For positive integers
![n,](/media/m/5/f/2/5f26ebab144fee216fbc733cb1fa2f2b.png)
the numbers
![f(n)](/media/m/d/3/e/d3e47283bffbbf24c97f0c6474d5a82d.png)
are defined inductively as follows:
![f(1) = 1,](/media/m/3/2/b/32b27b113680d523c0453c7e58adfa74.png)
and for every positive integer
![f(n+1)](/media/m/7/3/6/736ef77774d03102b4d56272205ce96c.png)
is the greatest integer
![m](/media/m/1/3/6/1361d4850444c055a8a322281f279b39.png)
such that there is an arithmetic progression of positive integers
![a_1 < a_2 < \ldots < a_m = n](/media/m/1/0/8/1086e2711ac0b1c4c09447553fc79eb8.png)
for which
Prove that there are positive integers
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
and
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
such that
![f(an+b) = n+2](/media/m/0/1/e/01ed449d41a7ea49ffac8f8a126da5a3.png)
for every positive integer
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For positive integers $n,$ the numbers $f(n)$ are defined inductively as follows: $f(1) = 1,$ and for every positive integer $n,$ $f(n+1)$ is the greatest integer $m$ such that there is an arithmetic progression of positive integers $a_1 < a_2 < \ldots < a_m = n$ for which
$$f(a_1) = f(a_2) = \ldots = f(a_m).$$
Prove that there are positive integers $a$ and $b$ such that $f(an+b) = n+2$ for every positive integer $n.$