Throw a die three times, and get maximum number of different sums.

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The IBM Ponder This problem for July 2013 throws an 8 sided die 3 times, and can get 120 possible different positive integer sums. If all the faces have positive integer sides, what is the lowest possible value for the highest face? For an $n$ sided die, the maximal number of different sums is a tetrahedral number. Here are my best results for d3-d7.

d3: 1, 2, 5 -- 10 sums
d4: 1, 2, 8, 12 -- 20 sums
d5: 1, 2, 16, 19, 24 -- 35 sums
d6: 1, 3, 12, 27, 43, 46 -- 56 sums
d7: 1, 2, 8, 51, 60, 79, 83 -- 84 sums

The d8 is the contest problem, so don't post that answer. I'm curious about d9, d10, and so on. I used a sieve method, but it doesn't scale up well. There is a related question in polynomials

Total[Sign[CoefficientList[Expand[(1 + x^2 + x^8 + x^12)^3 ], x]]]  

The above 4 term binary polynomial has 20 coefficients in its cube, which is maximal.

Is there an elegant way to find d9, d10, d11, d12, and so on?

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George Sicherman (of Sicherman dice fame) sent me his best results for d9 and d10

1 2 19 93 133 162 200 204 210

1 4 5 14 47 156 216 270 332 347

These involved a large search, no nice method for finding these is known yet.