The Dartboard Sequence
The arrangement of the numbers around the circumference of a standard
dart board is as shown below
20 1 18 4 13 6 10 15 2 17 3 19 7 16 8 11 14 9 12 5
Oddly enough, no one seems to know how this particular arrangement was
selected. It evidently dates back over 100 years, but I've been unable
to find any simple criterion that uniquely singles out this arrangement
as the best possible in any quantitative sense. It may be just an
accident of history that this particular arrangement has been adopted
as the standard dart board format.
However, it got me thinking about possible criteria for choosing a
circular arrangement of the first n positive integers. I think the
most natural idea, in order to get as "flat" a distribution as possible,
is to minimize the sum of the squares of each k consecutive terms.
For example, setting k = 3 the standard dard board sequence gives
(20+1+18)^2 + (1+18+4)^2 + (18+4+13)^ + ... + (5+20+1)^2 = 20478
In contrast, if you arrange the numbers by just inter-weaving the
largest and smallest numbers like this
20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10
the resulting sum of squares of each 3 consecutive elements is 20510,
so the standard dart board is, in this sense, a more flat distribution.
Both of these arrangements are much more flat than the natural sequence
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
which has a sum of 72000. In general, for each permutation of the
integers from 1 to n we can compute the sum of squares (or any other
power) of every k consecutive terms.
This raises the question: Which permutation gives the smallest sum,
and which gives the largest? Although the standard dart board gives
a fairly low sum, it is not the lowest for n=20,k=3 (nor is it the
lowest for k=2 or k=4). I more or less randomly checked several other
arrangements, and the lowest sum of squares of 3 consecutive terms I
found was 20050 for the following arrangement
7 18 4 9 20 2 13 17 1 15 12 3 16 6 5 19 10 8 11 14
Subsequently, Ben Kibel found a better sequence
20 1 12 17 3 13 14 4 15 11 6 16 8 7 18 5 9 19 2 10
which yields a total of 19874. Does anyone know of a better one?
I.e., a circular arrangement of the integers 1 through 20 such that
the sum of squares of every 3 consecutive terms is less than 19874?
In general if H(n,k) and L(n,k) are the highest and lowest sums of
squares of every k consecutive elements in a circular arrangement
of the first n positive integers, are the values of H(n,k) and L(n,k)
well known and/or easily computed?
When considering the possible ways of "optimizing" the arrangement
of the numbers 1 through 20 on a dart board, one proposal was to
minimize the sum of the squares of every sum of two consecutive
numbers. In general, I think the minimum sum of squares of every
sum of two consecutive numbers in a cyclical arrangement of the
integers 1 through N is
S_min(N) = N^3 + 2N^2 + 2N - j
where j is 1 if N is odd, and j is 2 if N is even. For the particular
case N=20 this formula gives a minimum sum of 8838.
For even N the minimum arrangement has the odd and even numbers
restricted to separate halfs of the cycle, as illustrated below
for N=20
1 3 5 7 9 10 8 6 4 2
19 17 15 13 11 12 14 16 18 20
For odd N the minimum arrangement is very simple, as shown below
for N=19.
1 3 5 7 9 11 13 15 17 19
18 16 14 12 10 8 6 4 2
This raises some interesting questions. Given any circular
arrangement of the integers 0 through n-1, let S denote the sum
of squares of every sum of two contiguous numbers, and let v(n)
denote the number of distinct values of S for all n! possible
arrangements. Following is a table of the number of distinct
values of v(n)
n v(n)
----- ---------
1 1
2 1
3 1
4 3
5 8
6 21
7 43
8 69
9 102
10 145
The sequence of integers v(n) does not appear to have been considered
before. (It doesn't appear in Sloane's Encyclopedia of Integer
sequences.) Hugh Montgomery, A. M. Odlyzko, and Bjorn Poonen developed
a very nice approach to this problem, showing that the general term
with n>6 is given by
/ (n^3 - 16n + 27)/6 if n is odd
v(n) = (
\ (n^3 - 16n + 30)/6 if n is even
A whole family of interesting sequences can be produced by
generalizing the definition as follows: Given any circular
arrangement of the integers 0 through n-1, let S denote the sum
of the qth powers of every sum of k contiguous numbers. Then
let v(q,k,n) denote the number of distinct values of S for all
possible arrangements.
With this nomenclature, the previous sequence is denoted as v(2,2,n).
Of course, we have v(1,k,n) = 1 for all k and n, because the sum of
the 1st powers is independent of the arrangement. We also have
v(q,1,n) = 1 because the sum of any fixed power of the individual
numbers is also independent of the arrangement. Also, for fixed
values of q and n, the function v is PERIODIC in k.
Another generalization is to add some constant integer j to each
of the numbers 0 to n-1. Thus, the general function has four indices,
v(q,k,j,n). Notice that v is independent of j for q<3, but for larger
values of q, j becomes significant. Can v(q,k,j,n) be expressed in
closed form as a function of the indices? Which other integer
sequences are contained in this family? Which continuous functions
(e.g., sin(x), cos(x), exp(x), etc) can be approximated by sequences
of this form?
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