Integer Ordered Triplets – Mattia Giuri's bizarre blog

Distinguishing cases, find the number of ordered triplets (x, y, z) of positive integers such that

$x+y+z = n$

Let $S_n$ be the number of such triplets for n. We can get a first recursive formula by noticing that if $x + y + z = n$ and $a, b, c > 1$ then $(x-1) + (y-1) + (z-1) = n - 3 \land (x-1), (y-1), (z-1) > 0$ .

From this we deduce that

$S_n = S_{n-3} + \lfloor \frac{n-1}{2} \rfloor$

as one can easily check that the cases where at least one of $x, y, z$ numbers equals 1 is the integer part of n-1 divided by 2.

Also, notice there is a bijection between $x_1 + y_1 + z_1 = n \land x_1 \leq y_1 \leq z_1$ and the ordered triplets $x + y + z = n \land x < y < z$ given by $x = x_1, y = y_1 + 1, z = z_1 + 2$

Hence, we can apply Burnside’s Lemma on $S_n$ with the group of permutations of 3 elements.

In total we have ${{n-1} \choose 2}$ elements, which we can divide into the conjugacy classes of: all different (1), exactly two are equal (2), exactly 3 are equal if possible (3). Only the identity fixes any element in 1, 2 transpositions fix the elements in 2, all 6 transformations fix the element in 3.

If n is not divisible by 3 we see, using the bijection between $S_{n-3}$ and the conjugacy class 1 (just multiply by 6, the number of possible permutations) we get the recursive formula: