(Insert Name Here) Sequence

First consider $N\leq3$.

• For $N=1$, the definition gives $A=0<S=1$.
• For $N=2$, the only adjacent OR is $1\mathbin{\mathrm{OR}}2=3=S$.
• For $N=3$, the bitwise OR of any two distinct values is $3$, so $A=6=S$.

Thus, no valid permutation exists in these three cases.

Now suppose $N\geq4$. It is enough to print

$$P=(1,2,3,\ldots,N).$$

For every $i$, we have $i\mathbin{\mathrm{OR}}(i+1)\geq i+1$. Hence,

$$A\geq\sum_{i=1}^{N-1}(i+1)=S-1.$$

The first term is $1\mathbin{\mathrm{OR}}2=3$, which is one greater than the value $2$ used in this lower bound. Since the value $4$ exists, the term $3\mathbin{\mathrm{OR}}4=7$ is three greater than the corresponding lower-bound value . Therefore,

$$A\geq(S-1)+1+3>S.$$

The time complexity is $O(N)$, equal to the output size, and the additional space complexity is $O(1)$.

Solution written by GPT6