I was "playing with $\pi$" trying to look at it in different numeral systems and it's not so hard to obtain $\pi$ base $2$ or $3$ or even $\varphi=\frac{\sqrt{5}+1}{2}$, using Maclaurin series of $\tan^{-1}$ at $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{3}}$ and $\cos^{-1}$ at $\frac{\sqrt{5}+1}{4}$ respectively, to an arbitrary precision.
In base 3 there are at least two systems of digits: $\{0,1,2\}$ and $\{0,1,-1\}$ and the latter looks "more symmetric" to me.
So I wonder

if there's $a_n$ such that $\pi=\sum\limits_{n=-1}^{\infty} a_ne^{-n}$ where $a_n\in\{0,1,-1\}\ \forall n$
or how to obtain these $a_n$ by computing.

The few values of $\pi_e$ (i.e. $a_n$) might be 1,0;1,0,1,0,1,-1,1,1,1,0,1,0,-1,0,0,-1,1,0,-1,1,0,-1,0,-1,0,-1,0,-1,0,0,1,...
(yes, this representation is ambiguous).
But I'm stuck to find a $\pi_e$ as a series, maybe I should rearrange $e^x$ Maclaurin series somehow?
P.S. I've found $\pi$ base $e$ using $\{0,1,2\}$ digits https://oeis.org/A050948 but it doesn't help much.

  • $\begingroup$ What do you mean by $\pi$ base 2? or 3? $\endgroup$ Commented Dec 18, 2015 at 0:09
  • $\begingroup$ Have you tried using permutation? I'm not good enough with functions/summations to give you an answer. $\endgroup$ Commented Dec 18, 2015 at 0:15
  • 1
    $\begingroup$ The best I was able to do was: $$\pi-e-\frac{1}{e}-\frac{1}{e^3}-\frac{1}{e^5}+\frac{1}{e^6}-\frac{1}{e^7}-\frac{1}{e^8}-\frac{1}{e^9}-\frac{1}{e^{11}}+\frac{1}{e^{13}}+\sum _{k=17}^{\infty } \frac{1}{e^k}=-9.5070539522 \cdot 10^{-9}$$ $\endgroup$
    – Yuriy S
    Commented Jan 16, 2017 at 23:31


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