# Does every record of the arithmetic derivative of natural numbers occur at a practical number?

Consider the arithmetic derivative of natural numbers, as defined here.

By this definition, for every integer $n>1$, with canonical prime factorization $p_1^{a_1}p_2^{a_2}...p_{\omega(n)}^{a_{\omega(n)}}$, where $\omega(n)$ is the number of distinct prime factors of $n$, we have a positive arithmetic derivative $n'$ such that $$n'=n\sum_{i=1}^{\omega(n)}\dfrac{a_i}{p_i}$$

My question is, does the sequence of $n$ such that $n'>m'$ for every $m<n$ consists only of practical numbers i.e. $n=1$ or $n>1$ such that $p_1=2$, and $$p_i\leq 1+\sigma(p_1^{a_1}p_2^{a_2}...p_{i-1}^{a_{i-1}})$$ for every $i\in[2,\omega(n)]$? It is certainly not true that every practical number belongs to this sequence, however.

This inequality requires a practical number to have its factorization weighted, in a sense, towards smaller prime factors. This is a characteristic we might expect to be exhibited by the records of $n'$ as well, since large exponents over small primes make the greatest contribution to the sum portion of $n'$. I've confirmed that the first $250$ records occur only when $n$ is a practical number. Can we determine if this is true in general? I don't see an obvious way to put these pieces together.

Edit: A131117 is the OEIS sequence for the records of $n'$ and has a link to the $250$ terms I tested.

• For a moment I thought they might be products of primorials (oeis.org/A025487) which is a tighter restriction, but 640 is the first one that is not. I'll have to keep thinking about it. – half-integer fan Dec 2 '13 at 2:38
• @half-integerfan A possible tighter restriction is $p_i\leq 2p_1^{a_1}p_2^{a_2}...p_{i-1}^{a_{i-1}}$ for every $i\in[1,\omega(n)]$. It holds for at least the first $250$ terms. – Jaycob Coleman Dec 2 '13 at 7:14

Before I get to the equations in depth, let me observe that writing out the prime factorizations of the arithmetic derivative records I see that they have very high powers of two and only a few additional factors. Since the high powers of two mean that for a practical number there are many primes that could be the second largest prime, informally it appears very likely that your conjecture is true.

The following is not rigorous and I will not formally justify some statements, but I believe it provides a framework that you could extend for a more complete proof.

Consider that every power of two will set a new record for the arithmetic derivative, since in $$N' = N \cdot \sum_{p|N} \dfrac{a_i}{p_i}$$ $p_1 = 2$ has both the largest possible value of $\dfrac1{p_i}$ and the largest possible $a_i$ in a given range. For $$N=2^e \implies N'=e \cdot 2^{e-1}$$ Thus any other record N in the range $2^e \lt N \lt 2^{e+1}$ must have $N' \gt e \cdot 2^{e-1}$ at a minimum.

For the rest of this answer we will consider a number $N=2^a \cdot m$ where $m$ is any number with only odd prime factors. To be in the desired range we must have $$a=e- \lfloor \log_2 m \rfloor$$ I will define $\delta_m = \lfloor \log_2 m \rfloor$ so we have $$a=e-\delta_m$$ Now we have $$N' = a \cdot 2^{a-1} \cdot m + 2^a \cdot m' = 2^{a-1} \left(am+2m'\right)$$ Since we want N to be a record, we have $$2^{a-1} \left(am+2m'\right) \gt e \cdot 2^{e-1}$$ $$2^{e-\delta_m-1} \left((e-\delta_m)m+2m'\right) \gt e \cdot 2^{e-1}$$ $$(e-\delta_m)m+2m' \gt e \cdot 2^{\delta_m}$$ $$em-\delta_mm+2m' \gt e \cdot 2^{\delta_m}$$ $$-\delta_mm+2m' \gt e (2^{\delta_m} - m)$$ Noting that $2^{\delta_m} \lt m \implies 2^{\delta_m} - m \lt 0$, $$e \gt \dfrac{-\delta_mm+2m'}{2^{\delta_m} - m}$$ $$e \gt \dfrac{\delta_mm-2m'}{m - 2^{\delta_m}}$$ and thus $$a \gt \dfrac{\delta_mm-2m'}{m - 2^{\delta_m}} - \delta_m$$ $$a \gt \dfrac{\delta_m 2^{\delta_m}-2m'}{m - 2^{\delta_m}}$$

We will now only consider when $m$ is a single odd prime, thus $m'=1$. I believe the following argument can be extended to cover prime powers and products of multiple primes, and that the single prime case is the most limiting, but I have not formally done so.

The condition for a practical number is $$m \le \sigma (2^a) + 1 = 2^{a+1}$$ $$\log_2 m \le {a+1}$$ $$\log_2 m \le \dfrac{\delta_m 2^{\delta_m}-2}{m - 2^{\delta_m}} + 1$$ Now we can observe that $m \le 2^{\delta_m + 1} - 1 \implies m-2^{\delta_m} \le 2^{\delta_m}-1$ and we can substitute in the inequality: $$\log_2 m \le \dfrac{\delta_m 2^{\delta_m}-2}{2^{\delta_m}-1} + 1$$ $$\log_2 m \le \dfrac{\delta_m (2^{\delta_m}-1) + \delta_m -2}{2^{\delta_m}-1} + 1$$ $$\log_2 m \le \delta_m + 1 + \dfrac{\delta_m -2}{2^{\delta_m}-1}$$ But note that $\delta_m + 1 = \lceil \log_2 m \rceil$, so $$\log_2 m \le \lceil \log_2 m \rceil + \dfrac{\delta_m -2}{2^{\delta_m}-1}$$ which is always true as long as the remaining fraction is non-negative. Since $2^{\delta_m}-1$ is obviously positive, all we need is $$\delta_m -2 \ge 0$$ which is true for all $m \ge 4$.

Treating $m = 3$ as a special case, we have (hopefully) shown that all record arithmetic derivatives of the form $2^a \cdot m$ with m a single odd prime are practical numbers.