In how many ways one can arrange the numbers $0,1,2,\dots,n$ using each number exactly once so that no two adjacent numbers sum greater than $n+1$?

Question

Given the numbers $0,1,2,\dots,n$. In how many ways one can arrange the numbers using each number exactly once so that no two adjacent numbers sum greater than $n+1$?

I found the answers for small values of $n$. For $n=1$, the answer is $2$. For $n=2$, the answer is $6$. For $n=3$, the answer is $12$. But how to solve this problem for any $n$?

This is a generalization of this very recent Mathematics SE question from the Bangladesh Mathematical Olympiad and this other one with a different number. — Mark S., Jan 31 '22 at 14:27

score 1 · Accepted Answer · answered Jan 31 '22 at 16:51

1

See Number of Ways of Ordering Numbers.

The solution is $6^k$ for $n=2k$ and $2.6^k$ for $n=2k+1$.

answered Jan 31 '22 at 16:51

In how many ways one can arrange the numbers $0,1,2,\dots,n$ using each number exactly once so that no two adjacent numbers sum greater than $n+1$?

1 Answers1

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