Properties of square array of numbers min (i,j)

Question

I'm working on a pattern in the form below: $$ 1 1 1 1 1 1$$ $$ 1 2 2 2 2 2$$ $$ 1 2 3 3 3 3$$ $$ 1 2 3 4 4 4$$ $$ 1 2 3 4 5 5$$ $$ 1 2 3 4 5 6$$ If i took the diagonals $123321$ an upper and lower triangle formed. Whenever i subtract the sum of all elements in the lower triangle to the sum of all elements in the upper triangle, in every possible size of this array I always come up with a sum of triangular number. But I cant prove how this happened... I already work out in proving the other diagonal to be consecutive natural number and as well as the diagonals always form palindromic numbers... but im wondering how will i prove my problem about triangular number. I make used of minimum (i,j) as starting point. But I go to nowhere... any idea?

Answer 1

Let me see if I understand correctly.

You have an array like the following: \begin{matrix} 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2 & 2 & 2\\ 1 & 2 & 3 & 3 & 3 & 3\\ 1 & 2 & 3 & 4 & 4 & 4\\ 1 & 2 & 3 & 4& 5 & 5 \\ 1 & 2 & 3 & 4 & 5 & 6 \end{matrix}

Now you split it into an upper triangle and lower triangle: \begin{matrix} 1 & 1 & 1 & 1 & 1 & &&&& &&&& &\\ 1 & 2 & 2 & 2 & & &&&& &&&& &2\\ 1 & 2 & 3 & & & && \& && &&&&3& 3\\ 1 & 2 & & & & &&&& & & & 4& 4 & 4 \\ 1 & & & & & &&&& & &3&4 & 5 & 5\\ & & & & & &&&& & 2 &3 &4 & 5 & 6 \end{matrix} Now add all the elements in the lower triangle and call this $L$, add all the elements in the upper triangle and call this $U$. The difference $L-U$ can be computed by subtracting corresponding elements one at a time. What I mean is, first reflect the lower triangle like so: \begin{matrix} 1 & 1 & 1 & 1 & 1 & &&&& 6&5&4&3&2 &\\ 1 & 2 & 2 & 2 & & &&&& 5&5&4&3& & \\ 1 & 2 & 3 & & & && \& && 4 &4&4 && & \\ 1 & 2 & & & & &&&& 3& 3& & & & \\ 1 & & & & & &&&& 2& & & & & \\ & & & & & &&&& & & & & & \end{matrix}

This transformation is done by sending the $i$th row to the $7-i$th row, and the $j$th column to the $7-j$th column, so the entries in the triangle on the right are given by $\min(7-i,7-j)$. In general it will be $\min(n+1-i,n+1-j)$ if you start with an $n \times n$ grid. Now subtract element-by-element:

\begin{matrix} 6-1 & 5-1 & 4-1 & 3-1 & 2-1 & &&&& 5&4&3&2&1 &\\ 5-1 & 5-2 & 4-2 & 3-2 & & &&&& 4&3&2&1& & \\ 4-1 & 4-2 & 4-3 & & & && = && 3 &2&1 && & \\ 3-1 & 3-2 & & & & &&&& 2& 1& & & & \\ 2-1 & & & & & &&&& 1& & & & & \\ & & & & & &&&& & & & & & \end{matrix}

We can clearly see in the example that the sum of these elements will be a sum of triangular numbers, but we haven't proven it in general. Let's prove that the $k$th north-east diagonal contains the number $n-k$ (so in the example above, the first diagonal contains the number $5$, the second the number $4$, and so on). The $k$th diagonal consists of entries with coordinates $(i,j)$ such that $i+j=k+1$. If $i \leq j$, then $n+1-i \geq n+1-j$, so we have $$\min(n+1-i,n+1-j)-\min(i,j) = n+1-j-i = (n+1)-(i+j)= (n+1)-(k+1)=n-k.$$ Similarly, if $j \leq i$, then $n+1-j \geq n+1-i$, so $$\min(n+1-i,n+1-k)-\min(i,j)=n+1-i-j=(n+1)-(i+j)=(n+1)-(k+1)=n-k.$$

So, the $k$th north-east diagonal contains the number $n-k$, and it contains $k$ copies of it. So the sum of all the elements is: $$\sum_{k=1}^{n-1}(n-k)k=\sum_{k=1}^{n-1}k\sum_{i=1}^{n-k}1=\sum_{k=1}^{n-1}\sum_{i=1}^{n-k} k =\sum_{i=1}^{n-1}\sum_{k=1}^{n-i}k.$$ Now $\sum_{k=1}^{n-i}k$ is the $n-i$th triangular number, and we are summing these from $i=1$ to $i=n-1$. In other words, we get the sum of the first $n-1$ triangular numbers.

Answer 2

The element at location $(r,c)$ indeed has the value $\min(r,c)$.

The main diagonal is $(r,r)$, hence values $\min(r,r)=r$, linear.

The other diagonal, $(r,n+1-r)$, giving $\min(r,n+1-r)$, triangular.