Infinite series with binomial summation If I have an infinite series with first term two and for any other term:
$$a_n=\sum_{k=1}^{n-1}\binom{4n-2}{4k-1}a_ka_{n-k}$$
What is the general term formula?
 A: Let $y=\sum_{n=1}^{\infty} \frac{a_n x^n}{(4n-1)!}$. Then by applying $\sum_{n=2}^{\infty}$ to the recurrence, we obtain
$$
\sum_{n=2}^{\infty} \frac{a_n x^n}{(4n-1)!}(4n-1)=\sum_{n=2}^{\infty}\sum_{k=1}^{n-1} \frac{a_k a_{n-k} x^k x^{n-k}}{(4k-1)!(4(n-k)-1)!}.
$$
The above can be written as
$$
4xy'-y - \frac{a_1 x}2 = y^2.
$$
Applying $a_1=2$, we obtain
$$
4xy'=x+y+y^2.
$$
This is Riccati equation according to Wolfram Alpha. 
Following the steps of solving Ricatti equation, the substitution $\frac y{4x}=-\frac{u'}u$ yields 
$$
u''+\frac3{4x} u'+\frac1{16x}u=0.
$$
According to this input in Wolfram Alpha, we obtain the general solution
$$
u=C_1 x^{\frac18}J_{-\frac14} (\frac{\sqrt x}2) + C_2 x^{\frac18} J_{\frac14}(\frac{\sqrt x}2).
$$
This output of Wolfram Alpha can easily be justified by suitable substitutions. 
Due to our requirement that $\frac y{4x}$ needs to be analytic at $0$, the second part should not appear. Thus, 
$$
u=C_1 x^{\frac18}J_{-\frac14} (\frac{\sqrt x}2).
$$
Hence, we recover $y$.
$$
y=4x\left( - \frac{ \left( x^{\frac18} J_{-\frac14} ( \frac{\sqrt x}2)\right)'}{x^{\frac18}J_{-\frac14}(\frac{\sqrt x}2)}\right).
$$
Since $J_v'(x)=\frac12 (J_{v-1} (x)- J_{v+1}(x))$, we obtain 
$$
y=\frac{- \frac12 x^{\frac18}J_{-\frac14}(\frac{\sqrt x}2)- \frac12 x^{\frac58}J_{-\frac54}(\frac{\sqrt x}2)+\frac12 x^{\frac58}J_{\frac34}(\frac{\sqrt x}2)}{ x^{\frac18}J_{-\frac14}(\frac{\sqrt x}2)}.
$$
Then the power series expansion of the above solution will yield an expression for $a_n$. 
