Storming Arasaka

Let

$n=\prod_{i=1}^{r} p_i^{a_i}.$

Define:

• $\omega(n)=r$, the number of distinct prime factors.
• $\Omega(n)=a_1+a_2+\cdots+a_r$, the number of prime factors counted with multiplicity.

The answer is

$\boxed{\Omega(n)+\omega(n)-1}.$

Yeah, the whole problem collapses to counting prime factors. Very Arasaka: huge security system, password is basically factor(n).

Why We Need At Least This Many Layers

The first condition means layers respect strict divisibility. If $y$ is a nontrivial proper divisor of $x$, then $y$ must be in an earlier layer than $x$.

Now look at prime divisors of $n$.

Two different primes have gcd $1$, so they cannot be in the same layer. Also, a prime $p$ cannot share a layer with any composite divisor divisible by $p$, because then $p$ would be a proper divisor of that composite number and should be earlier. If the composite divisor is not divisible by $p$, their gcd is $1$ anyway.

So every prime divisor of $n$ sits alone in its own layer. Therefore the latest prime appears in layer at least $\omega(n)$.

Pick that latest prime, call it $p$. Starting from $p$, we can build a chain of divisors ending at $n$ by multiplying in one prime factor at a time:

$p \mid p\cdot q_1 \mid p\cdot q_1\cdot q_2 \mid \cdots \mid n.$

This chain has exactly $\Omega(n)$ divisors, because $n$ contains $\Omega(n)$ prime factors total, and the first divisor already contributes one of them.

Since every step is strict divisibility, each next divisor must be in a later layer. The first one, $p$, is already at layer at least $\omega(n)$, so the final divisor $n$ must be at layer at least

$\omega(n)+\Omega(n)-1.$

So no solution can use fewer than that many layers.

Why This Many Layers Are Enough

Now we show a construction with exactly $\Omega(n)+\omega(n)-1$ layers.

Order the distinct primes however you want:

$p_1,p_2,\ldots,p_r.$

Put each prime alone:

$L_i=\{p_i\}\quad\text{for }1\le i\le r.$

For every composite divisor $d$ of $n$, put it into layer

$\Omega(d)+r-1.$

This uses layers from $r+1$ up to $\Omega(n)+r-1$, so the total number of layers is exactly

$\Omega(n)+\omega(n)-1.$

The divisibility condition is fine:

• If a proper divisor $y$ of composite $d$ is prime, it is in one of the first $r$ layers.
• If $y$ is composite, then $\Omega(y)<\Omega(d)$, so its assigned layer is smaller.

Now we only need the gcd-chain condition inside each layer.

For layers after the first $r$, all divisors in the layer have the same value of $\Omega(d)=s$, where $s\ge 2$.

Group those divisors by their smallest prime factor. Inside one group, every number shares that smallest prime, so the group can be ordered in literally any way.

To connect two consecutive nonempty groups, suppose their smallest primes are $p_i$ and $p_j$ with $i<j$. Since $s\ge 2$, there exists some divisor in the $p_i$-group that is also divisible by $p_j$: start with $p_i p_j$ and fill the remaining $s-2$ prime factors using whatever exponents are available from $n$. This is possible because the layer itself exists at size $s$.

Put such a divisor at the end of the $p_i$-group. The next group consists of numbers divisible by $p_j$, so the transition has gcd greater than $1$.

Thus every layer can be chained properly. Lower bound met, construction valid, answer proven.

Algorithm

For each $n$:

1. Factorize it.
2. Count total exponents: $\Omega(n)$.
3. Count distinct primes: $\omega(n)$.
4. Print $\Omega(n)+\omega(n)-1$.

Since $n\le 10^6$, a smallest-prime-factor sieve is more than enough.