How can I express $3 \cdot 7 \cdot 11 \cdots (4n+3)$ in terms of factorial?

Question

This is the work I have done so far:

$\prod_{k=0}^n(4k+3) = \frac{(4n)!}{2^n(2n)!}\cdot\prod_{k=0}^n\frac{1}{4k+1}$.

I would really appreciate a clever trick how to reduce the latter product that involves $\frac{1}{4k+1}$ ;)

Answer 1

These are Quadruple factorial numbers, see OEIS A008545 for the formula. The sequence starts with $$ 1, 3, 21, 231, 3465, 65835, 1514205, 40883535, 1267389585, $$

$$ 44358635475, 1729986783525, \cdots $$