MATHCOUNTS Exercise – Convolution of Non-zero Squares

Solution (a) By filling numbers in to the grid, we have the following $4\times 4$ grid: $$ \begin{array}{|c|c|c|c|} \hline 4 & 6 & 6 & 4 \\ \hline 6 & 10 & 10 & 6 \\ \hline 6 & 10 & 10 & 6 \\ \hline 4 & 6 & 6 & 4 \\ \hline \end{array} $$ Therefore the answer to the questions is $\boxed{104}$.

Solution (b) Assume the sum is $S$. There are $10^2$ squares of $1\times 1$, producing a partial sum of $1^2\times 10^2$; and $9^2$ squares of $2\times 2$, producing a partial sum of $2^2 \times 9^2$; and so on; and finally, there is $1^2$ square of $10\times 10$, producing a partial sum of $10^2 \times 1^2$. Therefore, $$S=\sum_{i=1}^{10}i^2\cdot(10-i+1)^2=2\cdot(1^2\cdot 10^2 + 2^2\cdot 9^2 + 3^2\cdot 8^2+ 4^2\cdot 7^2 + 5^2\cdot 6^2)$$ $$=2\cdot(100+324+576+784+900)=2\cdot 2684=\boxed{5368}$$

Note: For a $n\times n$ grid, the sum is $$S(n)=\sum_{i=1}^{n}i^2\cdot(n-i+1)^2$$ The above is the convolution of non-zero squares, producing OEIS sequence A033455, with the following closed form: $$\boxed{S(n)=\dfrac{(n+1)((n+1)^4-1)}{30}=\dfrac{n(n+1)(n+2)(n^2+2n+2)}{30}}$$

Challenge: In stead of a grid of squares, what is the answer for a grid consisting of equilateral triangles, such as the following: