probably, yeah

Assuming F(k, n) = sum(Xik * C(n + i)) for i from 1 to k where Xik is coefficient of C(n + i) in F(k, n)

F(n, k + 1) = sum(F(t, k) * C(n — t)) for t from 0 to n (Combine first k factors while summing)

= sum(Xik * sum(C(S + i) * C(n — S)) for S from 0 to n) for i from 1 to k (Just written by expanding)

= sum(Xik * (C(n + i + 1) — (sum(C(j) * C(n + j)) for j from 0 to i — 1)) for i from 1 to k (Negation comes as some of the terms were missing from product(using formula for K = 2)

So I am getting somewhat recursive way of getting the coefficients, maybe implementation can help us get few more F's to make some observation.

One silly implementation later:

K = 2
1 * C[n+1]
K = 3
1 * C[n+2] + -1 * C[n+1]
K = 4
1 * C[n+3] + -2 * C[n+2]
K = 5
1 * C[n+4] + -3 * C[n+3] + 1 * C[n+2]
K = 6
1 * C[n+5] + -4 * C[n+4] + 3 * C[n+3]
K = 7
1 * C[n+6] + -5 * C[n+5] + 6 * C[n+4] + -1 * C[n+3]
K = 8
1 * C[n+7] + -6 * C[n+6] + 10 * C[n+5] + -4 * C[n+4]
K = 9
1 * C[n+8] + -7 * C[n+7] + 15 * C[n+6] + -10 * C[n+5] + 1 * C[n+4]
K = 10
1 * C[n+9] + -8 * C[n+8] + 21 * C[n+7] + -20 * C[n+6] + 5 * C[n+5]
K = 11
1 * C[n+10] + -9 * C[n+9] + 28 * C[n+8] + -35 * C[n+7] + 15 * C[n+6] + -1 * C[n+5]
K = 12
1 * C[n+11] + -10 * C[n+10] + 36 * C[n+9] + -56 * C[n+8] + 35 * C[n+7] + -6 * C[n+6]
K = 13
1 * C[n+12] + -11 * C[n+11] + 45 * C[n+10] + -84 * C[n+9] + 70 * C[n+8] + -21 * C[n+7] + 1 * C[n+6]
K = 14
1 * C[n+13] + -12 * C[n+12] + 55 * C[n+11] + -120 * C[n+10] + 126 * C[n+9] + -56 * C[n+8] + 7 * C[n+7]
...

A pretty clear Pascal's triangle pattern, but... in each column, the numbers seem to be shifted down by double the normal amount! This could very well be related to the other formula for Catalan numbers:

$$$C_{n} = \binom{2n}{n} - \binom{2n}{n-1}$$$

You can use generating functions to brute-force this.

A simpler way is to use the first relation (which is often used to derive the generating function $$$f$$$) in the form: $$$x f(x)^2 - f(x) + 1 = 0$$$. This resembles a second order recurrence. You can now do things like this:

1. $$$f(x)^3 = f(x) \cdot f(x)^2 = f(x) \cdot (f(x) - 1) / x = (f(x)^2 - f(x)) / x = ((1 - x)f(x) + 1) / x^2$$$.
2. $$$f(x)^4 = (f(x)^2)^2 = (f(x)^2 - 2f(x) + 1)/x^2$$$ etc.

Basically, you note that the $$$k$$$-variable identity is the convolution of the generating function with itself $$$k$$$ times. To find a general formula, a more direct way is to just expand:

The RHS gives a linear function in the coefficients of $$$f(x)$$$ which are just the Catalan numbers.