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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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