Asymptotics for $f(n+1) = (a n + b) f(n) + c \sum_{i=0}^{n} f(i) f(n-i)$?

Question

I was wondering about

$$f(0) = 1$$

$$f(n+1) = (a n + b) f(n) + c \sum_{i=0}^{n} f(i) f(n-i)$$

for given positive integers $a,b,c$.

How fast does this grow ?

Such equations occur often and are somewhere between the recurrence equation of factorial and of the Catalan numbers.

In fact I can get a closed form for them. But that might not be the easiest way to get a growth rate.

Getting the closed form directly relates to why this occurs; it is related to combinatorics, calculus and differential equations.

Just like we get the generating taylor series for the Catalan numbers we get a similar equation here :

$$c(x) = 1 + c x c(x)^2 + a x c'(x) + b x c(x)$$

For instance look at this calculus problem and related comment I found here on MSE :

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An integral involving Airy functions $\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx$

https://math.stackexchange.com/a/525096/39261

$$\mathcal{K}(3n)=\frac{\pi^2}{6\cdot64^n}a_{2n},$$ where $a_n$ is the sequence defined recursively as follows: $$a_0=1,\ \ a_{n+1}=(6\,n+4)\,a_n+\sum\limits_{i=0}^n a_i\,a_{n-i}$$

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Here $a=6,b=4,c=1$

and we get the generating taylor series $c(x)$ :

$$c(x) = 1 + x c(x)^2 + 6 x c'(x) + 4 x c(x)$$

We can solve this differential equation and get "this monster" :

$$c(x) = -(6 (k_1 (-2/3 e^{-(2 x)/3} x^{7/6} U(23/24, 13/6, (2 x)/3) - 23/36 e^{-(2 x)/3} x^{7/6} U(47/24, 19/6, (2 x)/3) + 7/6 e^{-(2 x)/3} x^{1/6} U(23/24, 13/6, (2 x)/3)) - 1/6 e^(-(2 x)/3) x^{1/6} (4 x L_{-47/24}^{13/6}((2 x)/3) + (4 x - 7) L_{-23/24}^{7/6}((2 x)/3))))/(-k_1 e^{-(2 x)/3} x^{7/6} U(23/24, 13/6, (2 x)/3) - e^{-(2 x)/3} x^{7/6} L_{-23/24}^{7/6}((2 x)/3))$$

where $L$ is the associated Laguerre polynomial and $U$ is the confluent hypergeo function of the second kind.

But it seems nontrivial to me how fast this function (the taylor coefficients of $c(x)$ ) grows ? What are good asymptotics ?

Do we need to resort to general techniques or is this specific case open for a specific strategy ?

Can we get around needing to solve the differential equation for the generating function ? Since it seems not so easy to me ?

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so far the main question.

I couldnt help wondering about fractional derivatives. And ramanujans master theorem.

Since we seem to have

An integral involving Airy functions $\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx$

$K(3n)$ is given by the n'th coefficient of $c(x)$.

Maybe it is true that

$K(s) =$ the s'th coefficient of $c(x)$ in the sense of fractional derivatives.

This is just a comment. I am aware of different ideas of fractional calculus ...