What are better approximations to $\pi$ as algebraic though irrational number? I know that $\pi \approx \sqrt{10}$, but that only gives one decimal place correct. I also found an algebraic number approximation that gives ten places but it's so cumbersome it's just much easier to just memorize those ten places.

What's a good approximation to $\pi$ as an irrational algebraic number (or algebraic integer if possible) that is easier to memorize than the number of places it gives correct?

EDIT: Algebraic number preferably of low degree, such as $2$ or $3$ (quadratic or cubic).
(expanding my comments)

Let's start with the fraction $\;\dfrac{355}{113}\,$ easy to remember with something like :
"doubling the odds to be near the pi" (whatever this may mean...).

It is easy to find starting with the continued fraction of $\pi$ and stopping just before the (relatively) large term $292$ :
\begin{align}
\pi&amp;=[3; 7, 15, 1\color{#00ff00}{, 292, 1, 1, 1, 2, 1, 3, 1, 14,\cdots}]\\
\pi&amp;\approx \frac{355}{113}\approx 3.141592\color{#808080}{035}\\
\end{align}

My next step will be to compute the continued fractions of the first powers of $\pi\,$ and stop the expansion before the first large term (as previously) to get :

$$\frac {355}{113},\; \sqrt{\frac{227}{23}},\; \sqrt[3]{31},\;\sqrt[4]{\frac{2143}{22}},\;\sqrt[5]{306},\cdots, \;\sqrt[11]{294204},\cdots$$

After the first power the most interesting term was the fourth :
\begin{align}
\pi^4&amp;=[97; 2, 2, 3, 1\color{#00ff00}{, 16539, 1, 6, 7, 6,\cdots}]\\
\pi^4&amp;\approx \frac{2143}{22}\\
\pi&amp;\approx \sqrt[4]{\frac{2143}{22}}\approx 3.14159265\color{#808080}{258}\\
\end{align}

Mnemonic : think at "three ways to reverse two two" that I'll note $\pi\approx \sqrt{+ \negthickspace+/}$ :


Power way: two $\sqrt{}\;$ to reverse the double squaring $^2$ $^2$.
Incremental way : reverse two times two terms of the $2\times 2$ terms $\;\underbrace{12}\underbrace{34}$
Divide by $\,22$.


This solution is interesting because of the large (omitted) $16539$. Should we incorporate this term in the c.f. then the next numerator and denominator would have around $4$ additional digits (since $\log_{10}(16539)\approx 4.2\;$ and from the method to obtain the next fraction in the first link).
The precision will be better with this supplementary term (say $4.3$ digits more) but we needed $4+4$ more digits for this. Without this term we used $4+2=6$ digits for a result of $10$ digits (excellent), with this term we have $8+6=14$ digits for a result of $14$ digits (average for a c.f.).

Searching the largest terms at the beginning of a c.f. (excluding the first non-zero term) should thus be rather interesting! Unfortunately c.f. coefficients as large as $16539$ are rather uncommon.

This result was found by Ramanujan and is given too by Mathworld with many others.
$$-$$
Some additional results :

A palindrome for the fractional part of $\pi$ : $\frac 1{\large{\sqrt[5]{17571}}}\approx 0.1415926\color{#808080}{48}$ (with two more terms this becomes $\sqrt[5]{\dfrac{296}{5201015}}\approx 0.141592653589\color{#808080}{63}$). Another one : $\;\dfrac 1{\sqrt[8]{6189766}} \approx 0.141592653\color{#808080}{64}$.

We may too search continued fractions $\dfrac{\log\pi}{\log n}\,$ to obtain :
\begin{align}
7^{10/17}&amp;\approx 3.141\color{#808080}{35}\\
6^{23/36}&amp;\approx 3.1416\color{#808080}{09}\\
7^{58701/99785}&amp;\approx  3.1415926535\color{#808080}{9651}\\
\end{align}

Other random solutions perhaps nearer to OP's question (with some usual c.f. for reference) :
\begin{align}
\frac{22}7 &amp;\approx  3.14\color{#808080}{2857}\\
\frac{8.5^2}{23} &amp;\approx  3.141\color{#808080}{30}\\
\sqrt[3]{31}&amp;\approx 3.141\color{#808080}{38}\\
\sqrt{51}-4  &amp;\approx  3.141\color{#808080}{428}\\
\sqrt{4508}-64   &amp;\approx  3.141\color{#808080}{64}\\
4-\sqrt{\frac {14}{19}} &amp;\approx 3.141\color{#505050}{60}\color{#808080}{49}\\
7-\left(\frac{55}{28}\right)^2 &amp;\approx  3.1415\color{#808080}{816}\\
1+\left(\frac{60}{41}\right)^2 &amp;\approx  3.1415\color{#808080}{82}\\
\sqrt{14434}-117 &amp;\approx  3.1415\color{#808080}{83}\\
2+\sqrt[17]{9.5} &amp;\approx  3.14159\color{#808080}{78}\\
5-\sqrt[5]{22+\frac{1}6} &amp;\approx  3.14159\color{#808080}{62}\\
\sqrt{\frac{1961}2}-19 &amp;\approx  3.1415\color{#808080}{898}\\
2+\sqrt[8]{\frac{75}{26}} &amp;\approx  3.141592\color{#808080}{19}\\
\frac{355}{113} &amp;\approx 3.141592\color{#808080}{92}\\
\sqrt[11]{294204} &amp;\approx 3.1415926\color{#808080}{36}\\
\left(\sqrt{\frac {1731}{76}}-3\right)^2 &amp;\approx 3.1415926\color{#808080}{65}\\
\sqrt{6}+\sqrt[3]{\frac {61}{184}}&amp;\approx 3.1415926\color{#808080}{45}\\
\sqrt{35}-\sqrt[3]{\frac{6215}{291}} &amp;\approx 3.14159265\color{#808080}{266}\\
\sqrt[4]{\frac{2143}{22}}&amp;\approx 3.14159265\color{#808080}{258}\\
5-\sqrt[11]{913+\frac 16} &amp;\approx  3.141592653\color{#808080}{37}\\
\sqrt{5}+\sqrt[4]{\frac{2323}{3455}} &amp;\approx 3.141592653\color{#808080}{436}\\
\sqrt{4508-\frac 1{153}}-64 &amp;\approx  3.1415926535\color{#808080}{28}\\
\sqrt[4]{\frac{788453}{95}}-\sqrt{41} &amp;\approx  3.1415926535\color{#808080}{918} \\
\sqrt[4]{\sqrt{\frac{1087906}{63}}-34}&amp;\approx 3.14159265358\color{#808080}{876}\\
\frac{5419351}{1725033}&amp;\approx 3.141592653589\color{#808080}{815}\\
\sqrt{7}+\sqrt[8]{\frac{94680}{25912921}} &amp;\approx 3.141592653589793\color{#808080}{309}\\
\sqrt{\sqrt{\frac{10521363651}{311209}}-174} &amp;\approx 3.141592653589793238\color{#808080}{01}\\
\frac{21053343141}{6701487259}&amp;\approx 3.141592653589793238462\color{#808080}{38}\\
\sqrt{\sqrt{\frac{20448668456155}{3958899937}}-62}  &amp;\approx 3.14159265358979323846264338\color{#808080}{5}\\
\sqrt{12}-\sqrt[3]{\frac{626510899334}{18676834489131}} &amp;\approx 3.1415926535897932384626433832\color{#505050}{80}\color{#808080}{4}
\end{align}

We could too use the integer relation algorithms as in Will Jagy's answer or this one but this seems more cumbersome for this problem.
It is not a low degree polynomial, but easy to remember for sure. 
We know that $$\frac{1}{1+x}=1-x+x^2-x^3+\cdots$$
Substituting $x^2$, $$\frac1{1+x^2}=1-x^2+x^4-x^6+\cdots$$
But we also know that $\int\frac1{1+x^2}dx=\arctan x$. 

So let us integrate both sides (from $x=0$ to $x=y$), $$\arctan y=y-\frac{y^3}3+\frac{y^5}{5}-\frac{y^7}{7}+\cdots$$
Substitute $y=1$ and we get $$\pi=4(1-\frac13+\frac15-\frac17+\cdots)$$.