How many triangles with integral side lengths are possible, provided their perimeter is $36$ units?

My approach:

Let the side lengths be $a, b, c$; now,

$$a + b + c = 36$$

Now, $1 \leq a, b, c \leq 18$.

Applying multinomial theorem, I'm getting $187$ which is wrong.

Please help.

Solution 1:

The number of triangles with perimeter $n$ and integer side lengths is given by Alcuin's sequence $T(n)$. The generating function for $T(n)$ is $\dfrac{x^3}{(1-x^2)(1-x^3)(1-x^4)}$. Alcuin's sequence can be expressed as

$$T(n)=\begin{cases}\left[\frac{n^2}{48}\right]&n\text{ even}\\\left[\frac{(n+3)^2}{48}\right]&n\text{ odd}\end{cases}$$

where $[x]$ is the nearest integer function, and thus $T(36)=27$. See this article by Krier and Manvel for more details. See also Andrews, Jordan/Walch/Wisner, these two by Hirschhorn, and Bindner/Erickson.

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