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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 $n\geq 3$.

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    $\begingroup$ $i.$ is almost surely correct. I get the correct result within hundreds of decimal digits. $\endgroup$
    – Peter
    Apr 24, 2020 at 11:58
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    $\begingroup$ 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. $\endgroup$
    – Peter
    Apr 24, 2020 at 12:31
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    $\begingroup$ I could post the pari/gp-code as a partial answer. $\endgroup$
    – Peter
    Apr 24, 2020 at 12:34
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    $\begingroup$ In fact, I got this precision : ... E-35676 , this cannot be a coincidence ! $\endgroup$
    – Peter
    Apr 24, 2020 at 14:12
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    $\begingroup$ Interesting! Found your conjectures i) and ii) here arxiv.org/abs/1907.00205 $\endgroup$
    – Quintec
    Apr 24, 2020 at 14:52

1 Answer 1

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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.

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