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A sequence of positive integers is defined by $a_1=a\geq 2$ and $$a_{n+1}=a_n+f(a_n)$$ for $n\geq 1$, where $f(k)$ is the largest proper divisor of $k$. For which positive integer $d$ is it true that no matter the value of $a$, there will be some term $a_n$ divisible by $d$?

For example, for $d=2$ this is true, because if $a_1$ is odd then $f(a_1)$ is also odd, and so $a_2=a_1+f(a_1)$ is even. For $d=3$, this is less clear. It might be possible that $a_i\equiv f(a_i)\equiv 1\pmod 3$ for $i$ odd and $a_i\equiv f(a_i)\equiv 2\pmod 3$ for $i$ even.

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Partial answer. (1). For $d=3$: If any $a_n$ is even the largest proper divisor of $a_n$ is $a_n/2$ and $a_{n+1}=a_n+a_n/2=3(a_n/2)$ is divisible by $3.$ If any $a_n$ is odd then $f(a_n)$ is odd so $a_{n+1}$ is even, so $a_{n+2}$ is divisible by $3.$

(2).Consider the cae $a_0=2:$ Then $a_1=3, a_2=4,a_3=6,a_4=9,a_5=12,...$ When $a_n=2^p3^q$ with $p>0$ we have $a_{n+j}=2^{p-j}3^{q+j}$ for $1\leq j< p,$ and $a_{n+p}=3^{q+p}.$ When $a_n=3^q$ with $q>0$ we have $a_{n+1}=2^{2}3^{q-1}$.

So $d$ must be of the form $2^p3^q$ with $p,q$ not both $0 $. I dk whether all such values are possible.