About $f(x y) = f(x-1) f(y-1)$

Question

Consider the naturals. $f(n)$ maps naturals to naturals. ( positive integers is meant with naturals )

Whenever $z$ is composite we use

$$ f(z) = f(x y) = \max f(x-1) f(y-1)$$

More precisely

$$f(z)={\max{f(x−1)f(y−1)}∣x,y>1,z=xy}$$

For instance $f(12) = \max [ f(1) f(5) , f(2) f(3) ]$.

We desire Condition 1 for all integer $ n $ :

$$f(n+1) > f(n)$$

It appears not so simple to me to find a solution for $f$ ... Condition 1 USUALLY breaks before $n$ reached 200.

As for the solutions - assuming they exist - how fast do they grow ?

What is the SMALLEST possible value for $f(1),f(2),f(3) $ ?

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Analogue questions for

$$ g(z) = g(x y) = \max g(x-2) g(y-2) $$

Answer 1

Take $x=y$ and then substitute $x$ with $x+ 1$.

We get

$$f(x^2 + 2x + 1) = f(x)^2$$ And $$f(x^2 + 2x+1) > f(x^2 + x) > f(x^2)$$

Hence $$f(x^2) << f(x)^2$$

So $f(x) - f(x-1)$ is decreasing on average towards $0$ so $f$ can not be strictly increasing over the naturals.

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Notice $t(xy) = t(x+1)t(y+1)$ does not suffer this problem.

I think $t$ grows like $ c x + C_2 ln(x) \sqrt x$.

Answer 2

The first question one has to ask himself is: what kind of degrees of freedom do we have in the construction of such a function $f$?

Let $P = \{1\}\cup\{\text{primes}\}$. Then $f(p)$ for $p\in P$ can only appear in relations as a factor of $f(\text{a bigger number})$, and $f(n)$ can be expressed as a product of values of $f(p)$ for $p\in P$. Therefore, any such $f$ is completely determined by its values on $P$, which can be chosen freely.

I think that this should allow you to prove that condition $1$ cannot be satisfied (but I don't have a formal argument right now).

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If the defining condition also has to be satisfied when $z$ is prime (and non zero, else it becomes really messy...), then $$f(p) = f(p-1)f(0)$$ and $$f(1) = f(0)^2,$$ so everything is determined by $f(0)$, and we have $f(n) \ge f(0)f(n-1)$, so that by induction condition $1$ is automatically satisfied. But I don't think we want to consider this, as it is trivial...