The following sequence has its 2nd, 6th, 10th... terms 2 more than the term preceding them; the 3rd, 7th, 11th... terms are twice the preceding term, the 4th, 8th, 12th... terms are 2 less than the preceding term, and finally, the 5th, 9th, 13th... terms are the result of dividing the preceding term by 2.

$$
6,8,16,14,7,9,18,16,8, \ldots
$$

With which sequence numbers does 1970 appear, and with which sequence numbers does 1971 appear?
The $(4k+1)$-th term of the sequence, denoted by $c$, is defined according to the formation rule as follows:

\[
\begin{array}{llll}
a_{4 k+1}=c, & a_{4 k+2}=c+2, & a_{4 k+3}=2 c+4, & a_{4 k+4}=2 c+2, \\
a_{4 k+5}=c+1, & a_{4 k+6}=c+3, & a_{4 k+7}=2 c+6, & a_{4 k+8}=2 c+4, \\
a_{4 k+9}=c+2, & \ldots & &
\end{array}
\]

We observe that by dividing our sequence into 4 subsequences based on the remainder when the index is divided by 4 (i.e., 1, 2, 3, or 0), each subsequence is an arithmetic sequence with differences of 1, 1, 2, and 2, respectively. The first terms of these subsequences are 6, 8, 16, and 14. Thus, the number 1970 appears as a term in all 4 subsequences, while 1971 appears only in the first two.

In the first subsequence, $a_{4 k+1}=6+k$, and $a_{4 k+1}=1970$ when $k=1964$, which gives $4 k+1=7857$. For 1971, it appears at index 7861.

Similarly, in the second subsequence, $4 k+2=4 \cdot 1962+2=7850$, and 1971 appears at index 7854.

In the third and fourth subsequences, $a_{4 k+3}=16+2 k$ and $a_{4 k}=14+2(k-1)$, respectively. From these, $k=977$ and $k=979$, and the term equal to 1970 has indices 3911 and 3916. Summarizing:

\[
\begin{gathered}
1970 \text { is the term with indices } 3911, 3916, 7850 \text {, and } 7857, \\
1971 \text { is the term with indices } 7854 \text { and } 7861.
\end{gathered}
\]

László Hannák (Miskolc, Földes F. Gymnasium, 3rd grade)