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Yesterday I posted this conjecture, but then deleted it thinking it was false. Turns out Python doesn't define $a^b$ as a^b, but rather as a**b.


Conjecture: Denote by $G$ Catalan's constant, then $$G=\cfrac{1}{1+\cfrac{1^4}{8+\cfrac{3^4}{16+\cfrac{5^4}{24+\cfrac{7^4}{32+\cfrac{9^4}{40+\ddots}}}}}}$$

Given the connection $G$ has with the number $8$ shown here, as well as this continued fraction reaching nearly the first five decimal places of $G$ after around $200$ iterations (vinculums), I am confident this is true.

However, I do not know how to code a continued fraction on Python or Pari/GP (a friend of mine gave it a go, but also to no avail) up to an iteration $n$ without having to write it out manually, which is really tedious. Here is some python code from a friend, coding this fraction up to $12$ iterations to be $\approx 0.9151$, reaching the first three decimal places of $G$.

The only 'local' behaviour that I can say about continued fractions is that most of them are convergent, and that they all converge via oscillation at each iteration. But, more importantly, I'd like to know that if this be true, can it be shown from here that $G$ is irrational (or even transcendental, if you are willing)? I am aware this is an unsolved problem, which was what inspired me to write $G$ in another closed form.

Any thoughts?

Thank you in advance.

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    $\begingroup$ This CF can be found in D. Bowman and J. Mc Laughlin, Polynomial continued fractions, Acta Arithmetica 103 (4) 2002, 329–342, see the bottom of page 2 here: wcupa.edu/sciences-mathematics/mathematics/jMcLaughlin/… $\endgroup$ – Gary Apr 11 at 11:16
  • $\begingroup$ @Gary D'oh! Another continued fraction of mine, rediscovered once again! Guess I just have to think of something crazier. Thanks Gary for your comment. You can post that as an answer if you like :) $\endgroup$ – Mr Pie Apr 11 at 11:18
  • $\begingroup$ Following the pattern $\zeta(3)$ in Mathematica Speak is 8/7 (1/(1 + ContinuedFractionK[-(2 n - 1)^6, (2 (n + 1) - 1)^3 + (2 (n) - 1)^3, {n, 1, [Infinity]}])) $\endgroup$ – James Arathoon Apr 11 at 13:59
  • $\begingroup$ "most of them are convergent": all continued fractions are convergent. $\endgroup$ – TonyK Apr 13 at 0:11
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    $\begingroup$ Allow me to suggest to unaccept the currently accepted answer and then accept @EulerwasHere's answer $\endgroup$ – Ewan Delanoy Apr 18 at 11:56
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The accepted answer is misleading. The continued fraction may well be found in that reference, but this is not a result from 2002, but rather a trivial consequence of Euler's continued fraction formula from 1748. You should take a look at the wikipedia page:

https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula

Euler's continued fraction formula $$a_0 + a_0 a_1 + \ldots + a_0 \cdots a_n = \frac{a_0}{\displaystyle{1 - \frac{a_1}{\displaystyle{1 + a_1 - \frac{a_2}{\ldots (1 + a_{n-1}) - \frac{a_n}{1 + a_n}}}}}}$$

Now exactly as in the worked example in the wikipedia page for $\tan^{-1}(x)$, you get the completely formal identity: $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}} {(2n+1)^2} = x + x \left(\frac{-x^2}{3^2}\right) + x \left(\frac{-x^2}{3^2}\right) \left(\frac{-3^2 x^2}{5^2}\right) + x \left(\frac{-x^2}{3^2}\right) \left(\frac{-3^2 x^2}{5^2}\right)\left(\frac{-5^2 x^2}{7^2}\right)+ \ldots$$ $$=\frac{x}{\displaystyle{1 + \frac{x^2}{\displaystyle{9 - x^2 + \frac{(9x)^2}{25 - 9 x^2 + \displaystyle{ \frac{(25 x)^2}{49 - 25 x^2 + \ldots }}}}}}}$$

The case $x=1$ is your example. You can plug in $x=i$ if you want to get a continued fraction for $\pi^2/8$.

There are literally thousands of completely trivial continued fractions on can create in this way; take any infinite sum and just formally write out the corresponding Euler continued fraction, clearing denominators in the obvious way. None of those should be considered anything more than a corollary of Euler's result (given the evaluation of the initial sum). Of course, in this case, the evaluation of the initial sum is that it is $G$ by definition.

(And no, this doesn't give anywhere near good enough convergents to say anything about the rationality or otherwise of $G$.)

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    $\begingroup$ The Eulerian CF is definitely a very interesting thing; see e.g. this older thread. $\endgroup$ – J. M. isn't a mathematician Apr 12 at 1:02
  • $\begingroup$ Perhaps I should accept this answer instead. I will wait a couple days until I am able to give this a bounty reward for its very clean and thorough answer, subsequent to giving it the checkmark. Very well done! :) $\endgroup$ – Mr Pie Apr 12 at 2:28
  • $\begingroup$ However, are you sure substituting $x=i$ would yield a real number such as $\pi^2\div 8$? This would inevitably mean that $\pi^2\cdot (8i)^{-1}$ is a real number, which it isn't. $\endgroup$ – Mr Pie Apr 12 at 2:30
  • $\begingroup$ @Mr, I think his mention of the arctangent is making it more complicated than necessary. All you need to apply the Eulerian CF is that $$-\prod_{k=0}^n \left(-\left(\frac{2 k-1}{2 k+1}\right)^2\right)=\frac{(-1)^{n+2}}{(2 n+1)^2}$$ and note the correspondences in Euler's formula. $\endgroup$ – J. M. isn't a mathematician Apr 12 at 6:18
  • $\begingroup$ Substituting $x = i$ the LHS is $i \cdot \pi^2/8$ and the RHS is $i$ times a real continued fraction. @J.M.isn'tamathematician the point of referencing arctan is that it has almost the identical Taylor series and that specific example is worked out on the wikipedia page. Of course it only has to do with the ratios of the coefficients but it helps to point to a particular place on the wikipedia page where the OP can look. $\endgroup$ – EulerWasHere Apr 12 at 15:47
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This continued fraction can be found in the paper D. Bowman and J. Mc Laughlin, Polynomial continued fractions, Acta Arithmetica 103 (4) 2002, 329–342. See the bottom of page 2 here: https://www.wcupa.edu/sciences-mathematics/mathematics/jMcLaughlin/documents/4paper1.pdf

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  • $\begingroup$ How did you find this source, incidentally? I tried doing an internet search of something along the lines of "Catalan's constant continued fraction" and nothing like mine showed up. $\endgroup$ – Mr Pie Apr 11 at 11:22
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    $\begingroup$ I did the same and found first the paper arxiv.org/pdf/1806.03346.pdf where the CF appears (p. 9, eq. (4)) and is attributed to Bowman and Laughlin. $\endgroup$ – Gary Apr 11 at 11:30
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    $\begingroup$ Man, I should have researched more thoroughly. $\endgroup$ – Mr Pie Apr 11 at 11:34
  • $\begingroup$ I must admit I'm having trouble understanding how you transform Eq. 1.4 in D. Bowman and J. Mc Laughlin to the continued fraction for Catalan's Constant. i.e. in Mathematica [1/(1 + ContinuedFractionK[(2 n - 1)^4, (2 (n + 1) - 1)^2 - (2 (n) - 1)^2, {n, 1, [Infinity]}])] $\endgroup$ – James Arathoon Apr 11 at 14:12
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    $\begingroup$ @JamesArathoon In the document it converts the series $G=\sum\limits_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}$ into a continued fraction (cfrac) by Euler's continued fraction. It then uses some algebra to rewrite the result as a simpler looking continued fraction which, in terms of cfracs, is referred to as an "equivalence transform" because the actual result doesn't change (e.g. like simplifying $6/10$ to $3/5$). To explicitly write this out is some work, as the entire process involves lots of substitution and is messy. $\endgroup$ – Mr Pie Apr 11 at 14:45
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Too long for a comment.

However, I do not know how to code a continued fraction on Python or Pari/GP (a friend of mine gave it a go, but also to no avail) up to an iteration $n$ without having to write it out manually, which is really tedious. Here is some python code from a friend, coding this fraction up to $12$ iterations to be $\approx 0.9151$, reaching the first three decimal places of $G$.

To address this part, in Python you can do quickly something like this:

from fractions import Fraction

n = 5
a = [1] + [8*(i + 1) for i in range(n)]
b = [1] + [(2*i + 1)**4 for i in range(n)]

x = Fraction(0, 1)
for ai, bi in zip(reversed(a), reversed(b)):
    x = bi / (ai + x)
print(x, float(x))
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  • $\begingroup$ There are many, many users on Math.SE... but it seems like the universe attracts me to the same users nearly all the time XD $\endgroup$ – Mr Pie Apr 11 at 14:48
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    $\begingroup$ @MrPie I know what you mean :) Typically if questions have similar tags, it will lure similar people who have those in watched. $\endgroup$ – Sil Apr 11 at 16:09
  • $\begingroup$ @MrPie : I have also noticed fairly strong time-of-day bias in the set(s) of common respondents. $\endgroup$ – Eric Towers Apr 11 at 19:40

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