Expressing a positive integer as a sum of 3 positive integers

Question

Let us denote the the number of ways in which a positive integer, $n$, can be expressed as a sum of $3$ positive integers (not necessary distinct) by $W_3(n)$.

$W_3(n)$ can be evaluated using any of the following formulae:

• $W_3(n)=\frac{1}{72}[6n^2-7-9(-1)^n+16\cos(\frac{2\pi}{3}n)]$, or

• $W_3(n)=\left \langle \frac{1}{12}n^2 \right \rangle$, where $\left \langle m \right \rangle$ is the nearest integer to $m$ if $m$ is not an integer, and it is $m$ if $m$ is an integer.

The nearest integer function is usually denoted by $[n]$, but I have used this notation in the first formula of $W_3(n)$.

$W_3(n)$ is usually denoted by $P(n,3)$, but I used $W$ for the number of WAYS.

The two formulae above can be found in http://mathworld.wolfram.com/PartitionFunctionP.html .

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How can we prove that $\frac{1}{72}[6n^2-7-9(-1)^n+16\cos(\frac{2\pi}{3}n)]=\left \langle \frac{1}{12}n^2 \right \rangle$?

Answer 1

Note that $\, W_3(n) = a(n-3)\,$ where $\,a(n)\,$ the OEIS sequence A001399. As noted in the sequence entry formulas: $$a(n) = 1 + a(n-2) + a(n-3) - a(n-5). \tag1$$ In order to prove your two formulas you just need to prove that they have the same $5$ initial values and satisfy recursion equation $(1)$. One way to do this is to define an operator $\,L\,$ on sequence as follows:

$$ L[a(n)] := a(n) - a(n-2) - a(n-3) + a(n-5). \tag2 $$ Equation $(1)$ can be stated as $\, L[a(n)] = 1.\,$ Your first formula is a sum of four terms and since $\,L\,$ is linear you can prove that it takes the first term to one and the rest to zero. For the second formula you can show each residue class modulo $6$ satisfies equation $(1)$.

Answer 2

$\frac{1}{72}[6n^2-7-9(-1)^n+16\cos(\frac{2\pi}{3}n)] =\left \langle \frac{1}{12}n^2 \right \rangle $

Let $n = 6m+k, 0 \le k \le 5$.

$\begin{array}\\ LHS &=\dfrac{6n^2-7-9(-1)^n+16\cos(\frac{2\pi}{3}n)}{72}\\ &=\dfrac{6(6m+k)^2-7-9(-1)^{6m+k}+16\cos(\frac{2\pi}{3}(6m+k))}{72}\\ &=\dfrac{6(36m^2+12mk+k^2)-7-9(-1)^{k}+16\cos(4\pi m+\frac{2\pi k}{3})}{72}\\ &=3m^2+mk+\dfrac{6k^2-7-9(-1)^{k}+16\cos(\frac{2\pi k}{3})}{72}\\ &=3m^2+mk+\dfrac{(0, 0, 0, 72, 72, 144)}{72}\\ &=3m^2+mk+(0, 0, 0, 1, 1, 2)\\ \end{array} $

$\begin{array}\\ RHS &=\left \langle \frac{1}{12}n^2 \right \rangle\\ &=\left \langle \frac{(6m+k)^2}{12} \right \rangle\\ &=\left \langle \frac{36m^2+12mk+k^2}{12} \right \rangle\\ &=\left \langle 3m^2+mk+\frac{k^2}{12} \right \rangle\\ &=3m^2+mk+\left \langle \frac{k^2}{12} \right \rangle\\ &=3m^2+mk+\left \langle (0, \frac1{12}, \frac13, \frac34, \frac43, \frac{25}{12}) \right \rangle\\ &=3m^2+mk+ (0, 0, 0, 1, 1, 2)\\ \end{array} $

So LHS = RHS.