Continued Fraction using all Perfect Squares What is known about the infinite continued fraction
$$1 + \cfrac{1}{4 + \cfrac{1}{9 + \cfrac{1}{16 + \cdots}}} $$
whose terms include all perfect squares in order?
Do we have a closed form expression for the value of this number? Is it known to be transcendental, or satisfy any other interesting properties?
 A: Every irrational number $\ > 1\ $ is expressed in a unique way as a simple infinite continued fraction
$$a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cdots}}} $$
where every $\ a_k\ $ is a positive integer.
(It follows that the continued fraction from the OP's question is irrational).
And no other real number $\ > 1\ $ (i.e. no such rational number) ca be represented as a simple infinite continued fraction.
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REMARK 1  One can easily formulate similar theorems for all positive irrational numbers or even for all non-zero irrational numbers, as well as for all irrational numbers between $\ 0\,$ and $\,1$. Each of such spaces (under the induced Euclidean topology) is homeomorphic to the topological (Tikhonov) Cartesian power $\ \Bbb Z^\Bbb Z.$

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It is also very well known that the OP's fraction is not a quadratic irrationality because quadratic irrationalities are represented by eventually (including pure) periodic sequences $\{a_n\}_{n=1}^\infty$.

REMARK 2   The characterization of the purely periodic continued fraction was quite hard; it was done by Evariste Galois himself! (not to many people no about it but number-theorist -- ok, I am not one of them :) ).

A: Here are my two cents. (Incomplete solution)
The idea is to use the theory of convergents. Given a continued fraction with the representation $[a_0;a_1,a_2,...]$, one may write a series of rational approximations to the truncated continued fraction
$$\dfrac{p_n}{q_n}:=a_0 + \dfrac{1}{a_1 + \dfrac{1}{a_2 + \dfrac{1}{... + \dfrac{1}{a_n}}}}$$
which are given by the recursive relation $A_n = a_nA_{n-1} + A_{n-2}$, valid for $n\ge -1$, where the initial conditions are $p_0 = a_0,p_{-1} = 1$, and $q_0 = 1, q_{-1} = 0$. Substituting $a_n = n^2$ for your continued fraction, it seems that we are looking at the recursive relation, for example for the numerators, given by
$$p_{n} = n^2p_{n-1} + p_{n-2}.$$
Your question reduces to the question of finding a closed form for this discrete ordinary differential equation. I would suggest checking out Lukas' theorem which holds for relations of the form $a_n = \alpha a_{n-1} + \beta a_{n-2}$ such as the Fibonacci sequence for example and many others.
The idea is then that since the set of solutions to the equation
$$A_{n} = n^2A_{n-1} + A_{n-2}$$
form a two dimensional vector space, by applying the initial conditions of $p$ and $q$ you can construct a closed form for the two. Typically, you would have one solution decaying and one going to infinity. Then if you found
$$p_n = \alpha D(n) + \beta E(n),\quad q_n = \gamma D(n) + \delta E(n)$$
where $D(n)$ is the decaying function, then your continued fraction would be precisely equal to
$$\dfrac{\beta}{\delta}.$$
As I said, this is incomplete as the question is what are the solutions to the recursive relations $A_n = n^2A_{n-1} + A_{n-2}$. Comparing with Lucas' Theorem, to solve this equation, you would essentialy want to know how to solve the following ODE
$$f(x) = x^2f'(x) + f''(x).$$
All I can tell you is that it is a Strum-Liouville equation and maybe some analyst can pick up the ball from here :)
