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A362634
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Let (k_1, ..., k_m) be the partition with Heinz number n (i.e., n = Product_{i=1..m} prime(k_i)) and consider a set of fair dice with k_1, ..., k_m faces numbered 1, ..., k_1 + ... + k_m (in any order). The outcome of a roll of the dice determines an ordering of them. a(n) is the minimum difference of the number of outcomes resulting in the most common ordering and the number of outcomes resulting in the least common ordering.
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2
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0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 2, 2, 0, 2, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 2, 2, 1, 0, 1, 0, 2, 0, 2, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 2, 1, 0, 2, 0, 3, 0, 3, 0, 1, 0, 0, 2, 3, 0, 2, 0, 1, 2, 1, 0, 2, 1, 0, 0
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OFFSET
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1,27
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COMMENTS
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In other words, a(n)/(Product_{i=1..m} k_i) is the minimum difference of the largest and smallest probabilities of the orderings of the dice.
a(n) = 0, with n = Product_{i=1..m} prime(k_i), means that there exists a set of m dice with k_1, ..., k_m faces, such that all m! orderings are equally likely, i.e., a permutation-fair set of dice. A necessary condition for a(n) = 0 to hold is that Product_{i=1..m} k_i be divisible by m!. This condition is not sufficient; for n = 45 = prime(2)^2*prime(3), for example, 2*2*3 is divisible by 3! but a(n) = 2 (see example).
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LINKS
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James Grime and Brady Haran, Go First Dice, Numberphile video, 2023.
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FORMULA
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a(2^k) = 1 for k >= 2 (corresponding to a set of k 1-sided "dice").
a(p) = 0 if p is prime (corresponding to a single die).
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EXAMPLE
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For two 2-sided dice and one 3-sided die, the best we can do is to number the faces (2,6), (3,5), and (1,4,7), respectively. We then have the following 12 possible outcomes and orderings:
d1 d2 d3 | ordering
---------+---------
2 3 1 | d3<d1<d2
2 3 4 | d1<d2<d3
2 3 7 | d1<d2<d3
2 5 1 | d3<d1<d2
2 5 4 | d1<d3<d2
2 5 7 | d1<d2<d3
6 3 1 | d3<d2<d1
6 3 4 | d2<d3<d1
6 3 7 | d2<d1<d3
6 5 1 | d3<d2<d1
6 5 4 | d3<d2<d1
6 5 7 | d2<d1<d3
The maximum frequency for an ordering is 3 (for d1<d2<d3 and d3<d2<d1) and the minimum frequency is 1 (for d1<d3<d2 and d2<d3<d1), so a(prime(2)^2*prime(3)) = a(45) = 3-1 = 2. (This is the unique way of numbering the faces so that the difference of the maximum and minimum frequencies is 2.)
There exists a permutation-fair set of three dice with 2, 4, and 6 faces, so a(prime(2)*prime(4)*prime(6)) = a(3*7*13) = a(273) = 0. (See Harshbarger link.)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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