This all started for me with Sicherman dice. They are only a few decades old (first discovered, as far as I've been able to tell, in the late '70s). Sicherman (among others) considered 2d6 and wondered if there were any other way to label their faces with integers that preserves the familiar bilinear statistics resulting from rolling them. If we look at the possible rolls, considering the sum of the numbers showing, die A 1 2 3 4 5 6 die B +---+---+---+---+---+---+ 1 | 2 | 3 | 4 | 5 | 6 | 7 | +---+---+---+---+---+---+ 2 | 3 | 4 | 5 | 6 | 7 | 8 | +---+---+---+---+---+---+ 3 | 4 | 5 | 6 | 7 | 8 | 9 | +---+---+---+---+---+---+ 4 | 5 | 6 | 7 | 8 | 9 |10 | +---+---+---+---+---+---+ 5 | 6 | 7 | 8 | 9 |10 |11 | +---+---+---+---+---+---+ 6 | 7 | 8 | 9 |10 |11 |12 | +---+---+---+---+---+---+ we get a frequency histogram 2 * 3 ** 4 *** 5 **** 6 ***** 7 ****** 8 ***** 9 **** 10 *** 11 ** 12 * Sicherman wondered if there were any other way to label the dice faces with other integers that results in the same frequency histogram. It turns out there are an infinite number of ways of doing this, but if we eliminate changes of the "add 1 to each number on die A and subtract 1 from each number on die B" sort by requiring that each die's minimum number is 1, the answer is unique (up to swapping the dice and reordering each die's faces, of course): die A has 1 2 2 3 3 4 and die B has 1 3 4 5 6 8. This can easily be verified by simple exhaustive search; counting to 11^12=3138428376721 is impractical, but it is easy to cut the search space down in various ways (as a rudimentary example, fix one face of each die at 1 and eliminate 1 as a possible value for all other faces, and it shrinks to 10^10), and it doesn't take much of that before exhaustive search is within reach of even a modest desktop. As a (past) role-playing gamer, I am very aware that six is not the only possible number of sides for a die. I got curious about the analogous questions for other die sizes. My program was easily adapted to other die sizes, including sizes that are difficult to realize physically (such as d7), and I rapidly found there is a lot of structure hiding beneath this apparently simple question, and the answer above is intimately related to the ways the number 6 can be factored. (It turns out to be closely related to factoring polynomials; stick with me and we'll get there.) In particular, consider a d6 as a cross product of a d2 and a d3 (with suitable labelings). There are two ways to do this: 1---2---3 1---3---5 | | | | | | 4---5---6 2---4---6 There's no need to draw out the whole matrix, though; if we note the size and increment of the arithmetic progressions in question 1x3 2x3 1---2---3 1---3---5 3x2 | | | 1x2 | | | 4---5---6 2---4---6 (where NxM means a progression of M elements with difference N), the description of the progressions captures the labeling. To avoid having to constantly subtract and add 1, below, I'm going to switch to 0-based numbering, so that a d6 has numbers 0 through 5 and the histogram of two of them has numbers 0 through 10. This is strictly a convenience change; by adding and subtracting 1 at suitable times everything works just as well without it, just slightly more annoyingly. Sicherman's alternative labeling becomes 0 1 1 2 2 3 on one die and 0 2 3 4 5 7 on the other. Here's that diagram renumbered that way: 1x3 2x3 0---1---2 0---2---4 3x2 | | | 1x2 | | | 3---4---5 1---3---5 Now, let's make a small change: 1x3 2x3 0---1---2 0---2---4 1x2 | | | 3x2 | | | 1---2---3 3---5---7 All I've done here is swap the "3x2" with the "1x2", then recompute the numbers based on the new progressions - and Sicherman's alternative labeling (in its 0-origin form) has appeared as if out of nowhere. We can even do things a bit more radical: 1x2 2x3 0---1 0---2---4 1x2 | | 2x3 | | | 1---2 2---4---6 | | | 4---6---8 We now have a d4 and a d9 instead of 2d6, but, upon working out the frequencies, 0 2 2 4 4 4 6 6 8 0 0 2 2 4 4 4 6 6 8 1 1 3 3 5 5 5 7 7 9 1 1 3 3 5 5 5 7 7 9 2 2 4 4 6 6 6 8 8 10 we see the same histogram of sums reappear. We can also swap things around different ways, eg 1x2/3x2 on the d4 and 1x3/2x3 on the d9 1x2 1x3 0---1 0---1---2 3x2 | | 2x3 | | | 3---4 2---3---4 | | | 4---5---6 which leads to dice 0 1 3 4 and 0 1 2 2 3 4 4 5 6, or 3x2/3x2 and 1x3/1x3 3x2 1x3 0---3 0---1---2 3x2 | | 1x3 | | | 3---6 1---2---3 | | | 2---3---4 0 3 3 6 0 1 1 2 2 2 3 3 4 or even go with a d3 and a d12 (because 12 has three factors instead of two, the diagram grows a third dimension): 1x3 1x3 0---1---2 0---1---2 3x2 |\ |\ |\ 3-\-4-\-5 \ \ \ \ \ \ \ 3x2 \ 3---4---5 \| \| \| 6---7---8 0 1 2 0 1 2 3 3 4 4 5 5 6 7 8 There clearly is some underlying structure going on here. I spent a good deal of time investigating this from this point of view, computing (via exhaustive search of various sorts) alternative labelings for d12, d30, and eventually even d210 (210 being interesting because it is the smallest number with four distinct prime factors). The details are not important here, because once I started looking at it this way I found a point of view that links it directly to a pre-existing field of combinatorics. Consider the generating series for a traditional d6 2 3 4 5 6 [1] x + x + x + x + x + x The generating series for the sum of 2d6 is of course the square of this, or 2 3 4 5 6 7 8 9 10 11 12 [2] x + 2 x + 3 x + 4 x + 5 x + 6 x + 5 x + 4 x + 3 x + 2 x + x This can, naturally, be factored back into two copies of the original generating polynomial. But it can also be factored as 2 3 4 3 4 5 6 8 [3] (x + 2 x + 2 x + x ) (x + x + x + x + x + x ) which corresponds to Sicherman dice. Simplifying by dividing everything by x (which corresponds to subtracting 1 from all the dice labels), we can write the generating polynomial for a 0-orgin d6 as 2 3 4 5 [4] 1 + x + x + x + x + x This can be factored in two ways (among others) 2 4 [5] (1 + x) (1 + x + x ) 2 3 [6] (1 + x + x ) (1 + x ) which correspond to the two ways of diagramming shown above. This of course allows us to factor [2] (or, rather, its 0-origin form) as 2 3 2 4 [7] (1 + x) (1 + x + x ) (1 + x ) (1 + x + x ) and choosing to multiply out the left two terms and the right two terms gives us 2 3 2 3 4 5 7 [8] (1 + 2 x + 2 x + x ) (1 + x + x + x + x + x ) which corresponds exactly to the 0-origin Sicherman labeling. Thus, finding Sicherman-style labelings is isomorphic to finding factorizations of generating polynomials. As the Wikipedia page for "Sicherman dice" points out, the theory of cyclotomic polynomials is of use here; one factorization cited on that page amounts to writing a 0-origin d6 as the product of three factors: 2 2 [9] (1 + x) (1 + x + x ) (1 - x + x ) The last of these factors does not correspond to a die; viewed as a die, it is a d1 with one face labeled 0, one face labeled 2, and minus one face labeled 1. However, nothing says we have to use an irreducible factorization like that rather than the other factorizations (into non-irreducible polynomials) like [7]. It's not hard to see that this business of swapping axes around, as outlined above when generating a d4 and a d9 with the statistics of 2d6, is equivalent to writing generating polynomials as products of polynomials with coefficients all 1 or 0 and exponents forming arithmetic progressions. Incidentally, these can always be written as fractions; for example, 6 4 2 x - 1 [10] x + x + 1 = ------ 2 x - 1 The generating polynomial for an MxN axis is (X^(M*N)-1)/(X^M-1). When working with diagrams, I found that the ways of writing a single dN corresponded to the ways of building N up by multiplying its factors together. For example, a d12 can be diagrammed three different ways: 4x3 2x3 1x3 1---5---9 1---3---5 1---2---3 1x2 |\ |\ |\ 1x2 |\ |\ |\ 3x2 |\ |\ |\ 2-\-6-\10 \ 2-\-4-\-6 \ 4-\-5-\-6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x2 \ 3---7--11 6x2 \ 7---9--11 6x2 \ 7---8---9 \| \| \| \| \| \| \| \| \| 4---8--12 8--10--12 10--11--12 which correspond to the three orders of multiplying 2, 2, and 3 together to yield 12: 1->2->4->12 (2,2,3); 1->2->6->12 (2,3,2); and 1->3->6->12 (3,2,2). In terms of generating polynomials, this makes sense; the first of the above three, for example, corresponds to writing 12 2 4 12 x - 1 x - 1 x - 1 x - 1 [11] ------- = ------ * ------ * ------- 2 4 x - 1 x - 1 x - 1 x - 1 When dealing with diagrams, I noticed something. Working with d6s, I wondered why I couldn't make anything work with axes like 2x2 or 3x3 that don't appear in any of the ways of drawing the original dice. (Considering a d2 bearing 0 2 and a putative d18, it's easy to see that, quite aside from all other considerations, there's no way to fill in the other numbers on the d18 that works - the odd sums never work out.) From the generating polynomial point of view, this means that x^2 + 1 either doesn't divide the 0-origin analog of [2] - [2] divided by x^2 - or the quotient involves terms with negative coefficients, which make no sense as die labelings. (In this case, as it happens, it's the former; division leaves a remainder of 2x.) I formed two conjectures based on my work with diagrams: Conjecture A: any dice that "work" can be obtained by factoring the original dice and recombining them (as outlined above). Conjecture B: an axis labeled MxN is possible only when M*N divides the original die size (6, here). Conjecture A clearly implies conjecture B. In terms of generating polynomials, these amount to saying that the generating polynomial for the joint roll of the standard dice admits only certain factorizations - those based on (x^(M*N)-1)/(x^M-1) factors, as sketched above. I see no obvious reason this should be so. What little exhaustive search I've been able to do indicates that these conjectures are true, but exhaustive search becomes impractical long before numbers large enough to really hold interest here are reached. I did exhaustive search on 3d6 and found that the only alternative labelings consisted of a Sicherman labeling plus a standard d6. This fits with conjecture A; there is no way to factor 3d6 and reassemble the factors to form three six-siders that doesn't involve at least one standard d6. I did exhaustive search on some of the smaller prime sizes, such as d5 and d7, and found that they have no Sicherman-style alternative labelings. This also follows from cyclotomic polynomial theory, since (x^N-1)/(x-1) - the generating function for a dN - is irreducible whenever N is prime, so the product of two such cannot be factored any other way at all, never mind any other way that actually corresponds to dice. I'm going to go back to drawing diagrams, since polynomial factoring theory does not, as far as I know, have anything corresponding to the constraints we are imposing. These constraints basically mean nonnegative coefficients, not necessarily in the factors, but in the products resulting after multiplying them back into generating polynomials for (putative) dice. The next composite size is 8. The basic diagram is 1x2 0---1 2x2 |\ |\ 2-\-3 \ \ \ \ \ 4x2 \ 4---5 \| \| 6---7 which of course can be drawn any of various ways. This means our prime dice, to coin a term, are two each of 1x2, 2x2, and 4x2; omitting the "x2", which is always the same when working with the d8, and writing only the other numbers, we can get various ways of reassembling these prime dice into two d8s, or a d4 and a d16, or even a d2 and a d32. This is, however, the first size for which three dice do anything new; 3d8 can be factored and reassembled as (to name just one possibility) 1x2/1x2/1x2 + 2x2/2x2/2x2 + 4x2/4x2/4x2, or 0 1 1 1 2 2 2 3 - 0 2 2 2 4 4 4 6 - 0 4 4 4 8 8 8 12. (There are of course a lot of other ways of assembling three d8s out of those factors, many of which don't involve any standard d8s.) I'll be confining myself to considering only two base dice for most of the rest of this, though. The d9 has nothing new to offer. We can do 1x3/1x3 and 3x3/3x3, we can do d3/d27, and that's about it here. The d10 is mildly interesting, because its factors are unequal, but it doesn't really have anything to say we didn't get from the d6. The d12 begins to get interesting. 12 factors as 2x2x3, leading to diagrams for the base dice as shown above. If we restrict ourselves to plans that generate two 12-siders, we can diagram the resulting (Sichermanoid) dice as Ax3 Dx3 0---x---x 0---x---x Bx2 |\ |\ |\ Ex2 |\ |\ |\ x-\-x-\-x \ x-\-x-\-x \ \ \ \ \ \ \ \ \ \ \ \ \ Cx2 \ x---x---x Fx2 \ x---x---x \| \| \| \| \| \| x---x---x x---x---x I then did exhaustive search on A,B,C,D,E,F values taken from divisors of 12 (1,2,3,4,6). To cut down on repetitions, I required that if ABC is different from DEF, that ABC be lower; I also required that B<=C and E<=F (since swapping E and F, or B and C, does not change anything important). Twenty-three sets produce values that work, ie, that give the same frequency table as two traditional d12s; when looked at from the point of view of the labels on the dice, there are only eight distinct labelings - most of them correspond to more than one set of A-F values. Specifically: die 1 die 2 A B C D E F 0 1 1 2 2 2 3 3 3 4 4 5 0 3 4 6 7 8 9 10 11 13 14 17 1 1 2 4 3 6 0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 1 1 3 2 6 6 0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 1 1 3 4 2 6 0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 2 1 1 2 6 6 0 1 1 2 2 3 3 4 4 5 5 6 0 2 4 6 6 8 8 10 10 12 14 16 2 1 1 4 2 6 0 1 1 2 2 3 6 7 7 8 8 9 0 2 3 4 5 6 7 8 9 10 11 13 1 1 6 2 3 6 0 1 1 2 2 3 6 7 7 8 8 9 0 2 3 4 5 6 7 8 9 10 11 13 1 1 6 4 2 3 0 1 1 2 4 5 5 6 8 9 9 10 0 2 2 4 4 6 6 8 8 10 10 12 4 1 1 2 2 6 0 1 1 2 4 5 5 6 8 9 9 10 0 2 2 4 4 6 6 8 8 10 10 12 4 1 1 4 2 2 0 1 2 2 3 3 4 4 5 5 6 7 0 1 4 5 6 7 8 9 10 11 14 15 1 2 3 4 1 6 0 1 2 2 3 3 4 4 5 5 6 7 0 1 4 5 6 7 8 9 10 11 14 15 2 1 2 4 1 6 0 1 2 2 3 4 6 7 8 8 9 10 0 1 3 4 4 5 7 8 8 9 11 12 1 2 6 4 1 3 0 1 2 3 3 4 4 5 5 6 7 8 0 1 2 6 6 7 7 8 8 12 13 14 1 3 3 1 6 6 0 1 2 3 3 4 4 5 5 6 7 8 0 1 2 6 6 7 7 8 8 12 13 14 2 1 3 1 6 6 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 1 3 6 1 3 6 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 1 3 6 2 1 6 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 1 3 6 4 1 2 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 2 1 6 1 3 6 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 2 1 6 2 1 6 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 2 1 6 4 1 2 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 4 1 2 1 3 6 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 4 1 2 2 1 6 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 4 1 2 4 1 2 Looking at these, the first thing to note is that all these A/B/C/D/E/F sets satisfy the conjecture mentioned above. The next thing of interest is that each of them has exactly two 1s in it; this makes sense, in that another 1 would lead to too many 1s on the dice and thus too many 1s in the frequency table. The next thing of interest appears upon factoring the A-F values and looking at the merged set of factors (which we can do by multiplying them together and factoring the result): A B C D E F product factors 1 1 2 4 3 6 144 2 2 2 2 3 3 1 1 3 2 6 6 216 2 2 2 3 3 3 1 1 3 4 2 6 144 2 2 2 2 3 3 1 1 6 2 3 6 216 2 2 2 3 3 3 1 1 6 4 2 3 144 2 2 2 2 3 3 1 2 3 4 1 6 144 2 2 2 2 3 3 1 2 6 4 1 3 144 2 2 2 2 3 3 1 3 3 1 6 6 324 2 2 3 3 3 3 1 3 6 1 3 6 324 2 2 3 3 3 3 1 3 6 2 1 6 216 2 2 2 3 3 3 1 3 6 4 1 2 144 2 2 2 2 3 3 2 1 1 2 6 6 144 2 2 2 2 3 3 2 1 1 4 2 6 96 2 2 2 2 2 3 2 1 2 4 1 6 96 2 2 2 2 2 3 2 1 3 1 6 6 216 2 2 2 3 3 3 2 1 6 1 3 6 216 2 2 2 3 3 3 2 1 6 2 1 6 144 2 2 2 2 3 3 2 1 6 4 1 2 96 2 2 2 2 2 3 4 1 1 2 2 6 96 2 2 2 2 2 3 4 1 1 4 2 2 64 2 2 2 2 2 2 4 1 2 1 3 6 144 2 2 2 2 3 3 4 1 2 2 1 6 96 2 2 2 2 2 3 4 1 2 4 1 2 64 2 2 2 2 2 2 The first thing that appears is that the factor list has the same number of factors in every case. Conjecture A would explain this, but I see no other obvious explanation. That they aren't all the same is interesting; it may be related to conjecture A and the asymmetry of the various diagrams that yield standard d12s. d13, d14, and d15 have nothing new to offer. For the d16, using a 2x2x2x2 diagram, exhaustive search similar to that for 12 finds 10 die pairs: die 1 die 2 die 1 die 2 0 1 1 2 2 2 3 3 3 3 4 4 4 5 5 6 0 4 4 8 8 8 12 12 12 12 16 16 16 20 20 24 1 1 2 2 4 4 8 8 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 0 2 4 6 8 8 10 10 12 12 14 14 16 18 20 22 1 1 2 4 2 4 8 8 0 1 1 2 2 3 3 4 8 9 9 10 10 11 11 12 0 2 4 4 6 6 8 8 10 10 12 12 14 14 16 18 1 1 2 8 2 4 4 8 0 1 1 2 4 4 5 5 5 5 6 6 8 9 9 10 0 2 2 4 8 8 10 10 10 10 12 12 16 18 18 20 1 1 4 4 2 2 8 8 0 1 1 2 4 5 5 6 8 9 9 10 12 13 13 14 0 2 2 4 4 6 6 8 8 10 10 12 12 14 14 16 1 1 4 8 2 2 4 8 0 1 1 2 8 8 9 9 9 9 10 10 16 17 17 18 0 2 2 4 4 4 6 6 6 6 8 8 8 10 10 12 1 1 8 8 2 2 4 4 0 1 2 2 3 3 4 4 5 5 6 6 7 7 8 9 0 1 4 5 8 8 9 9 12 12 13 13 16 17 20 21 1 2 2 4 1 4 8 8 0 1 2 2 3 3 4 5 8 9 10 10 11 11 12 13 0 1 4 4 5 5 8 8 9 9 12 12 13 13 16 17 1 2 2 8 1 4 4 8 0 1 2 3 4 4 5 5 6 6 7 7 8 9 10 11 0 1 2 3 8 8 9 9 10 10 11 11 16 17 18 19 1 2 4 4 1 2 8 8 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 4 8 1 2 4 8 Another curious thing appears: the diagram step values always have exactly two each of 1, 2, 4, and 8, and each die labeling corresponds to exactly one set of numbers. Everything between 16 and 30 is either prime, has only two prime factors, or has at least two prime factors which are equal. None of these give us anything new, possibly excepting 24 (2*2*2*3). But 30 is the smallest number with three distinct prime factors, and it has some interest for us. Using a diagram like the one used above for d12, but larger A or D 0---x---x---x---x B |\ |\ |\ |\ |\ or x-\-x-\-x-\-x-\-x \ E |\ \|\ \|\ \|\ \|\ \ x-\-x-\-x-\-x-\-x \ \ \ \ \ \ \ \ \ \ \ \ \ C \ \ x---x---x---x---x or F \ \|\ \|\ \|\ \|\ \| \ x---x---x---x---x \| \| \| \| \| x---x---x---x---x Stock d30s can be generated by six A-B-C triples: 1-5-15, 1-10-5, 3-1-15, 6-1-3, 2-10-1, and 6-2-1. Interestingly, looking at the A values, I see 1, 1, 2, 3, 6, 6; the B values, 1, 1, 2, 5, 10, 10; the C values, 1, 1, 3, 5, 15, 15: in each case, 1, 1, X, Y, X*Y, X*Y, where X*Y is 30 divided by the size of the diagram in the corresponding dimension. I wonder if that is significant. I did a similar exhaustive search, finding 13 die pairs, corresponding to 69 distinct parameter sets (with de-duping rules similar, though not quite identical since all factors are different, to the ones used for the d12). Here are the die pairs and the parameter sets A-F that generate them: 0 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 7 7 7 8 8 9 - 0 5 6 10 11 12 15 16 17 18 20 21 22 23 24 25 26 27 28 29 31 32 33 34 37 38 39 43 44 49 1 1 3 6 5 15 1 1 3 6 10 5 1 2 1 6 5 15 1 2 1 6 10 5 0 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9 10 10 11 - 0 3 6 9 10 12 13 15 16 18 19 20 21 22 23 24 25 26 27 28 29 31 32 34 35 37 38 41 44 47 1 1 5 3 10 15 1 1 5 6 10 3 2 1 1 3 10 15 2 1 1 6 10 3 0 1 1 2 2 2 3 3 3 4 4 4 5 5 6 15 16 16 17 17 17 18 18 18 19 19 19 20 20 21 - 0 3 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32 34 37 1 1 15 3 5 15 1 1 15 3 10 5 1 1 15 6 5 3 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 - 0 2 4 6 8 10 12 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 33 35 37 39 41 43 1 5 1 2 10 15 1 5 1 6 2 15 3 1 1 2 10 15 3 1 1 6 2 15 0 1 1 2 2 3 3 4 4 5 10 11 11 12 12 13 13 14 14 15 20 21 21 22 22 23 23 24 24 25 - 0 2 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 33 1 10 1 2 5 15 1 10 1 2 10 5 1 10 1 6 2 5 0 1 1 2 2 3 6 7 7 8 8 9 12 13 13 14 14 15 18 19 19 20 20 21 24 25 25 26 26 27 - 0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 6 1 1 2 10 3 6 1 1 3 2 15 6 1 1 6 2 3 0 1 2 2 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9 10 10 11 11 12 13 - 0 1 6 7 10 11 12 13 16 17 18 19 20 21 22 23 24 25 26 27 28 29 32 33 34 35 38 39 44 45 1 2 5 6 10 1 2 1 3 6 10 1 2 2 1 6 10 1 0 1 2 2 3 3 4 4 4 5 5 6 6 7 8 15 16 17 17 18 18 19 19 19 20 20 21 21 22 23 - 0 1 5 6 6 7 10 11 11 12 12 13 16 17 17 18 18 19 22 23 23 24 24 25 28 29 29 30 34 35 1 2 15 6 5 1 0 1 2 2 3 4 4 5 6 6 7 8 8 9 10 15 16 17 17 18 19 19 20 21 21 22 23 23 24 25 - 0 1 3 4 6 7 9 10 10 11 12 13 13 14 16 17 19 20 20 21 22 23 23 24 26 27 29 30 32 33 2 1 15 3 10 1 0 1 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 16 17 - 0 1 2 6 7 8 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 33 34 35 39 40 41 1 5 3 6 1 15 3 1 3 6 1 15 3 2 1 6 1 15 0 1 2 3 3 4 4 5 6 7 10 11 12 13 13 14 14 15 16 17 20 21 22 23 23 24 24 25 26 27 - 0 1 2 5 6 6 7 7 8 11 12 12 13 13 14 17 18 18 19 19 20 23 24 24 25 25 26 29 30 31 1 10 3 6 1 5 0 1 2 3 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 16 17 18 19 - 0 1 2 3 4 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 35 36 37 38 39 1 5 5 1 10 15 2 5 1 1 10 15 3 1 5 1 10 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 1 5 15 1 5 15 1 5 15 1 10 5 1 5 15 2 10 1 1 5 15 3 1 15 1 5 15 6 1 3 1 5 15 6 2 1 1 10 5 1 5 15 1 10 5 1 10 5 1 10 5 2 10 1 1 10 5 3 1 15 1 10 5 6 1 3 1 10 5 6 2 1 2 10 1 1 5 15 2 10 1 1 10 5 2 10 1 2 10 1 2 10 1 3 1 15 2 10 1 6 1 3 2 10 1 6 2 1 3 1 15 1 5 15 3 1 15 1 10 5 3 1 15 2 10 1 3 1 15 3 1 15 3 1 15 6 1 3 3 1 15 6 2 1 6 1 3 1 5 15 6 1 3 1 10 5 6 1 3 2 10 1 6 1 3 3 1 15 6 1 3 6 1 3 6 1 3 6 2 1 6 2 1 1 5 15 6 2 1 1 10 5 6 2 1 2 10 1 6 2 1 3 1 15 6 2 1 6 1 3 6 2 1 6 2 1 (Incidentally, note these also fit conjecture B above; I haven't done the tedious work to check conjecture A.) Based on the results for the d12, I took each set A-F, multiplied them, and factored the product. Using . for 2, = for 3, and @ for 5, to produce more visually obvious patterns, here they are, with each line annotated with distribution of the factors (or, equivalently, their exponents in the factoring) and a graphic histogram of the number of sets that correspond to that line: . . . . = = 4 2 + . . . . = @ 4 1 1 +++ . . . . @ @ 4 2 + . . . = = = 3 3 +++ . . . = = @ 3 2 1 +++++ . . . = @ @ 3 2 1 +++++ . . . @ @ @ 3 3 +++ . . = = = = 4 2 + . . = = = @ 3 2 1 +++++ . . = = @ @ 2 2 2 +++++++++++++++ . . = @ @ @ 3 2 1 +++++ . . @ @ @ @ 4 2 + . = = = = @ 4 1 1 +++ . = = = @ @ 3 2 1 +++++ . = = @ @ @ 3 2 1 +++++ . = @ @ @ @ 4 1 1 +++ = = = = @ @ 4 2 + = = = @ @ @ 3 3 +++ = = @ @ @ @ 4 2 + Throwing away the actual factors and concentrating on the histograms and exponent distribution, I see only five different lines: 4 2 + 3 3 +++ 4 1 1 +++ 3 2 1 +++++ 2 2 2 +++++++++++++++ These look suspiciously related to the partitions of 6: 6 5 1 4 2 4 1 1 3 3 3 2 1 3 1 1 1 2 2 2 2 2 1 1 2 1 1 1 1 1 1 1 1 1 1 where partitions with more than three elements don't appear (they can't, because 30 has only three distinct prime factors). Question: Why don't the "6" and "5 1" partitions appear? Question: Why 6 factors? Because the 1s disappear, I don't really think it's because there are 6 axes (A through F); after deleting the 1s, there are only 4. (The d12 also has 6 factors; they're just simpler factors.) Question: Are the histogram counts related to anything? They are not just the ways of arranging those factors, not even after dividing by a constant; consider 4 2 6!/(4!2!)=15 + (1) 3 3 6!/(3!3!)=20 +++ (3) 4 1 1 6!/(4!1!1!)=30 +++ (3) 3 2 1 6!/(3!2!1!)=60 +++++ (5) 2 2 2 6!/(2!2!2!)=90 +++++++++++++++ (15) Also, in the first pair, 0 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6 6 7 7 7 8 8 9 0 5 6 10 11 12 15 16 17 18 20 21 22 23 24 25 26 27 28 29 31 32 33 34 37 38 39 43 44 49 the distribution of numbers on the "lower" is provocatively similar to the sort of frequency distribution we've been looking at all along. The discrete derivative of the second die 0 5 6 10 11 12 15 16 17 18 20 21 22 23 24 25 26 27 28 29 31 32 33 34 37 38 39 43 44 49 5 1 4 1 1 3 1 1 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1 3 1 1 4 1 5 feels as though it has something significant lurking. I see a pattern in that collapsing things that add up to 6 does something interesting: 5 1 4 1 1 3 1 1 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1 3 1 1 4 1 5 ---- ------- ---------- ------------- ------------- ---------- ------- ---- This is probably related to the cross-product that generated this die (which can be 6x5/5x3/15x2 or 6x5/10x3/5x2), but the relationship is obscure to me. The next number I would expect to hold any interest is 60; it's the first number with four prime factors and three _distinct_ prime factors. I don't expect it to hold much interest, though. That distinction belongs to 210, the smallest number with four distinct prime factors (2*3*5*7). Using a diagram like B or F 0---x---x---x---x C |\ |\ |\ |\ |\ or x-\-x-\-x-\-x-\-x \ G |\ \|\ \|\ \|\ \|\ \ A or E x-\-x-\-x-\-x-\-x \ \ . . . [7 copies] . . . \ \ \ \ \ \ \ \ \ \ \ D \ \ x---x---x---x---x or H \ \|\ \|\ \|\ \|\ \| \ x---x---x---x---x \| \| \| \| \| x---x---x---x---x I did the same kind of exhaustive search I did for 12 and 30. It took a significant time in this case; it involved counting to 2^32 (the eighth power of the number of divisors of 210, that number being 16), doing a nontrivial amount of work at each step, but fortunately I have a machine which turned out to be fast enough to complete the job in a matter of hours. (If I were convinced conjecture B were true I could have pruned the search substantially; conjecture A would allow so much pruning it hardly counts as search - that would be more accurately called `enumeration'.) This results in 230 dice pairs, derived from 1956 different A-H sets. All the sets conform to conjecture A, and thus of course conjecture B as well. From the generating polynomial perspective, this implies some things about the factorizations of a particular degree-418 polynomial. However, it is not clear that there aren't factors involving negative coefficients (thus not corresponding to dice) that end up resulting in valid dice by the time they get multiplied back out again.