A conjectured continued fraction for $\phi^\phi$

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.