An infinite number of boards are lined up, and the number $1$ is written on the first board.
Write the numbers on the board from the left, but the number written on the $n$th board can only be the sum of the two numbers written on the $1$st to $n-1$th boards. (It is allowed to draw two numbers from the same board and write the sum of the numbers.)
Let f(x) be the minimum position on the board where x can appear.
For example, $1,2,3,4,5$ are possible and $1,2,4,5$ are also possible and $3$th boards's number can only $3$ or $4$. From this, you can know $f(5)=4$.
I think it so hardly but I cannot think answer. I proved $f(mn) \leq f(m)+f(n)-1$. This is simply proved. If $(a_1 , a_2, ...,a_{f(n)}=n)$ and $(b_1 , b_2, ...,b_{f(m)}=m)$ are possible then $(a_1, a_2, ...,a_{f(n)}=n=nb_1, nb_2, ..., nb_{f(m)}=nm)$ are possible. So $f(nm) \leq f(n)+f(m)-1$.
I think If $n$ is prime, $f(mn) = f(m)+f(n)-1$.(Maybe be wrong) So I think we can find a function using the function values for all prime numbers.
Is my guess correct? How can I find the function $f$?