$[a_{i,j}]_{n\times n}=\left[ \begin{array}{ccc} 1&4&32&…\\ 2&16&…&…\\ 8&…&…&…\\ …&…&…&…\\ \end{array}\right]$, find $a_{3,7}$ [duplicate]

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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

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

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

• Do the indices start at $0$ or $1$? The former implies $a_{1,\,1}=16$, the latter $a_{1,\,1}=1$. – J.G. May 5 at 14:35
• Assume indices start at $1$, $\log_2 a_{ij} = \frac{(i+j-1)(i+j-2)}{2} + (j-1)$ – achille hui May 5 at 14:37
• @J.G. It start at $1$. – David May 5 at 14:38
• @achillehui Your answer is correct, it's $2^{42}$. How did you come up with that formula? – David May 5 at 14:47
• @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). – achille hui May 5 at 14:49

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. $$2^{42}$$.

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 $$2^{3+4+5+6+7+8+9}=2^{42}.$$

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.