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I've looked on Math.SE and Wikipedia so far and found nothing. I thought I found what I was looking for, but it turned out to be Euler's Totient function.

Taking the canonical representation of an integer $n$:

$$n=p_0^{a_0}\cdot p_1^{a_1}\cdots p_r^{a_r}$$

Is there a standard notation for $\sum_{i=0}^r a_r$? Are there any known theorems dealing directly with this quantity aside from those dealing with the Totient function? Thank you!

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    $\begingroup$ It's $\Omega(n)$, the number of all prime factors of $n$, in contrast to $\omega(n)$, which is the number of distinct prime divisors of $n$. $\endgroup$ – Daniel Fischer Sep 25 '13 at 17:51
  • $\begingroup$ @DanielFischer: perfect, thank you! $\endgroup$ – abiessu Sep 25 '13 at 17:53
  • $\begingroup$ @DanielFischer: wait a moment though, how does this usage relate to the one shown here? $\endgroup$ – abiessu Sep 25 '13 at 17:56
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    $\begingroup$ As for a theorem dealing with it, the Riemann hypothesis is equivalent to the statement $\sum_{k \le n}{(-1)^{\Omega(k)}} = O(n^{\frac{1}{2}+\epsilon})$ for all $\epsilon \gt 0$. $\endgroup$ – Dan Brumleve Sep 25 '13 at 17:57
  • $\begingroup$ @abiessu Not related. The same symbol used for different things. $\endgroup$ – Daniel Fischer Sep 25 '13 at 17:58
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The comments provided the requested notation accurately, namely, that if we have

$$n=p_0^{a_0}\cdot p_1^{a_1}\cdots p_r^{a_r}\tag 1$$

then we use $\Omega(n)=\sum_{i=0}^ra_i$ to denote the count of prime factors of $n$ including multiplicity. I believe I have even found a reasonable arithmetic function composition that can be used to "calculate" $\Omega(n)$ for any given $n$, although I introduce a "new" function (which can probably best be considered an "anti-derivative" of an existing function) in order to arrive at this information.

To begin, note that the Mangoldt function is defined as

$$\Lambda(n)=\left\{\begin{array}{ll}\log p & \text{if $n=p^k$ for prime $p$ and integer $k$}\\0 & \text{otherwise}\end{array}\right.$$

Now branch our "new" function from here as the arithmetic "anti-derivative" of the Mangoldt function, which is to say that the Mangoldt function will be the arithmetic derivative of our function:

$$\eth(n)=\left\{\begin{array}{ll}\log_np=\frac 1k & \text{if $n=p^k$ for prime $p$ and integer $k$}\\0 & \text{otherwise}\end{array}\right.$$

Now the sum over the distinct prime divisors of $n$ works like this:

$$\omega(n)=\sum_{d\mid n}\mu(d)\eth(d)$$

Getting $\Omega(n)$ is a little trickier, since we have $\eth(p^k)=\frac 1k$ for primes $p$ when we actually want $1$ for each of these values. But this can be solved with multiplication by another sum:

$$\Omega(n)=\sum_{d\mid n}\left(\mu(d)\eth(d)\sum_{d^k\mid n}1\right)$$

Using our definition, we can ask some questions about our function; for instance:

$$\sum_{d\mid n}\eth(d)=\sum_{i=0}^rH_{a_i}$$

where $H_n$ is the $n$th Harmonic Number, being the sum $\sum_{i=1}^n\frac 1i$. This is slightly counter-intuitive when we consider that $\log n=\sum_{d\mid n}\Lambda(d)$ and that $\eth(n)={\Lambda(n) \over \log n}$.

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