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The following are some conjectures of mine that I have discovered empirically. The last three conjectures are true if the first four are true, and vice versa.

i. $$e=3-\cfrac{1}{4-\cfrac{2}{5-\cfrac{3}{6-\ddots}}}$$ ii. $$\cfrac{e}{e-2}=4-\cfrac{1}{5-\cfrac{2}{6-\cfrac{3}{7-\ddots}}}$$ iii. $$\cfrac{e}{2(3-e)}=5-\cfrac{1}{6-\cfrac{2}{7-\cfrac{3}{8-\ddots}}}$$ iv. $$\cfrac{e}{3(3e-8)}=6-\cfrac{1}{7-\cfrac{2}{8-\cfrac{3}{9-\ddots}}}$$ v. Let $c_1(x)=5-\cfrac{1}{6-\cfrac{2}{7-\ddots}}$ and $c_2(x)=4-\cfrac{2}{5-\cfrac{3}{6-\ddots}}$. Then, $$\cfrac{c_1(x)}{c_2(x)}=\cfrac e2$$ vi. Let $c_3(x)=6-\cfrac{1}{7-\cfrac{2}{8-\ddots}}$ and $c_4(x)=5-\cfrac{2}{6-\cfrac{3}{7-\ddots}}$. Then, $$\cfrac{c_3(x)}{c_4(x)}=\cfrac e{3(e-2)}$$ vii. $$\cfrac{c_1(x)c_4(x)}{c_2(x)c_3(x)}=3\bigg(\cfrac e2-1\bigg)$$

Can these conjectures be proven/disproven, particularly either the first four or last three? If they are true, it appears the function $$f(n)=n-\cfrac{1}{n+1-\cfrac{2}{n+2-\cfrac{3}{n+3-\ddots}}}$$ is expressed through $e$, at least seemingly for natural $n\geq 3$.

metamorphy
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Mr Pie
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    $i.$ is almost surely correct. I get the correct result within hundreds of decimal digits. – Peter Apr 24 '20 at 11:58
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    I established the recursion for $i.$, but I could only numerical approve that the result is actually $\ e\ $. The precision is so high that there is no doubt anymore that $i.$ is actually correct. – Peter Apr 24 '20 at 12:31
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    I could post the pari/gp-code as a partial answer. – Peter Apr 24 '20 at 12:34
  • @Peter thanks heaps! Yes, that would glady be a valid answer that I'm looking for. I just found a site that verifies the first two cfracs after looking at ur comment, so now it boils down to the next two or so. But thank you so much for coding it - hundreds of decimal digits sounds pretty damn convincing! – Mr Pie Apr 24 '20 at 13:08
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    In fact, I got this precision : ... E-35676 , this cannot be a coincidence ! – Peter Apr 24 '20 at 14:12
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    Interesting! Found your conjectures i) and ii) here https://arxiv.org/abs/1907.00205 – Quintec Apr 24 '20 at 14:52
  • Have you seen this already? – J. M. ain't a mathematician Apr 27 '20 at 01:13
  • @J.M.isn'tamathematician thanks to you, I have now! Me and a friend have also conjectured a closed form for $f(n)$ :) – Mr Pie Apr 27 '20 at 09:29
  • vii follows trivially from v and vi. – 2'5 9'2 Apr 09 '22 at 16:14
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    Once I fiddled with this as well and found a nice table of related such results. At the moment I'm not at home and cannot put a .pdf file to my webspace; an older .htm file however is accessible. See https://go.helms-net.de/math/divers/GenContFracRationalE.htm (Hope, that the internal links to formulae work, otherwise one has to wait a couple of days until I'm back home. – Gottfried Helms Jun 10 '22 at 14:30

1 Answers1

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For $z\in\mathbb{C}$ with $\Re z>1$ we have (as a particular case of this) $$\cfrac1{z-\cfrac1{z+1-\cfrac2{z+2-\cfrac3{z+3-\ddots}}}}=I(z):=\int_0^1(1-x)^{z-2}e^{-x}\,dx$$ so that $f(n)=1/I(n)$; let's get it an alternative "generating function" way.

Fix any $z\in\mathbb{C}$; then the LHS equals $\lim\limits_{n\to\infty}P_n/Q_n$, where $R_n:=\big(P_n,Q_n\big)$ are defined by $$R_0=(0,1),\quad R_1=(1,z),\quad R_{n+1}=(z+n)R_n-nR_{n-1}\qquad(n>0)$$ Let $R_n=n!F_n$, then $(n+1)F_{n+1}=(z+n)F_n-F_{n-1}$ (for $n>0$).

Here, it's easy to show that $F(t)=\sum_{n=0}^\infty F_n t^n$ converges for $|t|<1$ and satisfies $$F'(t)-F_1=z\big(F(t)-F_0\big)+tF'(t)-tF(t).$$ The solution of this ODE, with the obvious condition $F(0)=F_0$, is $$F(t)=(1-t)^{1-z}e^t\left(F_0+(F_1-zF_0)\int_0^t(1-\tau)^{z-2}e^{-\tau}\,d\tau\right).$$

Now the behavior of $F_n$ as $n\to\infty$ follows from the singularity analysis of $F(t)$ as $t\to1$ (a good reference here is Analytic Combinatorics by P. Flajolet and R. Sedgewick); for $\Re z>1$, the integral converges at $t=1$, and the result of this analysis is that $$\lim_{n\to\infty}n^{-z}F_n=\frac1{\Gamma(z-1)}\lim_{t\to1^-}(1-t)^{z-1}F(t)=e\frac{F_0+(F_1-zF_0)I(z)}{\Gamma(z-1)}.$$ Recalling that $F_0=(0,1)$ and $F_1-zF_0=(1,0)$, we get the claimed $\lim\limits_{n\to\infty}P_n/Q_n=I(z)$.

The case of integer $z$ leads to an expression using subfactorials: $$I(n+2)=(-1)^n n!\left(-\frac1e+\sum_{k=0}^n\frac{(-1)^k}{k!}\right)=(-1)^n(!n-n!/e)\underset{n>0}{=}\langle\!\langle n!/e\rangle\!\rangle,$$ where $\langle\!\langle x\rangle\!\rangle=\big|x-\lfloor x\rceil\big|$ is the distance from $x$ to the nearest integer.

metamorphy
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