An Intriguing Triangle

Harold Cooper

18th August 2020

I recently noticed that the triangle with $n$ th row and $k$ th column:

Δ (n, k) = k (n - k)

…is closely related to the binomial coefficients $(3 n)$ and $(4 n)$ .

Here are the first several rows of the triangle¹, along with their sums ∑ and cumulative sums ∑∑:

n              𝚫               ∑    ∑∑
            1               1     1
          2   2             4     5
        3   4   3          10    15
      4   6   6   4        20    35
    5   8   9   8   5      35    70
              ...

(We omit all the zero entries, at $k = 0$ , $k = n$ , and $n < 2$ .)

Theorem 1

Note that the binomial coefficients $(3 3) = 1$ , $(3 4) = 4$ , $(3 5) = 10$ , $(3 6) = 20$ , … look a lot like the ∑ column of the sums of the rows. Sure enough:

k = 1 \sum n k (n - k) = (3 n + 1)

Proof

Using the facts² that $\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$ and $\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ , we have:

k = 1 \sum n k (n - k) = n k = 1 \sum n k - k = 1 \sum n k^{2} = n \frac{n ( n + 1 )}{2} - \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = n (n + 1) (\frac{n}{2} - \frac{2 n + 1}{6}) = n (n + 1) \frac{3 n - ( 2 n + 1 )}{6} = \frac{n ( n + 1 ) ( n - 1 )}{6} = (3 n + 1)

■

Theorem 2

Note that the binomial coefficients $(4 4) = 1$ , $(4 5) = 5$ , $(4 6) = 15$ , $(4 7) = 35$ , … look a lot like the ∑∑ column of the cumulative sums of the rows. Sure enough:

m = 1 \sum n k = 1 \sum m k (m - k) = m = 1 \sum n (3 m + 1) = (4 n + 2)

And in fact, more generally for any $k$ :

m = 0 \sum n (k m) = (k + 1 n + 1)

Proof

(Also proved in the Lean theorem proving language, for fun.³)

(n = 0)

If $k = 0$ , $\sum_{m = 0}^{0} (0 m) = (0 0) = 1 = (1 1)$

Otherwise, $k > 0$ so $\sum_{m = 0}^{0} (k m) = (k 0) = 0 = (k + 1 1)$

(n > 0)

Supposing by induction that $\sum_{m = 0}^{n - 1} (k m) = (k + 1 n)$ , we have:

m = 0 \sum n (k m) = (k n) + m = 0 \sum n - 1 (k m) = (k n) + (k + 1 n)

But from Pascal’s triangle we know $(k n) + (k + 1 n) = (k + 1 n + 1)$

■

Visual Interpretation

In the context of Pascal’s triangle, $\sum_{m = 0}^{n} (k m) = (k + 1 n + 1)$ says that you can sum a diagonal to find the value of the entry below and to the right.

For $n = 3$ , $k = 1$ for example, $\sum_{m = 0}^{3} (1 m) = 0 + 1 + 2 + 3 = 6 = (2 4)$ is illustrated by marking the relevant entries of Pascal’s triangle:

        1  +0
+1   0
+2   1   0
+3   3   1   0
 4  =6   4   1   0

Notice that the triangle is just the antidiagonals of a multiplication table:

1  2  3  4  5  .  .  .
2  4  6  8 10
3  6  9 12 15
4  8 12 16 20
5 10 15 20 25
.              .
.                 .
.                    .

It’s also known as OEIS A003991. ↩

$\sum_{k = 1}^{n} k$ has many derivations, including this clever visual one.

$\sum_{k = 1}^{n} k^{2}$ can be derived by expanding $(k - 1)^{3}$ as illustrated here. ↩

theorem two {n k : ℕ} : ∑ m in range n, m.choose k = n.choose (k+1) :=
begin
  induction n with n hn,
  { refl, },
  { rw [finset.sum_range_succ_comm, hn],
    refl, },
end

(Note that range n only goes up to $n - 1$ , so this equation uses $(k + 1 n)$ instead of $(k + 1 n + 1)$ .)

You can interact with this code in the Lean web editor. ↩