Have I discovered a new significance to a previously discovered constant? I've been interested in infinite sums for a while, though I have no formal education of them. I was messing around with repeated division and addition (e.g. 1 + (1 / (1 + (1 /...)))) I then plugged fibonacci numbers into the above pattern, and as I calculated more and more layers, the result converged to around 1.39418655, which I just now found out was a constant called Madachy's constant.  However, from the research that I did on the constant (I found very little), it doesn't seem to be related to infinite series or Fibonacci numbers at all. Have I found a new way to calculate this number? Does it give it any more significance? 
 A: Did you see this? fq.math.ca/Scanned/6-6/madachy.pdf It looks like it is exactly your calculation.
Friendly edit: This is in Fibonacci Quarterly, Volume 6 No. 6 page 385, as seen at the cover of that issue.
A: (Too long for a comment.) The OP's continued fraction which uses the Fibonacci numbers $F_n$ is
$$\mu =1+\cfrac{1}{2 + \cfrac{3} {5 + \cfrac{8} {13 + \cfrac{21} {34+\ddots}}}}=1.3941865\dots $$
which, as pointed out by Cheerful Parsnip, was previously investigated by Joseph Madachy and whose constant is designed by OEIS A130701 as $\mu$.
However, there is continued fraction whose terms are also Fibonacci numbers and has a known closed-form,
$$\frac{F_{n+2}}{F_{n+1}} =1^2+\cfrac{(1\cdot1)^2} {1^2+ \cfrac{(1\cdot2)^2} {2^2 + \cfrac{(2\cdot3)^2} {3^2 + \cfrac{(3\cdot5)^2} {5^2 + \cfrac{(5\cdot8)^2} {F_n^2 + \cfrac{(F_{n}\cdot F_{n+1})^2}  {F_{n+1}^2+\ddots} }}}}}$$
If we truncate it at the $n$th term, we get that ratio. Note $\lim \frac{F_{n+1}}{F_{n}} = \phi$ as $n\to\infty$. 
P.S. It would be nice if $\mu$ has a closed-form in terms of the golden ratio $\phi$ as well. 
A: The continued fraction you are calculating will converge to the golden ratio. It is most certainly a significant constant having history spanning several millennia.
