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For those of you unfamiliar with the concept of Sicherman dice, the idea is that you can have an alternate arrangement of numbers on the sides of a pair six-sided dice such that when the pair is rolled together, the probability curve for the total remains the same. Following the instructions in the justification section on the wiki page makes it pretty easy to find alternate configurations for larger n-sided dice.

My question is, is it possible to have a Sicherman trio of sorts, where three non-standard n-sided dice rolled together are equivalent to three standard n-sided diced rolled together. You can accomplish this trivially by taking Sicherman pairs and adding a standard die, but I'm not interested in that solution. I've tried with 8 and 10-sided dice, but there wasn't a non-trivial solution. Is there a better way than brute-force to find a n where a Sicherman Trio of n-sided dice is possible?

Dice sides must be positive integers.

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The solutions depend on the factorization of $x\cdot \frac{x^n-1}{x-1}$, where we are interested in factors that are $1$ when $x=1$, or two factors with same vaue as $x=1$ (or similar conincidences). – Hagen von Eitzen Nov 30 '12 at 19:00
up vote 3 down vote accepted

I think there are the following 8-face triples: $$ (1, 2, 2, 2, 3, 3, 3, 4), (1, 3, 3, 3, 5, 5, 5, 7), (1, 5, 5, 5, 9, 9, 9, 13)$$ $$(1, 2, 2, 3, 3, 4, 4, 5), (1, 3, 3, 5, 5, 7, 7, 9), (1, 2, 5, 5, 6, 6, 9, 10)$$ $$(1, 2, 3, 3, 4, 4, 5, 6), (1, 2, 2, 3,5, 6, 6, 7), (1, 3, 5, 5, 7, 7, 9, 11)$$

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Ah, I was staring right at it too. I understand your comment now. That makes it easy to find additional sets. Thanks! – EvilAmarant7x Nov 30 '12 at 19:30

To expand upon Hagen's comment: the generating function for three rolls of an n-sided dice is $\displaystyle R_{n,3}(x) = \left(\sum_{k=1}^nx^k\right)^3 = \frac{\left(x(x^n-1)\right)^3}{(x-1)^3}$. The factors of $x^n-1$ are known as the Cyclotomic Polynomials; more specifically, the relation is $\displaystyle x^n-1=\prod_{k|n}\Phi_k(x)$, where $\Phi_1(x) = x-1$, $\Phi_2(x) = x+1$, $\Phi_3(x)=x^2+x+1$, etc. For instance, for an eight-sided die $x^8-1$ factors as $(x-1)(x+1)(x^2+1)(x^4+1)$; this means that the generating function for three eight-sided dice (modulo the factors of $x$ that change your faces from starting at 1 to starting at 0) breaks down as $R_{8,3}(x) = (x+1)^3(x^2+1)^3(x^4+1)^3$. Note that $\Phi_2(1) = 2$, $\Phi_4(1) = 2$ and $\Phi_8(1) = 2$; this means that the product evaluates to $8^3=64$ at $x=1$. For a (sub)-generating function $g(x)$ to represent an eight-sided die is the same thing (as Hagen notes in his comment) as requiring $g(1)=8$; you also need all the coefficients positive (since it's hard to have -1 faces of a die in the number you want!), but because all the coefficients of your factors are positive that's trivial here. In this case this means that you can split your 9 factors into any three groups of three and come up with a valid solution; the original d8s correspond to the factorization as $\bigl((x+1)(x^2+1)(x^4+1)\bigr)\cdot\bigl((x+1)(x^2+1)(x^4+1)\bigr)\cdot\bigl((x+1)(x^2+1)(x^4+1)\bigr)$. Breaking them down as, e.g., $\bigl((x+1)(x+1)(x^2+1)\bigr)\cdot\bigl((x^2+1)(x^2+1)(x^4+1)\bigr)\cdot\bigl((x^4+1)(x^4+1)(x+1)\bigr)$ yields $\bigl(x^4+2x^3+2x^2+2x+1\bigr)\cdot\bigl(x^8+2x^6+2x^4+2x^2+1\bigr)\cdot\bigl(x^9+x^8+2x^5+2x^4+x+1\bigr)$, which corresponds to the solution of dice with faces $(5, 4, 4, 3, 3, 2, 2, 1)$, $(9, 7, 7, 5, 5, 3, 3, 1)$ and $(10,9,6,6,5,5,2,1)$ that Hagen mentions in his answer.

For the d10 case, the relevant polynomials are $\Phi_2(x) = x+1$, $\Phi_5(x) = x^4+x^3+x^2+x+1$, and $\Phi_{10}(x) = x^4-x^3+x^2-x+1$. Since $\Phi_2(1) = 2$, $\Phi_5(1) = 5$ and $\Phi_{10}(1)=1$ you'll need to have exactly one factor of $\Phi_2$ and one factor of $\Phi_5$ in every product; you can then try distributing the three $\Phi_{10}$ factors among your generating functions, but since you need to keep coefficients positive you can't put all three factors on one function (doing the multiplication with Wolfram Alpha gives $\Phi_2\Phi_5\Phi_{10}^3=1-x+2 x^2-2 x^3+3 x^4-x^5+2 x^6+x^8+x^9+2 x^{11}-x^{12}+3 x^{13}-2 x^{14}+2 x^{15}-x^{16}+x^{17}$ with many negative coefficients). You can split them (2, 1, 0) - but this corresponds to having one 'trivial' die rolled along with two Sicherman d10s.

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