What function gives this sequence? Recently, I have been trying to find what function $f(n)$, for natural $n$, gives the following sequence:
$$1, 5, 105, 4725, 363825, 42567525, \cdots $$
It turns out that $\frac{d^{2n}}{dy^{2n}} (1-y^2)^{-5/2}|_{y=0}$, generates it. However, is there any function $f(n)$ that gives the same?
So far, it has been kind of hard to spot a pattern:
$$1, 5, \underbrace{3 \cdot 5 \cdot 7}_{105}, \underbrace{3^3 \cdot 5^2 \cdot 7}_{4725}, \underbrace{3^3 \cdot 5^2 \cdot 7^2 \cdot 11}_{363825}, \underbrace{3^5 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13}_{42567525}$$
Any suggestions would be very much appreciated! Thanks
 A: As mentioned by others, this is an example of binomial series.
$$ (1-x^2)^{-X/2} = 1 + X\frac{x^2}{2!} + 3!! X(2+X)\frac{x^4}{4!} + \\
5!! X(2+X)(4+X)\frac{x^6}{6!} + \cdots.  $$
This implies that the exponential power series
coefficient is
$$ (2n\!-\!1)!! \prod_{k=0}^{n-1}
 (2k\!+\!X)= (2n\!-\!1)!!
\frac{(X\!+\!2n\!-\!2)!!} {(X\!-\!2)!!}. $$
where the $\,(...)!!\,$ denotes a
double factorial. When $\,X=5\,$
OEIS sequence A051577 is the double factorial quotient.
A: as mentioned in the comments
$(1-y^2)^{-5/2}= \sum \limits_{n=0}^\infty \binom{-5/2}{n}(-1)^n y^{2n}$
differentiating $2n$ times the general form would look like $\binom{-5/2}{n}(-1)^n (2n)!$
Edit
I tried the first $5$ values with Maple, they match your sequence.
you can also simplify the generalized binomial coefficient to get:
$f(n) = \frac{1}{3} \cdot \frac{1}{2^{2n+1}} \cdot \frac{(2n)! (2n+3)!}{n!(n+1)!}$ for $n=0,1,...$
Edit #2
Also note how the prime factorizations of the numbers in your sequence contain no powers of $2$, that is because of the denominator $2^{2n+1}\cdot n! \cdot (n+1)!$
A: hint
Observe that
$$(1-y^2)^\frac{-5}{2}=(1+y)^\frac{-5}{2}(1-y)^\frac{-5}{2}$$
$$=f(y).g(y)$$
and use Leibnitz formula which gives the $ n^{\text{th}} $ derivative of a product.
$$(f.g)^{(2n)}(y)=$$
$$\sum_{k=0}^{2n}\binom{n}{k}f^{(k)}(y)g^{(2n-k)}(y)$$
