While studying the Bessel's differential equation of order zero:
$$x^2y'' + xy' + x^2y = 0$$
I came across a recurrence formula that caught my eye:
$$(2)^2(4)^2(6)^2...(2n)^2 = 2^{2n}(n!)^2$$
I am trying just for fun to find an analogous recurrence for multiples of an odd number.
I am particularly trying it for multiples of 3:
$$(3)^2(6)^2(9)^2...(3n)^2$$
EDIT I meant to the power of three, my apologies.
$$(3)^3(6)^3(9)^3...(3n)^3$$
But formulas like $3^{2n+1}(n!)^3$ or $3^{2n+1}((n+1)!)^3$ don't work (note I am trying starting from $n=0$).
$3^26^29^2\cdots(3n)^2$ is the square of $A=(3)(6)(9)\cdots(3n)$. But $$A=(3\times1)(3\times2)(3\times3)\cdots(3\times n)=3^n(1)(2)(3)\cdots n =3^n n!.$$ So $A^2=3^{2n}n!^2$.