Difficulty with a Combinatorial Math problem I was doing this question from a national mathematical olympiad and although I couldn't solve it but I found something but don't know how to progress.
For a positive integer $N$, let $T(N)$ denote the number of arrangements of the
integers $1,2,\dots,N$ into a sequence $a_1,a_2\dots,a_n$ such that $a_i>a_{2i}$ for all $i$ with $1\le i<2i\le N$
and $a_i>a_{2i+1}$ for all $i$ with $1\le i<2i+1\le N$. For example, $T(3)$ is $2$, since the possible
arrangements are 321 and 312. If $K$ is the largest non-negative integer so that $2^K$ divides $T(2^n-1)$, show that $K=2^n-n-1$.
I was able to build this relationship. Is it correct? If yes how do I progress further? If not what is the explicit formula for $T(2^n-1)$?
$$T(2^n-1)=2^{2^{(n-2)}}\binom{2^n-2}{2^{n-1}-1}$$
but in the step-by-step solution, they gave this recurrent relation
$$T(2^n-1)=T(2^{n-1}-1)^2\binom{2^n-2}{2^{n-1}-1}$$

Your relationship is wrong; e.g. for $n = 2$ it suggests $T(3) = 4$, when as above the correct result is 2.
The recurrence relation is formed by considering each sequence $a_1, \dots, a_{2^n-1}$ as the array representation of a heap. From the tree representation, we see that the number of possible heaps of size $2^n-1$ is equal to the number of ways to divide the numbers $1, \dots, 2^n-2$ between the left and right sub-trees, times the number of ways to arrange the numbers within the sub-trees.
This directly gives the quoted recurrence
$$T(2^n-1)=T(2^{n-1}-1)^2\binom{2^n-2}{2^{n-1}-1}.$$
The result then follows by induction, noting that $2^{2^n-1}$ is the largest power of two to divide $(2^n)!$.
You don't need to find the explicit formula for $T$, but from the recurrence, you can see that it's
$$T(2^n-1)=\prod_{i=1}^{n}\binom{2^{i}-2}{2^{i-1}-1}^{2^{n-i}}.$$
