A131514 Number of ways to design a set of three n-sided dice (using nonnegative integers) such that summing the faces can give any integer from 0 to n^3 - 1.
1, 1, 1, 15, 1, 71, 1, 280, 15, 71, 1, 3660, 1, 71, 71, 5775, 1, 3660, 1, 3660, 71, 71, 1, 160440, 15, 71, 280, 3660, 1, 20365, 1, 126126, 71, 71, 71, 415185, 1, 71, 71, 160440, 1, 20365, 1, 3660, 3660, 71, 1, 6387150, 15, 3660, 71, 3660, 1, 160440, 71, 160440
Offset: 1
Keywords
Examples
a(4)=15 because we can choose any of the following 15 configurations for our three dice: [ {0, 1, 2, 3}, {0, 4, 8, 12}, {0, 16, 32, 48} ], [ {0, 1, 2, 3}, {0, 4, 16, 20}, {0, 8, 32, 40} ], [ {0, 1, 2, 3}, {0, 4, 32, 36}, {0, 8, 16, 24} ], [ {0, 1, 4, 5}, {0, 2, 8, 10}, {0, 16, 32, 48} ], [ {0, 1, 4, 5}, {0, 2, 16, 18}, {0, 8, 32, 40} ], [ {0, 1, 4, 5}, {0, 2, 32, 34}, {0, 8, 16, 24} ], [ {0, 1, 8, 9}, {0, 2, 4, 6}, {0, 16, 32, 48} ], [ {0, 1, 8, 9}, {0, 2, 16, 18}, {0, 4, 32, 36} ], [ {0, 1, 8, 9}, {0, 2, 32, 34}, {0, 4, 16, 20} ], [ {0, 1, 16, 17}, {0, 2, 4, 6}, {0, 8, 32, 40} ], [ {0, 1, 16, 17}, {0, 2, 8, 10}, {0, 4, 32, 36} ], [ {0, 1, 16, 17}, {0, 2, 32, 34}, {0, 4, 8, 12} ], [ {0, 1, 32, 33}, {0, 2, 4, 6}, {0, 8, 16, 24} ], [ {0, 1, 32, 33}, {0, 2, 8, 10}, {0, 4, 16, 20} ], [ {0, 1, 32, 33}, {0, 2, 16, 18}, {0, 4, 8, 12} ].
Links
- William P. Orrick, Table of n, a(n) for n = 1..10000
- M. Krasner and B. Ranulac, Sur une propriété des polynomes de la division du cercle, Comptes Rendus Académie des Sciences Paris, 240:397-399, 1937.
- Matthew C. Lettington and Karl Michael Schmidt, Divisor Functions and the Number of Sum Systems, arXiv:1910.02455 [math.NT], 2019.
Programs
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SageMath
@cached_function def R3(i,j,k): if i > 1 and j==1 and k==1: return(1) elif j > 1 or k > 1: divList = divisors(i)[:-1] return(sum(G3(d,j,k) for d in divList) + sum(B3(d,j,k) for d in divList)) @cached_function def G3(i,j,k): if i==1 and j > 1 and k==1: return(1) elif i > 1 or k > 1: divList = divisors(j)[:-1] return(sum(R3(i,d,k) for d in divList) + sum(B3(i,d,k) for d in divList)) @cached_function def B3(i,j,k): if i==1 and j==1 and k > 1: return(1) elif i > 1 or j > 1: divList = divisors(k)[:-1] return(sum(R3(i,j,d) for d in divList) + sum(G3(i,j,d) for d in divList)) def a3(n): if n == 1: return(1) else: return(R3(n,n,n) / 2) # William P. Orrick, Jan 26 2023
Formula
Recurrence: a(1) = 1. For n > 1, a(n) = r(n,n,n) / 2 where r(i,1,1) = g(1,j,1) = b(1,1,k) = 1 for all i, j, k > 1, r(i,j,k) = Sum_{d|i,dWilliam P. Orrick, Jan 26 2023
Extensions
Terms a(16) and beyond from William P. Orrick, Jan 26 2023
Comments