Continued Fraction using all Perfect Squares

Question

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?

Answer 1

Would you like $$\frac{138064447330372928950478420048463661504907828497126087600678688613823206940 422174}{1110477976182011935299028840335867265791530004627377300038704099001263 79105352933}$$ obtained after $35$ levels. Its decimal representation is $$1.24328847839971564408249654539442949923120026119744688506649745988163 032233825$$ which is not recognized by inverse symbolic calculators but, thanks to a friend of mine who enjoys this kind of problems is "close" to $$\frac{\exp\left(-\frac{10}{11}+\frac{35}{11 e}+\frac{57 e}{11}+\frac{49}{11 \pi }-\frac{18 \pi }{11} \right)\, \pi ^{\frac{4-39e}{11}}}{\sin ^{\frac{9}{11}}(e \pi ) }$$ corresponding to a relative error of $1.72\times 10^{-20} \text{ %}$

Answer 2

This is not an answer, but I cannot fit this in a comment. I'll remove this when it's not useful.

I wrote a code to numerically compute this. Go to https://sagecell.sagemath.org and enter this code:

t = continued_fraction([i ^ 2 for i in range(1, 1 + 500)]) # 500 levels
x = t.value().n(100000) # 100000 significant digits
print (x) # Decimal of 500 levels

Change the value 500 to a larger number for more levels, though you might need to change the $100000$ as well.

First few digits: (matches with the 600th level)

1.24328847839971564408249654539442949923120026119744688506649745988163032233825342145964981561218559538890467348297701381456087507250067681969369422287394481106823198525320924225098067697793784504053388839947325461765969991600416411663928784095005020600017176700932548190583234985348139858549897037249626487595101212224593274130148983922702655057821224816791808680516525373955799210324551761696924319169880868859867394373255507204036873905460245016380602140886971188932788792004534073572611777381862869537164754562257622055586200098962299400115788330686330366349692355506114874528531989148443788939713051110708718763499836700885424205347399404711205719025462322085656886062558109392836051726411871492635286791717082160974321477110902114255736671637367782211494704866808671214003103134264294526344163295510488028137066937962453903624712552782295199344460880013595284144054791164042024679560460069025003399104291075994866032492192680289402243801395246804366648572804594835036565027012956457848519024683621071759138854168892152936131601393045918225297361539050001254865220428685955018761302830097810409603105996267899586104453972012058998671779368289467285453550404957934466484420769438142316194311386546598316496579775081672073060062318502089135148552587389753909412229458433174767777975925507523398094199059397469890781801547075171199998106106018355690780407908234632985702846260738629497384291103636574665555025023044984923355982351253355767746319065904960149568635879463751398856235096803786299543409250013752226144096131959751997496961228513717287891950784266647143422837734191074225842373022792510603871652815370739861465958025081223972785667964221993904920774239800993089897415640045030651976293653978596239688796189356672841973461944224406222485237165933685215338328246965418663851771767016025733608304176229645795762169519518056131950947030623723715156671237757513236314813017765174652854414804459704778983840308236355458482186039701182366583821977275491568743188372032199686289186216846315221546449368928376288608392780822632928349666946023568458883521622471041331229744203333882400906017526808516258953901332656517503602022730289807020281449038190152461911466702688210790580820975642935599011794660455601260001262488904353784063224341997860484130953573461410423517702408264933194376007832029308421380987597133575841852513196469960405614112003845760361969054582688353236843506390988117926540512066533351583073802283028922608906654448662449054997495347325027233999796401254085959204252720843767920207799104869887641909309802874614799948020274500537148542809194708801514801275650396803610116643354860881194033219324043392559634749415341929458682483500024988236829244667869008730885020992781539100350964753606860542286406244822674909909096511713924803554889972628771512229241001266367571171782942501244731950206365757564271293930077461317892638806872857794787118618101894534315239368762309917444762674814034408808667064272939683133604067971556663209543191412897044483453793704784086966014311811427984571183731582044500653171023965354166173726605999750410259987178221411565219833571583646012434645410877998831097126426133022455440361397221606803160676840154875764713859289978683334924874812719239333459437327106481196547663762181322316890734498257255023063639045876242543382646774397248740883774622532481200089909953414322445317902407816185139510058816237228001451320678273320698929989619182317548214164841576919852284035002961368185099285222958366993365832170685992893264181846279722383519726362574882868847034823224829315592178714910265667288543693631300523569063128272379356587365322352703148911926725468239164802277365567194880593573754388524351582684345789895422232160786171441187363773334905390848403047587481578926134511772814005757130528683677651081667676932627827576617665022573874410745390880341108662804221361905655381031435163999823001657833327755210242936743621991472569045947564685591887622152883580360992431656812762677475894659540787689593277383518139895014434797569451537179920009707930708746075282453855994960954232820746522097507969820638141164117432428361090940429960665544051821446205554704164805241254378653682652254723855019850569931182778028110911895537501575145896862244139982948494435308482949491662745002005196944018205353794354017083085996488785569081915969582357820907666013388692800978962140796220310895633540740286312820178552956015336845371534604133185162559333014055080974211851967198670088950269832159882829258495227547916502961548819794467361143655606143125526916063299389171634645316974758731525019287

Time to think of the math :D

Answer 3

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 :) ).

Answer 4

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 :)