Permutation of the Year 2

Print the following permutation.

$$1,2026,3,2024,5,2022,\ldots,2025,2$$

Equivalently, for every integer $k$ ($0 \leq k \leq 1012$), print $2k+1$ and then $2026-2k$.

For every $k$, we have $(2k+1)+(2026-2k)=2027$. Also, for $0 \leq k < 1012$, we have $(2026-2k)+(2k+3)=2029$. Finally, $2+1=3$.

The numbers $2027$, $2029$, and $3$ are all prime, so every adjacent cyclic sum is prime. The odd numbers $1,3,\ldots,2025$ and the even numbers $2026,2024,\ldots,2$ are each used exactly once, so the printed sequence is a permutation of $(1,2,\cdots,2026)$.

Solution written by GPT5.5