Áp dụng liên tiếp đẳng thức trên ta có được:…Vậy ta có đẳng thức$$f(n,k)=f(n-1,k-1)+f(n-k,k)$$
$$\begin{equation}\label{dt1}f(n,k)=\sum_{i=0}^{\left\lfloor\frac{n-2}{k}\right\rfloor} f(n-1-ki,k-1)\end{equation}$$
Giờ ta sẽ thử dùng $(\ref{dt1})$ để tính $f(n,3)$
Ta có:
$f(n-1-3k,2)= \left\lfloor\frac{n-1-3k}{2}\right\rfloor $
Nên theo $(\ref{dt1})$
$f(n,3)=\sum_{k=0}^{\left\lfloor\frac{n-2}{3}\right\rfloor} \left\lfloor\frac{n-1-3k}{2}\right\rfloor$
$\;\;\;\; =\sum_{j=0}^{\left\lfloor\frac{n-2}{6}\right\rfloor} \left\lfloor\frac{n-1}{2}-3j\right\rfloor +\sum_{j=0}^{\left\lfloor\frac{n-5}{6}\right\rfloor} \left\lfloor\frac{n-4}{2}-3j\right\rfloor \;\;\; $(tách chẵn lẻ)
$\;\;\;\; = \left\lfloor\frac{n-1}{2}\right\rfloor \left\lfloor\frac{n+4}{6}\right\rfloor -\frac{3}{2} \left\lfloor\frac{n-2}{6}\right\rfloor \left\lfloor\frac{n+4}{6}\right\rfloor + \left\lfloor\frac{n-4}{2}\right\rfloor \left\lfloor\frac{n+1}{6}\right\rfloor -\frac{3}{2} \left\lfloor\frac{n-5}{6}\right\rfloor \left\lfloor\frac{n+1}{6}\right\rfloor $
Tới đây để nguyên vậy nhìn cho “nguy hiểm”
Thay $n=6p+(0,1,2,3,4,5)$, ta được:
$f(n,3)=\big(3p+(-1,0,0,1,1,2)\big) \big(p+(0,0,1,1,1,1)\big)-\frac{3}{2} \big(p+(-1,-1,0,0,0,0)\big) \big(p+(0,0,1,1,1,1)\big) + \big(3p+(-2,-2,-1,-1,0,0)\big) \big(p+(0,0,0,0,0,1)\big)- \frac{3}{2} \big(p+(-1,-1,-1,-1,-1,0)\big) \big(p+(0,0,0,0,0,1)\big) $
$=3p^2+p(-1,0,3,4,4,5)+(0,0,0,1,1,2) -\frac{3}{2}p^2 +\frac{3}{2}p(1,1,-1,-1,-1,-1) +3p^2+p(-2,-2,-1,-1,0,3) -\frac{3}{2}p^2+ \frac{3}{2}p(1,1,1,1,1,-1)$
$=3p^2+p(-3,-2,2,3,4,8)+\frac{3}{2}p(2,2,0,0,0,-2)+ (0,0,0,1,1,2)$
$=3p^2+p(0,1,2,3,4,5)+(0,0,0,1,1,2)$
$=\left\lfloor\frac{n^2+3}{12}\right\rfloor \;\;\;$(theo bài viết trên)