Let $$f(m)=(2n+1)((2n+1)^2-1^2)((2n+1)^2-3^2)\ldots((2n+1)^2-(2m-3)^2)/(2m-1)!$$
for some Positive integers $n,m$
we have to find the coefficients of $t^{1-k}$ quotient on polynomial Division of
$$nf(n+1)t^{n-1}+(n-1)f(n)t^{n-2}+(n-2)f(n-1)t^{n-3}+(n-3)f(n-2)t^{n-4}+\ldots$$
by the polynomial
$$f(n+1)t^{n}+f(n)t^{n-1}+f(n-1)t^{n-2}+f(n-2)t^{n-3}+f(n-3)t^{n-4}+\ldots$$
with respect to $t$
I am curious about this since when I attempted to do this I saw a Pattern for
k=2 coefficient=n
k=3 coefficient=(2n+1)/4
k=4 coefficient=3(2n+1)/16
k=5 coefficient=5(2n+1)/32
k=6 coefficient=35(2n+1)/256
k=7 coefficient=63(2n+1)/512
k=8 coefficient=231(2n+1)/2048
k=9 coefficient=429(2n+1)/4096
k=10 coefficient=6435(2n+1)/65536
I primarily noticed the pattern of power of two's in the denominator and attempted to fix them to form a more uniform format, The pattern of numerators was seen by me to follow a general form, which made me conjecture that this coefficient for some $k=p$ was $\binom{2p-5}{p-2}(2n+1)/2^{2p-4}$ except for p=2 where the binomial is invalid and we have it as $n$.
OEIS appears to confirm the sequence's conjectured general term.
I want to ask the correctness (if correct then proof) of this conjecture or how do we get the required coefficient?