\tfrac{n} {{k^j }}} \right\rfloor \cdot \mu \left( k \right)}$ but we want $\displaystyle {\bar A}$ and the rest follows. $\displaystyle \square$

At this point, note that every number that is free of squares, is free of cubes ... and so on. Further, if a number $\displaystyle x$ is free of cubes, but not of squares, then $\displaystyle E(x)=2$

That is: $\displaystyle \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n} {{k^{j + 1} }}} \right\rfloor \cdot \mu \left( k \right)} - \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n} {{k^j }}} \right\rfloor \cdot \mu \left( k \right)} = \left| {\left\{ {x \in \mathbb{Z}^ + /x \leqslant n{\text{ and }}E\left( x \right) = j} \right\}} \right|$

Thus we get: $\displaystyle \sum\limits_{k = 1}^n {E\left( k \right)} = \sum\limits_{j = 1}^\infty {\left( {\sum\limits_{k = 1}^\infty {\left( {\left\lfloor {\tfrac{n} {{k^{j + 1} }}} \right\rfloor - \left\lfloor {\tfrac{n} {{k^j }}} \right\rfloor } \right) \cdot \mu \left( k \right)} } \right) \cdot j}$ (This is already very suggestive)

Mmmm, here I've not been able to finish it formally -it gets quite nasty-, but I do think that: $\displaystyle \mathop {\lim }\limits_{n \to + \infty } \tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {E\left( k \right)} = {\tfrac{1} {{\zeta \left( {2} \right)}}} + \sum\limits_{j = 2}^\infty {\left( {\tfrac{1} {{\zeta \left( {j + 1} \right)}} - \tfrac{1} {{\zeta \left( j \right)}}} \right) \cdot j}$ -remember $\displaystyle \sum\limits_{k = 1}^\infty {\tfrac{{\mu \left( k \right)}} {{k^s }}} = \tfrac{1} {{\zeta \left( s \right)}}$ and $\displaystyle \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n} {k}} \right\rfloor \cdot \mu \left( k \right)} = \sum\limits_{k = 1}^n {\sum\limits_{\left. d \right|k} {\mu \left( d \right)} } = 1$ -