Arithmetic Derivatives (once again)

Question

anonymous · Answer

@Kainui I think we can calculate it.

anonymous · Answer

Fundamental theorem of arithmetic: $$\Large
n = \prod_{p|n}p^k
$$We will let: $$\Large
\eta= \prod_{p|n}p^{k-1} \
$$

anonymous · Answer

Finally we will let $$
\Large \lambda = \frac{n}{\eta}
$$

anonymous · Answer

For example's sake:$$
n =4804625 = 5^3\cdot 7\cdot 17^2\cdot 19
$$In this case: $$
\eta = 5^{3-1}\cdot 7^{1-0}\cdot 17^{2-1}\cdot 19^{1-1} = 5^2\cdot 17
$$Where as: $$
\lambda =5\times 7\times 17\times 19
$$

anonymous · Answer

Also, we can gives the primes an index $i$, such that the $i$th prime $p_i$ corresponds to its exponent $k_i$.

anonymous · Answer

The power rule for the arithmetic derivative gives us:$$
n' = \sum_{i} k_i\frac{n}{p_i}
$$

anonymous · Answer

Since $$
n = \lambda \eta
$$We can say: $$
n' =n\sum_i\frac{k_i}{p_i} = \eta\left(\sum_i\frac{k_i\lambda}{p_i}\right) = \eta u
$$I'm going to say that $$
u=\left(\sum_i\frac{k_i\lambda}{p_i}\right)
$$So that everything is accounted for.

anonymous · Answer

Remember that: $$
\eta \leq \gcd(n,n')\leq n
$$

anonymous · Answer

Now here is what I think can make this doable.  We don't need to worry about primes such that: $$\sqrt{5\times 10^{15}}\leq p\leq 5\times 10^{15}
$$

anonymous · Answer

The reason I think this is because for the prime to affect the $\gcd(n,n')$, it will either need to be to the power of $2$, or it will need to divide $u$ suth that $u$ increases its power.

anonymous · Answer

We'll have to calculate each $u$ anyway, so when we do we can just test whether it is a prime or not.

anonymous · Answer

$$
5\times 10^{7.5}
$$Is much less to deal with.

kainui · Answer

Ok this seems pretty reasonable, I like this. I don't really have time to play around with this now, but during the day I thought up some possible things to explore along with this that might cut our time down.