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The original problem has been discussed here, which involves a necklace of $p$ beads each of which can be one of $a$ colors and shows that $a^p - a$ must be a multiple of $p$ by classifying the necklaces which are cyclic permutations of each other as indistinguishable and a necklace of length $p$ has $p$ cyclic permutations leading to $\frac{a^p - a}{p}$ distinct necklaces. Please see the original posting for details.

Follow-up question: Since there are $a!$ ways to permute the $a$ colors, is $\frac{a^p - a}{p}$ divisible by $a!$ ?

Clarification: We should assume $a < p$ if not use $a \bmod p$. This will lead to such necklaces where all $a$ colors used.

Vectorizer
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  • No. If you try to partition necklaces according to permuted colors, not all classes will have the same number of elements -- some necklaces have fewer recolorings, e.g. if the necklace has only one color, it has $a$ recolorings. – Wojowu Sep 15 '17 at 16:08
  • Not all colors are used in any necklace, so you don't get equivalence classes all of sizes $n!$. – Thomas Andrews Sep 15 '17 at 16:10
  • @ThomasAndrews: Thanks for the response, please see the clarification I added to the question – Vectorizer Sep 15 '17 at 17:02
  • No, some necklaces still only use, say, $2$ colors, no matter what $a$ is. So, for example, when you have $a=4$ and $p=5$, you could use $2$ colors in a necklace $RRBBB$. Then permuting the colors (and rotating the necklace) gives only $4\times 3\times 5=120$ arrangements, which is not divisible by $5\cdot 4!=240$. – Thomas Andrews Sep 15 '17 at 17:13

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Suppose $a=4\,,p=3\,$ $$\frac{4^3-4}{3×4×3×2}$$ $$\frac{64-4}{4×3×3×2}$$ $$\frac{60}{4×3^2×2}$$ $$\frac{5}{6}$$ Hence disproved

The thing is, the necklaces you got may have such combinations where 1 color might have repeated less than $p$ times, $$$$(as you only discarded the ones which had 1 colour repeated $p$ times specifically)$$$$multiplying by $a!$ doesn't count those permutations

neonpokharkar
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Your mistake is assuming that just because $a<p$ it means every color is used in every necklace.

That's not true. $a^p-a$ counts necklaces that use more than one color, and less than or equal to $a$ colors.

For example, if you take $p=5$, $a=4$, with colors Red, Blue, Green, Purple, then: "Red, Red, Blue, Blue, Blue" is counted in the $4^5-4$ necklaces.

Under permutation of colors and rotating the necklace, this necklace is equivalent to $5\cdot 4\cdot 3$ other necklaces.

But $5\cdot 4\cdot 3=60\neq p\cdot a!=120$. And we see that $\frac{4^5-4}{5}=204$ is not divisible by $a!=24$.

Of course, when $a\geq p$ you also get the problem - if you use $p$ distinct colors, you get that any rotation is also a color permutation. So there are just $a(a-1)\cdots(a-p+1)$ elements in the equivalence class of all necklaces using $p$ distinct colors.

We can prove combinatorially that $a(a-1)$ is a divisor of $a^p-a$.

Thomas Andrews
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