How many consecutive integers have consecutive divisors?

For each positive integer $n$, let $f(n)$ be the number of positive integers pairs $(a,b)$ such that $a+i$ divides $b+i$ for $i = 1,2,...,n$. What is the growth rate of $f(n)$?

• Every pair for $n$ is also a pair for $n-1$ and $(a+1,b+1)$ is too, so $f(n)$ is strictly decreasing assuming it is finite. – Ross Millikan Nov 1 '17 at 4:05
• As written, it is always infinite because $(a,a)$ is always a pair for every $n$ and every $a$, right? – Alex S Nov 1 '17 at 4:15
• @AlexS: even simpler than my answer. I think you should make it an answer. I would upvote it. – Ross Millikan Nov 1 '17 at 4:17

For each $n$ there are infinitely many pairs $(a,b)$ satisfying the requirement. We can simply take $a=1$. This requires $n$ lines of $$b+1 \equiv 0 \pmod 2 \\b+2 \equiv 0 \pmod 3\\b+3 \equiv 0 \pmod 4$$ and so on. The Chinese remainder theorem guarantees an infinite number of solutions.
For every positive integer $a$, $(a,a)$ is a pair for all $n$. So $f(n)$ is always infinite.