I’ve always been fond of triangular numbers [1] and found the pattern they create beautiful.

```
.
. .
. . .
. . . .
. . . . .
. . . . . .
. . . . . . .
. . . . . . . .
. . . . . . . . .
. . . . . . . . . .
. . . . . . . . . . .
```

Embedded within are many a corporate logo:

Mitsubishi, Chase, Star of David, &c. All of these logos can easily be re-created using isometric graph paper.

I wanted to find out the relationships between the dots and the possible lines drawn between them. For example. If you have `T_n`

, how many dots exist as a recurrence relation? How many lines can one draw between the dots for `T_n`

?

```
o
/ \ (T_2 -> n_2 = 3)
o - o
```

number of dots: `d_i = d_{i-1} + i`

number of lines: `n_i = n_{i-1} + 3 * (i - 1)`

I wrote some of it up in CLISP to check it out:

```
;; calculates numer of edges in triangular graph
(defun nlines (i)
(let ((prev (- i 1)))
(if (eq i 0)
0
(+ (* 3 prev) (nlines prev)))))
;; returns number of nodes in graph
(defun ndots (i)
(if (eq i 0)
0
(+ (ndots (- i 1)) i)))
```

Now what if we looked at the triangular numbers with respect to a triangular array like Pascal’s Triangle? [2] That is, how many operations need to take place in order to calculate the entire triangle?

```
; calcs number of operations for triangular arrays based on n
;
; nops = 6, i = 3
; .
; / \
; . . (nops 3) == 6
; / \ / \
; . . .
;
(defun nops (i)
(if (eq i 0)
0
(let ((prev (- i 1)))
(+ (nops prev) (* 2 prev)))))
```

I brute-forced the limit by just putting large numbers in and found out that the ratio between the number of dots and lines approaches 1/3 as i approaches infinity and the number of dots to the number of operations (in a triangular array) approaches 1/2 as i approaches infinity.

```
\lim_{i\to\infty} \frac{ndots_I}{nlines_i} = \frac{1}{3}
\lim_{i\to\infty} \frac{ndots_I}{nops_i} = \frac{1}{2}
```

**Sources:**

http://en.wikipedia.org/wiki/Triangular_number

http://en.wikipedia.org/wiki/Pascal’s_triangle

To discuss this with the author, tweet @stephenbalaban.

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