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If $A=(a_{ij})_{n\times n}$, $a_{ij}=i\cdot j$ and

$B=(b_{ij})_{n\times n}$, $b_{ij}=\min(i,j)$.

How calculate a formula for $c_{ij}$, with $C=(c_{ij})_{n\times n}=AB$.

For example:

$n=2:$

$$C=\begin{pmatrix} 3&5\\ 6&10\end{pmatrix}$$

$n=3:$

$$C=\begin{pmatrix} 6& 11& 14\\ 12& 22& 28\\ 18 & 33 & 42\end{pmatrix}$$

$n=4:$

$$C=\begin{pmatrix} 10 & 19 & 26 & 30\\ 20 & 38 & 52 & 60\\ 30 & 57 & 78 & 90\\ 40 & 76 & 104 & 120\end{pmatrix}$$

$n=5:$

$$C=\begin{pmatrix} 15 & 29& 41 & 50 & 55\\ 30 & 58 & 82 & 100 & 110\\ 45 & 87 & 123& 150 & 165\\ 60 & 116 & 164 & 200 & 220\\ 75 & 145 & 205 & 250 & 275\end{pmatrix}$$

Note that $c_{ij}=i\cdot c_{1j}$. Is there a formula for the first row $c_{1j}$?

Any hint woul be appreciated.

felipeuni
  • 5,080
  • $c_{ik}=\sum_jij \min(j,k)=i\sum_jj \min(j,k)$ then break the sum into parts for $j<k$ and $j\ge k$ – Benjamin Wang Oct 02 '23 at 22:03
  • @BenjaminWang Thanks but Is there a formula without using the minimum function or by cases? – felipeuni Oct 02 '23 at 22:10
  • After analysing the cases you will have a cubic formula which won't involve cases. Alternatively you can look directly for patterns: Looking at the first row of $n=5$, the differences are $14,12,9,5$, and their differences are $2,3,4$ (consecutive integers). So there should be a cubic that you can fit. – Benjamin Wang Oct 03 '23 at 02:08
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    @BenjaminWang Thanks for your solution! – felipeuni Oct 03 '23 at 18:56

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