So in normal induction, we say that if $P(a)$ is true, and $P(n)\implies P(n+1)$, then $P(n)$ is true $\forall n\geq a$.
Then we have strong induction where you assume all preceeding values of the statement are true, or backwards induction where you show there are an infinite number of true statements and show that it holds backwards.
Here is a document that includes various forms of induction. I am however looking for a unique form of induction that is a mixture of these methods.
Assume that we know that $P(1)$ is true, and that $P(n)$ is true $\forall n\geq K$ where $K$ is some finite number. Now we only need to prove $P(n)$ for $1<n<K$.
My first thought was to "hop around" and show that $P(n)\land P(n+2) \implies P(n+1)$ and then have that fold in like an accordian when it hits the edges, however plugging this in to a truth table it seems to work in the situation where all are true, as well as when all are false, which is bad.
Does there exist any induction method that can take advantage of the fact that we are provably bounded on both sides?
Unfortunately, there is no such method in general. Maybe there are special cases in the literature for special problems.
For example, take an easy induction problem for $n$ large enough, like showing
$$2^n >n^3$$ for all $n\ge 0$. For $n=0, n=1$ it's clearly true. Now for $n=2$ through $n=9$ (where $2^9=512$ and $9^3=729$) it's false. But it's true for all $n\ge 10,$ and the obvious method of proof here is by induction.
There are many such problems, including serious research level problems where inbetween cases are just false, or if not impossible to prove, at least extremely hard to prove.