A362634 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.

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

Offset

1,27

Comments

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).

Links

Pontus von Brömssen, Table of n, a(n) for n = 1..1000

James Grime and Brady Haran, Go First Dice, Numberphile video, 2023.

Eric Harshbarger, Go First Dice.

Wikipedia, Go First Dice.

Formula

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).

Example

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.)

Crossrefs

Cf. A362633.

Sequence in context: A089799 A073464 A142242 * A050314 A316344 A122157

Adjacent sequences: A362631 A362632 A362633 * A362635 A362636 A362637

Keyword

nonn

Author

Pontus von Brömssen, Apr 29 2023

Status

approved