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exponents
For 
, define )
oto be the highest exponent to which a prime divides it. For instance, =E(36)=2)
. Prove that )
exists.
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Previous Result: The cardinal of: 
is } <br />
)
where 
is the floor function and )
is the Möbius function.
Proof
Consider: 
, the cardinal of this set is, by inclusion-exclusion: ^j } \right|x} \right\}} \right|} \pm ...<br />
)
But: 
thus note that we can write: } <br />
)
but we want 
and the rest follows. 
At this point, note that every number that is free of squares, is free of cubes ... and so on. Further, if a number 
is free of cubes, but not of squares, then =2)
That is: } - \sum\limits_{k = 1}^\infty {\left\lfloor {\tfrac{n}<br />
{{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|<br />
)
Thus we get: } = \sum\limits_{j = 1}^\infty {\left( {\sum\limits_{k = 1}^\infty {\left( {\left\lfloor {\tfrac{n}<br />
{{k^{j + 1} }}} \right\rfloor - \left\lfloor {\tfrac{n}<br />
{{k^j }}} \right\rfloor } \right) \cdot \mu \left( k \right)} } \right) \cdot j} <br />
)
(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: } =<br />
{\tfrac{1}<br />
{{\zeta \left( {2} \right)}}}<br />
+ \sum\limits_{j = 2}^\infty {\left( {\tfrac{1}<br />
{{\zeta \left( {j + 1} \right)}} - \tfrac{1}<br />
{{\zeta \left( j \right)}}} \right) \cdot j} <br />
)
-remember }}<br />
{{k^s }}} = \tfrac{1}<br />
{{\zeta \left( s \right)}}<br />
)
and } = \sum\limits_{k = 1}^n {\sum\limits_{\left. d \right|k} {\mu \left( d \right)} } = 1<br />
)
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