[ $\left.\begin{array}{c}\text { Induction (other) } \\ {\left[\begin{array}{l}\text { Irrational inequalities }\end{array}\right]}\end{array}\right]$

Given:

$$
a_{1}=1966, a_{\mathrm{k}}=\left[\sqrt{a_{1}+a_{2}+\cdots+a_{k-1}}\right]
$$

Find $a_{1966}$.
We will prove by induction that the initial segment of our sequence is $1966,44,44,45,45, .$. , where after the number 1966, each natural number greater than 43 appears exactly twice in a row, except for numbers of the form $74 \cdot 2^{k}$, which appear three times in a row (compare with the solution to problem 78594). Suppose we have already proved that the initial segment of our sequence has the required form $1966,44,44,45,45, . . n-1, n-1, n, n$. Let's find its next term. Take the smallest $k$ for which $74 \cdot 2^{k}>n$. The sum of the written numbers is $s=1966+2(44+45+. . n)+74+2 \cdot 74+. .2^{k-1} \cdot 74=(n-43)(n+44)+2^{k} \cdot 74+44^{2}=n^{2}+n+2^{k} \cdot 74$. If $n=2^{k} \cdot 74$, then $(n)^{2} \leq$ $s<(n+1)^{2}$, so the next term of the sequence is $n$. Otherwise, $n+1 \leq 2^{k} \cdot 74 \leq 2 n$, and $(n+1)^{2} \leq s<(n+2)^{2}$, so the next term of the sequence is $n+1$. The next terms are found similarly.

## Answer

1024