Clay Kitchings ::
EMAT 6600 :: Counting the Triangles II
http://jwilson.coe.uga.edu/EMT725/Bob/Count.Triangles.II.html
Count the triangles: Extend the sequence:
So far, we have 1, 5, 13, and 27. Why are we getting these
numbers? Are there ÒhiddenÓ patterns within these numbers?
LetÕs look at the n=4 case (the fourth one in the picture
above).
We shall methodically count the number of triangles of
various sizes based on what IÕll call the Òsize of the base.Ó For example, the
size of the base of the largest triangle for n=4 is 4. That one large triangle has a base size
of 4 for these purposes.
WeÕll create a table of values for the triangles that are
pointed up as well as down
For n = 4:
Base
Size (oriented ÒupÓ) |
Number
of Triangles |
4 |
1 |
3 |
3 |
2 |
6 |
1 |
10 |
This 1, 3, 6, 10 sequence is extremely likely to form the
entire triangular number sequence. At this point, I am at least suspicious that
triangular numbers might be involved.
Base
Size (oriented ÒdownÓ) |
Number
of Triangles |
4 |
0 |
3 |
0 |
2 |
1 |
1 |
6 |
If we add the number of triangles from each table, we do
indeed get 27 total distinct triangles.
Now, letÕs look at the n=5 case:
Base
Size (oriented ÒupÓ) |
Number
of Triangles |
5 |
1 |
4 |
3 |
3 |
6 |
2 |
10 |
1 |
15 |
Base
Size (oriented ÒdownÓ) |
Number
of Triangles |
5 |
0 |
4 |
0 |
3 |
0 |
2 |
3 |
1 |
10 |
So, for n = 5, we can count 48 distinct triangles.
LetÕs now consider n=6 and do a little more counting:
Base
Size (oriented ÒupÓ) |
Number
of Triangles (up) |
6 |
1 |
5 |
3 |
4 |
6 |
3 |
10 |
2 |
15 |
1 |
21 |
No surprises here with the triangles oriented Òup.Ó The issue is a question about those
triangles oriented Òdown.Ó
Base
Size (oriented ÒdownÓ) |
Number
of Triangles (down) |
6 |
0 |
5 |
0 |
4 |
0 |
3 |
1 |
2 |
6 |
1 |
15 |
Up + Down = 56 + 22 = 78
LetÕs continue to investigate by going to n = 7:
Base
Size (oriented ÒupÓ) |
Number
of Triangles (up) |
7 |
1 |
6 |
3 |
5 |
6 |
4 |
10 |
3 |
15 |
2 |
21 |
1 |
28 |
Base
Size (oriented ÒdownÓ) |
Number
of Triangles (down) |
7 |
0 |
6 |
0 |
5 |
0 |
4 |
0 |
3 |
3 |
2 |
10 |
1 |
21 |
So for n=7, we have 84 + 34 = 118
Consider also n=8:
It is no longer a mystery from my perspective concerning the
triangles that are oriented in the ÒupÓ direction. They are the sum of the first n triangular numbers. So in
this case, weÕre talking about the sum of the first 8 triangular numbers. In the n=8 case, the sum of the first 8
triangular numbers is 120.
What would be nice is to figure out what is going on with
the downward-pointing triangles.
Base
Size (oriented ÒdownÓ) |
Number
of Triangles (down) |
8 |
0 |
7 |
0 |
6 |
0 |
5 |
0 |
4 |
1 |
3 |
6 |
2 |
15 |
1 |
28 |
Total |
50 |
Conjecture: The number of triangles oriented down for a
triangle with base size of n will be some type of sum of odd numbered
triangular numbers. In the example of n=8, we see the triangular numbers 1, 6,
15, and 28, which are the first, third, fifth and seventh triangular numbers.
After going back and looking at the other trends for the
downward totals, I have the following conjecture (in two parts):
á it appears that if n is odd, then the number of downward
pointing down is the sum of the first even triangular numbers (second + fourth
+ É + (n-1)th )
á it appears that if n is even, then the number of downward
pointing triangles is the sum of the first odd triangular numbers (first,
third, fifth, É + (n-1)th )
In order to generate each of the
first n-even triangular numbers (even for the n-values such as n=2, n=4, n=6,
etc., for some natural number k), I substituted k=2n into the formula I knew
for triangular numbers: . Beginning with T(1),..., we obtain
the sequence of triangular numbers generated by even integers {3, 10, 21, 36,
55, É}.
In order to generate each of the
first n-odd triangular numbers (such as n=1, n=3, n=5É) we consider the
function where m=2n-1 for
natural numbers, n. This will generate the sequence of the triangular numbers
from odd n-values.
So, for now, we have a Òrough shellÓ of a formula for
counting all the triangles for the nth case:
Total number of triangles = Number of upward triangles +
number of downward triangles
Let T(n) be the number of total
triangles in the figure.
Then T(n) = (Sum of the first n
triangular numbers) + (Sum of the downward triangles)
This problem is still in progress. I wanted to document the work I have done so far with it. I
hope to complete it more as time goes on and I gain more insight into it.