# A recursively defined sequence

My question:

How does one determine the dependence upon $n$ of the sequence $(c_n)$ uniquely (well-)defined by $$c_0 = 1~~,$$ $$c_1 = \frac{1}{2}~~,$$ $$\sum_{k=0}^m(m-k+1)c_kc_{m-k+1} = 0 ~~~~\text{ for }~~~~m\ge1 ~~.$$ This sequence happens to be the Taylor series coefficients of a fairly common function, and so in this case the answer itself is already known and isn't the hard part (at anyone's request, I can add that here as well). The key is arriving at the solution using the above information, potentially by first constructing a more straightforward recurrence relation.

I arrived at this formula by using $m$-times differentiation on a function which may be represented by a first order nonlinear ODE, in order to generate relationships between the Taylor coefficients. Usually this method is used when the ODE is linear with polynomial coefficients in the independent variable, leading to linear recurrence relations with a fixed number of terms. However, I'd like to see if, by identifying or creating methods to solve implicitly defined recurrence relations like the above, some nonlinear ODEs may be tackled this way.

Apologies if something is grossly anomalous or misstated.

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I doubt there is an approach simpler than to encode $(c_n)_{n\geqslant0}$ as the coefficients of the series expansion of a function $u(x)=\sum\limits_{n\geqslant0}c_nx^n$ such that $u'(x)u(x)=c_1$ and $u(0)=c_0\gt0$, and to deduce from the recursion relations that $u(x)=\sqrt{c_0^2+2c_1x}$. –  Did Jul 1 '12 at 10:42
That does exactly recover the original function I used, $\sqrt{1+x}$. Of course the goal then, if possible, would be to recover these coefficients, ${1/2 \choose n}$, without appealing to the function or the original ODE itself, given that in other cases the relationship would not be so clear. –  Eugene Shvarts Jul 1 '12 at 10:50
There is a whole field devoted to the links between sequences and functions encoding them. In more complicated cases than the one in your post, the idea is not to avoid using generating functions, on the contrary. See generatingfunctionology, for examples of some rather sophisticated techniques. –  Did Jul 1 '12 at 11:15
That's excellent -- thank you for the link. Will be checking this out! –  Eugene Shvarts Jul 2 '12 at 5:26