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Moritz Firsching
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Answer: No this is not true.Answer: No this is not true.

At least if I am not mistaken, thenFor $(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$ is$m=5$, a counter example foris: $m=5$$(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$. In this case we have: $$\begin{align} U(1) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 4 & 4 & 3 & 3 & 2 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right) \\ U(2) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 4 & 4 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 3 & 2 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array}\right) \\ U(3) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 4 & 4 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 1 \end{array}\right) \\ U(4) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 4 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 4 & 5 & 4 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 3 & 3 & 2 \end{array}\right) \\ U(5) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 5 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 5 & 4 & 4 \\ 5 & 5 & 4 & 5 & 4 & 5 & 4 & 4 & 3 & 3 & 3 & 3 & 5 & 5 & 4 \end{array}\right) \end{align}$$

Each $U(k)$ has only $14$ entries $\geq k$.

To elaborate a bit:

  • For $m\leq4$ the statement in your question is true, as can be checked by complete enumeration of all cases.
  • For $m=5$ if you mod out cyclic permutations there are precisely 20 counter examples: $$\{(1, 1, 1, 1, 0, 2, 0, 2, 0, 0, 2, 0, 2, 1, 2), (1, 1, 1, 1, 0, 2, 0, 2, 0, 0, 0, 2, 2, 1, 2), (1, 1, 1, 1, 0, 2, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 1, 2, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 2, 1, 0, 2, 0, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 2, 1, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2, 1, 2, 0, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2, 1, 0, 2, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2, 0, 2, 1, 1, 2), (1, 1, 1, 0, 0, 2, 0, 0, 2, 2, 0, 2, 1, 1, 2), (1, 1, 0, 1, 2, 2, 1, 2, 2, 2, 0, 0, 1, 0, 0), (1, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 1, 0, 0, 0), (1, 0, 1, 0, 1, 2, 1, 0, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 0, 1, 0, 1, 2, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 0, 1, 2, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 0, 2, 1, 0, 0), (1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 2, 1, 2, 2, 2, 0, 1, 2, 0, 1, 0, 0), (1, 0, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 1, 0, 0)\}$$
  • I would expect that there counter examples for all $m\geq5$.

Answer: No this is not true.

At least if I am not mistaken, then $(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$ is a counter example for $m=5$. In this case we have: $$\begin{align} U(1) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 4 & 4 & 3 & 3 & 2 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right) \\ U(2) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 4 & 4 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 3 & 2 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array}\right) \\ U(3) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 4 & 4 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 1 \end{array}\right) \\ U(4) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 4 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 4 & 5 & 4 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 3 & 3 & 2 \end{array}\right) \\ U(5) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 5 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 5 & 4 & 4 \\ 5 & 5 & 4 & 5 & 4 & 5 & 4 & 4 & 3 & 3 & 3 & 3 & 5 & 5 & 4 \end{array}\right) \end{align}$$

Each $U(k)$ has only $14$ entries $\geq k$.

Answer: No this is not true.

For $m=5$, a counter example is: $(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$. In this case we have: $$\begin{align} U(1) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 4 & 4 & 3 & 3 & 2 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right) \\ U(2) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 4 & 4 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 3 & 2 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array}\right) \\ U(3) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 4 & 4 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 1 \end{array}\right) \\ U(4) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 4 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 4 & 5 & 4 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 3 & 3 & 2 \end{array}\right) \\ U(5) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 5 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 5 & 4 & 4 \\ 5 & 5 & 4 & 5 & 4 & 5 & 4 & 4 & 3 & 3 & 3 & 3 & 5 & 5 & 4 \end{array}\right) \end{align}$$

Each $U(k)$ has only $14$ entries $\geq k$.

To elaborate a bit:

  • For $m\leq4$ the statement in your question is true, as can be checked by complete enumeration of all cases.
  • For $m=5$ if you mod out cyclic permutations there are precisely 20 counter examples: $$\{(1, 1, 1, 1, 0, 2, 0, 2, 0, 0, 2, 0, 2, 1, 2), (1, 1, 1, 1, 0, 2, 0, 2, 0, 0, 0, 2, 2, 1, 2), (1, 1, 1, 1, 0, 2, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 1, 2, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 2, 1, 0, 2, 0, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 2, 1, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2, 1, 2, 0, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2, 1, 0, 2, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2, 0, 2, 1, 1, 2), (1, 1, 1, 0, 0, 2, 0, 0, 2, 2, 0, 2, 1, 1, 2), (1, 1, 0, 1, 2, 2, 1, 2, 2, 2, 0, 0, 1, 0, 0), (1, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 1, 0, 0, 0), (1, 0, 1, 0, 1, 2, 1, 0, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 0, 1, 0, 1, 2, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 0, 1, 2, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 0, 2, 1, 0, 0), (1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 2, 1, 2, 2, 2, 0, 1, 2, 0, 1, 0, 0), (1, 0, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 1, 0, 0)\}$$
  • I would expect that there counter examples for all $m\geq5$.
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Moritz Firsching
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  • 63
  • 88

Answer: No this is not true.

At least if I am not mistaken, then $(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$ is a counter example for $m=5$. In this case we have: $$\begin{align} U(1) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 4 & 4 & 3 & 3 & 2 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right) \\ U(2) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 4 & 4 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 3 & 2 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array}\right) \\ U(3) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 4 & 4 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 1 \end{array}\right) \\ U(4) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 4 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 4 & 5 & 4 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 3 & 3 & 2 \end{array}\right) \\ U(5) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 5 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 5 & 4 & 4 \\ 5 & 5 & 4 & 5 & 4 & 5 & 4 & 4 & 3 & 3 & 3 & 3 & 5 & 5 & 4 \end{array}\right) \end{align}$$

Each $U(k)$ has only $14$ entries $\geq k$.