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I built a Pascal Triangle, but instead of each value in a row being a sum of the two numbers above it I made each value the sum of the numbers above it divided by the product of those numbers. So, as an example the first three lines are 1, 1 1, 1 2 1, but the third line is 1 1.5 1.5 1, and the fourth is 1 1.6 1.35 1.35 1.6 1, etc.

The values appear to be converging (possibly $\sqrt2$?), but I cannot tell for certain if they do as I don't have a grid large enough to continue carrying out the values.

Is there an equation that I can plug in a row number and the element of that row to find what its value would be? I believe that I can use the expression $$n!/k!(n-k)!$$ for Pascal's Triangle, but I don't know how to modify the equation for this variation.

mkinson
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1 Answers1

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The limit deep inside the triangle is indeed $\sqrt 2$. If the limit is $L$, you iteration converges to $L=\frac {L+L}{L^2}$ with solution $L=\sqrt 2$. We can say more. The edge cells are all $1$ by definition. The cell inside the edge converges to $\phi=\frac {1+\sqrt 5}2 \approx 1.618$ Again you can find the limit by solving $L=\frac {1+L}{1 \cdot L}$ which reaches the value shown. The next one converges to $L=\frac {\phi + L}{\phi L}$ with solution about $1.356674$ You can continue to find the limit of each number of places in from the edge. If the limit of the next cell out is $A$, this cell converges to $L=\frac {A+L}{AL}$. These converge quickly to $\sqrt 2$.

$$\begin {array} {r r} \text{cells in}&\text{limit} \\ 0&1\\1&1.618034\\2&1.355674\\3&1.434666\\4&1.407504\\5&1.416462\\6&1.413465\\7&1.414463\\8&1.41413\\9&1.414241\\10&1.414204\\11&1.414217 \end {array}$$ We could express each of these with increasingly deeply nested radicals, but that doesn't seem useful

Ross Millikan
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