-2
$\begingroup$

This question is an exact duplicate of:

I found that $\left[ \begin{array}{ccc} 1&4&32&...\\ 2&16&...&...\\ 8&...&...&...\\ ...&...&...&...\\ \end{array}\right]=\left[ \begin{array}{ccc} 2^0&2^2&2^5&...\\ 2^1&2^4&...&...\\ 2^3&...&...&...\\ ...&...&...&...\\ \end{array}\right]$

Is there a quicker way to find $a_{3,7}$ than write down all the elements? Maybe predict the elements near $a_{3,7}$ ?

The indices start at 1

$\endgroup$

marked as duplicate by GNUSupporter 8964民主女神 地下教會, Dietrich Burde, Cameron Buie, Yanior Weg, Tianlalu May 7 at 9:19

This question was marked as an exact duplicate of an existing question.

  • 1
    $\begingroup$ Do the indices start at $0$ or $1$? The former implies $a_{1,\,1}=16$, the latter $a_{1,\,1}=1$. $\endgroup$ – J.G. May 5 at 14:35
  • $\begingroup$ Assume indices start at $1$, $\log_2 a_{ij} = \frac{(i+j-1)(i+j-2)}{2} + (j-1)$ $\endgroup$ – achille hui May 5 at 14:37
  • $\begingroup$ @J.G. It start at $1$. $\endgroup$ – David May 5 at 14:38
  • $\begingroup$ @achillehui Your answer is correct, it's $2^{42}$. How did you come up with that formula? $\endgroup$ – David May 5 at 14:47
  • 1
    $\begingroup$ @David look at mankind's answer, my logic is about the same as his (I have verified the answer is $2^{42}$ by explicitly construct the matrix in a CAS). $\endgroup$ – achille hui May 5 at 14:49
4
$\begingroup$

All elements on a right to left downward diagonal have indices adding to the same number. For instance, $a_{1,2}$ and $a_{2,1}$ are on the same diagonal and their indices add to three.

The element $a_{3,7}$ is on the diagonal with elements whose indices add to 10. This means there are $1+2+\cdots+8=36$ elements before $a_{9,1}$, which is thus $2^{36}$.

Then the sought after element is 6 elements after this one, i.e. $2^{42}$.

$\endgroup$
1
$\begingroup$

I would establish how the exponents increase as you move to the right along row $3.$ First, $2^3,$ then $2^7,$ then $2^{12}.$ Do you see a pattern? If not, you might want to think about how it progresses along the first two rows, as well. In the first row, we have $2^0,2^2,2^5,2^9,2^{14};$ in the second, $2^1,2^4,2^8,2^{13}.$


Added: As you've already accepted an answer, I'll go ahead and elaborate. The first entry in row $3$ is $2^3,$ next is $2^{3+4},$ then $2^{3+4+5},$ and so on. Thus, the seventh term is $2^{3+4+5+6+7+8+9}=2^{42}.$

$\endgroup$
0
$\begingroup$

General formula is: $$\log_2a_{i,j}=\frac{i(i-1)}{2}+\frac{(2i+j)(j-1)}{2}.$$

Derivation: The exponents of $a_{i,1}$ are triangular numbers: $$0,1,3,6,...,\frac{i(i-1)}{2}.$$

The exponents of $a_{i,j}$: $$\underbrace{\frac{i(i-1)}{2}}_{a_{i,1}},\underbrace{\frac{i(i-1)}{2}+(i+1)}_{a_{i,2}},...,\underbrace{\frac{i(i-1)}{2}+(i+1)+\cdots +(i+j-1)}_{a_{i,j}}=\\ \frac{i(i-1)}{2}+\frac{(i+1)+(i+j-1)}{2}\cdot (j-1)=\\ \frac{i(i-1)}{2}+\frac{(2i+j)(j-1)}{2},$$ which is the same as achille hui's answer in the comment above.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.