277. Combinatorial problem. Find the number of ordered sets $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of $n$ natural numbers, in which

$$
1 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} ; \quad a_{i} \leqslant i \quad(i=1,2, \ldots, n)
$$
277. Denoting the number of such sets by $A_{n}$ and setting $A_{0}=1$, we get for $n \geqslant 1$:

$$
A_{n}=\sum_{k=1}^{n} A_{k-1} \cdot A_{n-k} \cdot{ }^{\star}
$$

If $f(x)=\sum_{n=0}^{\infty} A_{n} x^{n}$, then from formula (1) it follows that $f(x)=$ $=1+x \cdot\{f(x)\}^{2}, f(0)=1$, so that

$$
f(x)=\frac{1-\sqrt{1-4 x}}{2 x}
$$

Expanding the right-hand side of equation (2) into a power series and equating the coefficients of like powers of $x$, we get for $n \geqslant 0$:

$$
A_{n}=\frac{1}{n+1} C_{2 n}^{n}=\frac{(2 n!)}{n!(n+1)!}
$$

[M. Bömer, A. M. M., 76, 193 (February 1969).]