As I previously explained, I am a "hobbyist" mathematician (see here or here); I enjoy discovering continued fractions by using various algorithms I have been creating for several years. This morning, I got some special cases for a rather heavy expression I was working on, finding the beautiful identities: $$ \phi^\phi \;=\; 2 + \displaystyle\frac{2\left(1-1/\phi\right)/\phi} {2\left(2-1/\phi\right) + \displaystyle\frac{3\left(1-2/\phi\right)/\phi} {3\left(2-1/\phi\right) + \displaystyle\frac{4\left(1-3/\phi\right)/\phi} {4\left(2-1/\phi\right) + \displaystyle\frac{5\left(1-4/\phi\right)/\phi} {5\left(2-1/\phi\right) + \;\ddots }}}} $$ Equivalently $$ \phi^\phi \;=\; 2 + \displaystyle\frac{2/\phi\left(\phi-1\right)} {2\sqrt5 + \displaystyle\frac{3\left(\phi-2\right)} {3\sqrt5 + \displaystyle\frac{4\left(\phi-3\right)} {4\sqrt5 + \displaystyle\frac{5\left(\phi-4\right)} {5\sqrt5 + \;\ddots }}}} $$ where $\phi$ is the golden ratio.

I was very excited to have found this $\phi^\phi$ formula and my question is about how original it is: did someone already see it or could I be confident in assuming it is new? Similarly $$ \phi^{2/\phi} \;=\; 2 + \displaystyle\frac{2\left(1-2/\phi\right)} {2\phi + \displaystyle\frac{3\left(1-3/\phi\right)} {3\phi + \displaystyle\frac{4\left(1-4/\phi\right)} {4\phi + \displaystyle\frac{5\left(1-5/\phi\right)} {5\phi + \;\ddots }}}} $$ By using a more compact notation (see at the end of this message for an explanation about how to read it), other identities for $\sqrt{3}$, $\sqrt[3]{4}$, $\sqrt[4]{5}$, etc. would be:

$$ \sqrt[a]{a+1} \;=\;2+ \displaystyle\operatorname*{K}_{n=1}^{\infty} \frac {\left(n+1\right)\left(1-an\right)/\left(a+1\right)} {\left(n+1\right)\left(1+a/\left(a+1\right)\right)} $$

These identities are special case of the following conjectured identity (of course, when asking whether the previous formulae are known or not, it also means the formula below should be new also): $$ \left(\displaystyle\frac{x}{x-1}\right)^{x-1} \;=\;2+ \displaystyle\operatorname*{K}_{n=1}^{\infty} \frac {\left(n+1\right)\left(x-n-1\right)/x} {\left(n+1\right)\left(x+1\right)/x} \tag{1}\label{1} $$ that I found from a more general conjecture being related to the following continued fraction: $$ g\left(k,x\right) = \displaystyle\operatorname*{K}_{n=1}^{\infty} \frac {\left(n+1\right)\left(k-n-k/x\right)/x} {\left(n+1\right)\left(1+1/x\right)} $$ for which I empirically noticed the relation (with $\textrm{B}$ the beta function): $$ \begin{array}{l} 2\;+\; g\left(\alpha,\xi\right) \;+\; g\left(\alpha,\displaystyle\frac{\xi}{\xi-1}\right) \\[8pt] \qquad\qquad =\; \alpha\left(\xi-1\right)^{\alpha/\xi-1}\left(\displaystyle\frac{\xi}{\xi-1}\right)^{\alpha-2} \textrm{B}\left(\alpha/\xi, \alpha-\alpha/\xi\right) \end{array} \tag{2}\label{2} $$ where it is easy to notice that the case $\alpha=\xi/\left(\xi-1\right)$ makes $g\left(\alpha,\xi\right)=0$ leading to an identity containing a single continued fraction (which can later be simplified as above).

The previous notation is the one I use; I find it convenient and it can be found for instance in Continued Fractions with Applications by Lorentzen & Waadeland, but I know that some people don't like it; it has to be read the following way: $$ a_0 + \operatorname*{K}_{n=1}^{\infty} \frac{b_n}{a_n} = a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \ddots}}} $$

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    $\begingroup$ Isn't there some continued fractions database to check these numbers against? Or maybe there are entries for such fractions in the OEIS? As for the algorithms for finding them, I am sure that googling will tell you about "scientific" methods from the literature to find such fractions. If your algorithm is fast and finds nice fractions accurately, you might consider advertising/publishing that (instead of just the "output" of your algorithm). $\endgroup$
    – TMM
    Apr 12, 2017 at 15:09
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    $\begingroup$ +1 I hope it is ok that I simplified your first two cfracs. Nice discovery, by the way. :) $\endgroup$ Apr 13, 2017 at 2:03
  • $\begingroup$ For simplifying cfracs, this post on Continued fraction manipulation is quite useful. $\endgroup$ Apr 13, 2017 at 2:09

1 Answer 1


Set $C_1=2-1/\phi$, then your CF can be writen as ($\phi^2=\phi+1$): $$ \phi^{\phi}=2+\textbf{K}^{\infty}_{n=1}\left(\frac{(n+1)\left(1-n/\phi\right)/\phi}{(n+1)C_1}\right)=\ldots=2+C_1\textbf{K}^{\infty}_{n=1}\left(\frac{\frac{1}{5}(n+1)(\phi-n)}{n+1}\right) $$ Now I use the following Theorem found in [1] p.369 $$ I=\int^{\infty}_{0}\frac{e^{-tz}}{\left(\cosh(t)+a\sinh(t)\right)^b}dt= $$ $$ =\frac{1}{z+ab+}\frac{1\cdot b(1-a^2)}{z+a(b+2)+}\frac{2(b+1)(1-a^2)}{z+a(b+4)+}\frac{3(b+2)(1-a^2)}{z+a(b+6)+\ldots}$$

with $\Re(a)>0$, I set $z=-ab$, $b=-\phi$, $1-a^2=\lambda$, and rewrite the last continued fraction as $$ \textbf{K}^{\infty}_{n=1}\left(\frac{(n+1)(-\phi+n)\lambda}{2a(n+1)}\right)=2 a \textbf{K}^{\infty}_{n=1}\left(\frac{(n+1)(-\phi+n)\lambda/(4a^2)}{(n+1)}\right). $$ Assuming that $\lambda/(4a^2)=-1/5$, then $(1-a^2)/(4a^2)=-1/5$, or equivalent $a=\sqrt{5}$ and hence $z=(5+\sqrt{5})/2$, then from Theorem $$ I\cdot b(1-a^2)-ab-2a-z=\textbf{K}^{\infty}_{n=1}\left(\frac{\frac{(n+1)}{5}(b+n)(1-a^2)}{z+a(b+2n+2)}\right)\\=2a\textbf{K}^{\infty}_{n=1}\left(\frac{\frac{(n+1)}{5}(\phi-n)}{(n+1)}\right)\tag1 $$ But from the conjecture $$ 2+C_1\textbf{K}^{\infty}_{n=1}\left(\frac{\frac{(n+1)}{5}(\phi-n)}{(n+1)}\right)=\phi^{\phi} $$ Hence we can find the next remarkable integral $I$. $$ I=\int^{\infty}_{0}e^{-t(2+\phi)}\left(\cosh(t)+\sqrt{5}\sinh(t)\right)^{\phi}dt=\frac{1}{2}\left(\phi^{\phi}-\frac{1}{\phi}\right)\tag2 $$ The integral and your cf expansion are equivalent.

[1]: H.S. Wall, ''Analytic Theory of Continued Fractions'', Van Nostrand, New York (1948).

Now the part of the proof of the conjecture:

The Euler continued fraction for ${}_2F_1(a,b;c;z)$ is (see [2] below): $$ c\frac{{}_2F_1(a,b;c;z)}{{}_2F_1(a,b+1;c+1;z)}=c+(b-a+1)z+\textbf{K}^{\infty}_{n=1}\left(\frac{-(c-a+n)(b+n)z}{c+n+(b-a+n+1)z}\right) \tag3 $$ Set $c-a=1$, $b=-\phi$, $c-1+(b-a)z=0$ in Euler's expansion, then $$ c\frac{{}_2F_1(a,b;c;z)}{{}_2F_1(a,b+1;c+1;z)}=z+1+\textbf{K}^{\infty}_{n=1}\left(\frac{(1+n)(\phi-n)z}{(n+1)(z+1)}\right)= $$ $$ =z+1+(z+1)\textbf{K}^{\infty}_{n=1}\left(\frac{(1+n)(\phi-n)z/(z+1)^2}{n+1}\right) $$ Solving now the equation $z/(z+1)^2=1/5$, we find $z=\frac{1}{2}(3-\sqrt{5})$, hence $$ c\frac{{}_2F_1(a,b;c;z)}{{}_2F_1(a,b+1;c+1;z)}=\frac{1-\sqrt{5}}{2}+\left(\frac{\sqrt{5}-1}{2}\right)^{-\phi}= $$ $$ =\frac{5-\sqrt{5}}{2}+\frac{5-\sqrt{5}}{2}\textbf{K}^{\infty}_{n=1}\left(\frac{\frac{(1+n)}{5}(\phi-n)}{n+1}\right) $$ From this identity the conjecture follows.

[2]: L.Lorentzen and H.Waadeland, "Continued Fractions with Applications", Elsevier Science Publishers B.V., North Holland, 1992.

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    $\begingroup$ I fixed your number tags. $\endgroup$ Apr 18, 2017 at 11:35
  • $\begingroup$ excellent answer ! $\endgroup$
    – G Cab
    Feb 15, 2020 at 22:11

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