How can one change this recurrence relation to be in terms of the first value of a?

Question

How does one put the recurrence relation:$$a_{2k}=\frac{(-1)^k}{4kl+2k(k+1)}a_{2k-2}$$ Where $l$ is any non-negative integer. In terms of $a_0$ so that:$$a_{2k}=\frac{(-1)^k}{f(k)}a_0$$

Answer 1

As you have $$ a_{2k}=\frac{(-1)^ka_{2k-2}}{2k(2l+k+1)} $$ you also get $$ (-1)^{k(k+1)/2}2^k\Gamma(k+1)\Gamma(2l+k+2)a_{2k} \\ =(-1)^{k(k-1)/2}2^{k-1}\Gamma(k)\Gamma(2l+k+1)a_{2k-2} $$ which gives you a constant, and thus is equal to the term for $k=1$, $$ …=\Gamma(2l+1)a_{0} $$