Approximate pi out of small numbers

Question

Given the numbers 1, 2, ..., n the puzzle is to try to make a number as close as possible to pi using only the four mathematical operations +, -, *, / and parentheses (brackets). You don't have to use all the numbers and you are not allowed to stick digits together to make new numbers for example.

• So if n = 1 the best answer is 1, clearly.
• For n = 2 it is 3. For n = 3 it is also 3 and for n = 4 it is 3 + 1/(2*4) = 3.125.
• For n = 8 you can do 3 + 1/(7 + 2/(4 * 8))= 355/113.

Give your best answer for n = 1 up to 20. That is, give a possibly different answer for each n up to 20.

Answer 1

Answer for n=20:

3 + 1/(7 + (6-5)/(15 + 2/((10-8) + (11-9)/(20*14+12+(17+16)/(13*4)))))

$$p = 3 + \cfrac{1}{ 7 + \cfrac{6-5}{ 15 + \cfrac{2}{ (10-8) + \cfrac{11-9}{ (20 \times 14 + 12) + \cfrac{17+16}{13 \times 4} } } } } $$

Uses all numbers except 18 and 19. Accuracy:

$$p = 3.141592653589\;81538324\cdots \\ \pi =3.141592653589\;79323846\cdots$$

Edit: Got the same fraction for n=19:

3 + 1/(7 + (10-9)/(15 + 2/((8-6) + (14-12)/(18*16+5-19/(13*4)))))

I got this expression using

the continued fraction expansion of $\pi$, truncated at 12 terms, i.e. $[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1]$. The next term is $14$, which makes $p$ accurate at its level.

Answer 2

n = 1: 1                                                        (error = 2.14e0)
n = 2: 1 + 2 = 3                                                (error = 1.42e-2)
n = 3: 3                                                        (error = 1.42e-1)
n = 4: 3 + 1 / (2 * 4)               = 25/8    = 3.125          (error = 1.66e-2)
n = 5: 3 + 1 / (2 + 5)               = 22/7    = 3.14285...     (error = 1.26e-3)
n = 6: 3 + (1 / 6 + 2 / 5) / 4       = 377/120 = 3.141666...    (error = 7.40e-5)
n = 7: (same as n = 6)
n = 8: 3 + (2 * 8) / (6 * 4 * 5 - 7) = 355/113 = 3.141592920... (error = 2.67e-7)
n = 9: (same as n = 8)

Answer 3

Making a community wiki so all of the answers are kept in one singular place.

$n$	Formula	Exact Value	Error	Credit
1	$1$	$1$	$2.14\text{E-}0$	Simd, 2012rcampion
2	$1+2=3$	$3$	$1.42\text{E-}1$	Simd, 2012rcampion
3	$3$	$3$	$1.42\text{E-}1$	Simd, 2012rcampion
4	$3+\frac1{2\times4}=3.125$	$\frac{25}{8}$	$1.66\text{E-}2$	Simd, 2012rcampion
5	$3+\frac1{2+5}$	$\frac{22}{7}$	$1.26\text{E-}3$	2012rcampion
6	$3+\frac{\frac16+\frac25}4$	$\frac{377}{120}$	$7.4\text{E-}5$	2012rcampion
7	$3+\frac{\frac16+\frac25}4$	$\frac{377}{120}$	$7.4\text{E-}5$	2012rcampion
8	$3+\frac{2\times8}{6\times4\times5-7}\approx3.14159292$	$\frac{355}{113}$	$2.67\text{E-}7$	Simd, 2012rcampion
9	$3+\frac{2\times8}{6\times4\times5-7}\approx3.14159292$	$\frac{355}{113}$	$2.67\text{E-}7$	2012rcampion
10	$3+\frac{1}{7+\frac{10}{4\cdot5\cdot8-\frac{2}{6\cdot9}}}$	$\frac{95828}{30503}$	$2.33\text{E-}8$	Tom Sirgedas
11	$\frac{11}{4}+\frac{2}{5+\frac{6+\frac{1}{9\cdot10-3}}{7\cdot8}} $	$\frac{312689}{99532}$	$2.91\text{E-}11$	Tom Sirgedas
12	$3+\frac{1}{7+\frac{2}{4\cdot8-\frac{11}{9\cdot10(6+12)-5}}} $	$\frac{1146408}{364913}$	$1.61\text{E-}12$	Tom Sirgedas
13	$3+\frac{5}{(6+11)2}-\frac{4}{(\frac{1}{7}+8)9\cdot10-\frac{13}{12}}$	$\frac{6565759}{2089946}$	$2.99\text{E-}13$	Tom Sirgedas
14	$3+\frac{2}{14+\frac{1}{8-\frac{9}{11(4(10\cdot12)+\frac{7}{13+\frac{6}{5}})}}}$	$\frac{133190959}{42395999}$	$5.20\text{e-}15$	Tom Sirgedas
15	$3+\frac{1}{2+15}+\frac{\frac{12}{7+\frac{10}{6\cdot8(14-\frac{9}{13})}}-\frac{4}{5}}{11}$	$\frac{85563208}{27235615}$	$8.60\text{e-}16$	Tom Sirgedas
16
17
18	$\frac{9+16+ (5*10 + (17/4)/(7*14-2/(12-6/(1+3+15)) ))/(13*(18+11))}{8}$	$\frac{411557987}{131002976}$	$1.9\text{e-}17$	Retudin
19
20	$3 + (11 + 6) / 12 / 10 + 18 / ((4 + 15) / 17 / 13 - ((20 * 5 * 16 * 19 - (1 / (2 / 7 - (9 + 14)))) * 8))$	$\frac{644339207239}{205099539720}$	$5.37\text{e-}18$	Dmitry Kamenetsky

Answer 4

Even better:
85563208 /27235615 (error 8.6E-16)
80143857/25510582 (error 5.4E-16)

3;7,15,1,292,1,1,1,2,1,3,1,15 = 85563208/(5 * 11 * 17 * 29129) error 8.6e-16
= 16/5+2/11-(4+2442/29129)/17
= 16/5+2/11-(4+(8+3)(1812+6)/(15+1)141013+9)/17
solution for 18

3;7,15,1,292,1,1,1,2,1,3,1,14 = 80143857/(2 * 31 * 479 * 859) error 5.4e-16
= 7/2 - 190 * (28+9/859)/(31 * 479)
= 7/2 - 19 * 10 * (11+17+9/(16x18x3-5)) / (15+16)/(6 * 8 * 10-1)
solution for 19 (14 unused)

= 16/5 + 2/11 +1/17 - 8711/29129
with:
8711 = (3x15x(14+6x(12+18)-19)
29129 = (4x7x8x10x13+9)

Better answer for n=19(+20) and 15(+):

$3 + \frac{9*13*7*(2*6*(19*11+8) +17)}{ 16*(1+12*14*15*(18*5+4))}$
$= 42208400/13435351$
error 1.61E-14

I looked at Bubbler's answer and the list supplied by ThePuzzler- or rather APuzzler to see if continued fractions can be beaten using a single divide of the form 3+a/b.
I tried 15160384 = 16*236881 because of the 16 where 236880 turned out to have a lot of divisors.

$3 + \frac{14*(2*15+11-6)-13}{ 5*9*10*12*(4+1)}$
$=84823/27000$
error 6.1E-8

Answer 5

Results

Exhaustive Search (n=10)

$3+\frac{1}{7+\frac{10}{4\cdot5\cdot8-\frac{2}{6\cdot9}}} \approx\pi -2.33\text{e-}08$

see 3+1/(7+10/((4*5*8)-2/(6*9))) on WolframAlpha

optimal:                        value          error
3+1/(7+10/((4*5*8)-2/(6*9)))    95828/30503    2.33e-08

((1/((10/((4*(5*8))-(2/(6*9))))+7))+3)   err=2.33567e-08  95828/30503
(4-(6/(7-(1/(9*((2/5)+((3/8)+10)))))))   err=5.96312e-08  85178/27113
((((4/((2*5)+(1/(6*10))))+(7/8))/9)+3)   err=6.22542e-08  135943/43272
((1/((5/(((4+6)*8)-((2/9)/10)))+7))+3)   err=8.13969e-08  79853/25418
(((1-(4/(9*(5*((2/(6*8))+10)))))/7)+3)   err=8.29516e-08  79498/25305
((1/((2/((4*8)-(5/(6*(9*10)))))+7))+3)   err=9.59077e-08  76658/24401

Smart Search (n=11) (nearly exhaustive?)

$\frac{11}{4}+\frac{2}{5+\frac{6+\frac{1}{9\cdot10-3}}{7\cdot8}} \approx\pi +2.91\text{e-}11$

((2/((((6+(1/((9*10)-3)))/8)/7)+5))+(11/4))   err=2.91434e-11  312689/99532
((10/(((9-((2/(7+((5/11)/4)))/6))*8)-1))+3)   err=1.22356e-10  208341/66317
((8/((1/(((4*(5*6))-(3/(10*11)))+9))+7))+2)   err=2.74029e-10  312334/99419
(((1/(2+5))-(6/((8*(9*((10-4)*11)))-7)))+3)   err=3.31628e-10  104348/33215
(((1-((10/(7+(4/(11-(5/8)))))/9))/6)+3)       err=5.77891e-10  103993/33102

Smart Search (n=12) (nearly exhaustive?)

$3+\frac{1}{7+\frac{2}{4\cdot8-\frac{11}{9\cdot10(6+12)-5}}} \approx\pi +1.61\text{e-}12$

((1/((2/((4*8)-(11/((9*(10*(6+12)))-5))))+7))+3)   err=1.61071e-12  1146408/364913
((1/(((5/(8-(2/(9*((11*12)-(6/4))))))/10)+7))+3)   err=8.71525e-12  833719/265381
(6/(((2+4)/((3/(10+(1/(8*(11*12)))))+5))+(7/9)))   err=1.8701e-11   3022542/962105
((((8/((((9+(2/(6+(10*11))))/12)/7)+5))-1)/4)+3)   err=2.91434e-11  312689/99532
((1/((2/((4*8)-(9/((6/5)+(10*(11*12))))))+7))+3)   err=3.14357e-11  521030/165849
((2/((5+10)-((7-(1/(9*((4*(11*12))-6))))/8)))+3)   err=4.80691e-11  1667793/530875
((12/((7/(10-(2/((1/(4*5))+6))))+3))-(8/(9*11)))   err=5.24358e-11  1355104/431343
((1/(((2/(8-(10/(11*((5*(9*12))-6)))))/4)+7))+3)   err=5.94227e-11  1042415/331811
((1/((2/((4*8)-(9/((6+(10*(11*12)))-5))))+7))+3)   err=7.18372e-11  937712/298483
(((5/(7-(11/((4*10)-(3/8)))))-(2/(9+(1/12))))*6)   err=7.23981e-11  729726/232279
((2-((11/((4/(7*9))+12))/10))*((5/3)-(1/(6*8))))   err=8.1021e-11   1146053/364800
((3/(8/9))-(7/((6-(1/(((4*11)-(2/10))*12)))*5)))   err=9.28817e-11  1980837/630520

Smart Search (n=13) (70%?? exhaustive)

$3+\frac{5}{(6+11)2}-\frac{4}{(\frac{1}{7}+8)9\cdot10-\frac{13}{12}} \approx\pi +2.99\text{e-}13$

((((5/(6+11))/2)-(4/((((1/7)+8)*(9*10))-(13/12))))+3)   err=2.9976e-13   6565759/2089946
(((13-(1/(10*((7*12)-(5/11)))))/4)-(8/(2+(6*(3+9)))))   err=4.04121e-13  4272943/1360120
((1/((((((9-2)/(12*((11*13)-6)))/4)+5)/(8*10))+7))+3)   err=6.01741e-13  11672421/3715447
(((2/(((11/5)+12)*((1/10)+13)))+(7/((6*9)-(4/8))))+3)   err=1.14264e-12  3126535/995207
((1/((5/(((13-((11/(9+(8*(10*12))))/4))*6)+2))+7))+3)   err=1.61071e-12  1146408/364913
(4-(6/(7-((11/(12-((3/(5+(8*13)))/2)))/((9*10)-1)))))   err=1.76081e-12  5106662/1625501
((((9/((10+(2/(8-(((7/12)/6)/4))))+(5/11)))+1)/13)+3)   err=2.28573e-12  11047043/3516383
((1/(((((5/(6+(11*((9*(10*12))-13))))+2)/4)/8)+7))+3)   err=2.64322e-12  8337545/2653923
(((1-((4/(5-((12-(13/(11-(2/(8*10)))))/9)))/7))/6)+3)   err=2.73692e-12  1980127/630294
((((((2/((9/8)+((7/(10*13))/5)))+1)/12)+6)/(4*11))+3)   err=3.36842e-12  4898321/1559184
((1/((2/((4*8)-(5/(9+(6*((10*12)+(11/13)))))))+7))+3)   err=3.47278e-12  6774100/2156263
((1/(((((9/(8-(4/((6*(11*13))-12))))+2)/5)/10)+7))+3)   err=3.90532e-12  3751913/1194271
((1/((2/((4*8)-(9/((10*(11*12))+((13-5)/6)))))+7))+3)   err=4.50839e-12  2813846/895675
(7/((2-(1/(5*((3*6)+(10+(11/(12*13)))))))+(4/(8+9))))   err=4.91474e-12  2605505/829358
((8/((1/(11*((7*(12*13))-(2/9))))+3))+((6-(5/4))/10))   err=4.93872e-12  40748593/12970680

note: my code prints the error out as 2.9976e-13 instead of 2.995199e-13 because double-precision math is accurate to only ~16 digits.

Smart Search (n=14) (and k=8)

$3+\frac{2}{14+\frac{1}{8-\frac{9}{11(4(10\cdot12)+\frac{7}{13+\frac{6}{5}})}}}\approx\pi-5.20\text{e-}15$

((2/((1/(8-(9/(11*((4*(10*12))+(7/((6/5)+13)))))))+14))+3)   err=4.88498e-15  133190959/42395999
((((2/(((9-(6/((8+11)*13)))*14)-(1/12)))+(7/5))/10)+3)       err=8.88178e-15  58466453/18610450
(2/((((3/(5*((9*((8*12)-14))-(1/11))))+(7/4))/10)+(6/13)))   err=2.08722e-14  42208400/13435351
((10+((3+13)/(6+((9/(5+14))/8))))/(4+(2/(7*((1/11)+12)))))   err=2.17604e-14  5419351/1725033
((10/3)-(11/(((7+((6+(1/13))/((5/9)+(2*14))))*8)-(4/12))))   err=2.22045e-14  5419351/1725033
((1/(((2/(4-(13/(((11+(9*(10*12)))*14)-5))))/8)+7))+3)       err=2.22045e-14  5419351/1725033
((2/((((7/((4*(6*(10*(5+(11*12)))))+(1/14)))+9)/8)+13))+3)   err=2.22045e-14  163414249/52016371
((((7/((10*(8-(1/11)))+(((13/14)/12)/4)))+6)/((5*9)-2))+3)   err=2.39808e-14  78997449/25145669
((8/(((1/(10-(((3/(4*6))+12)/((11*14)-(5/9)))))/13)+7))+2)   err=3.55271e-14  31369698/9985285
((2/(((((6/(11*(((4*10)-(7/12))*(14-1))))/5)+9)/8)+13))+3)   err=3.81917e-14  120059441/38216107
((2/((1/(8-(10/((6+(9*((4+((7/11)/5))*12)))*13))))+14))+3)   err=4.13003e-14  57320045/18245537
((6/(11*(12+(5/((10*(9+(8*14)))-7)))))+(3+((1/4)-(2/13))))   err=4.79616e-14  25950347/8260252
(((1-((5/((8*(6+(12/(13+(9/(10*(2+14)))))))-4))/11))/7)+3)   err=4.79616e-14  25950347/8260252
(((9-(((5-(1/(10*(12*((4/7)+14)))))/8)/(6+11)))/3)+(2/13))   err=4.88498e-14  203955211/64920960
((1/((6/((8*12)-((9/(10*(11+(2/((13*14)-5)))))/4)))+7))+3)   err=5.50671e-14  55339918/17615243
(4-(5/(6-((14/8)/(10-(2/(3+(11*(12+(1/((7+9)*13)))))))))))   err=6.66134e-14  20530996/6535219
((2/((5/((4*10)-(12/((((6/7)+9)*(11*13))-(1/8)))))+14))+3)   err=7.41629e-14  140069407/44585477
((2/((9/(8-(1/(12*((4*(10*11))+(7/(5+14)))))))+13))+3)       err=7.99361e-14  35642641/11345405
((1/((5/((8*10)-((9/((12-2)*(6+14)))-(4/(11*13)))))+7))+3)   err=8.52651e-14  50754286/16155591
((10/(((1-((6/((8*12)+(4/(9*13))))/5))/14)+7))+(2-(3/11)))   err=9.14824e-14  96089221/30586149  
((((2/((1/(10*(12-(11/((6*(13*14))-4)))))+5))+(7/8))/9)+3)   err=9.63674e-14  73890787/23520168
((1/(8-(((2/11)+12)/(13-(4/((5/7)+(10*((6*9)+14))))))))+3)   err=9.81437e-14  15111645/4810186

note: more rounding errors, and now duplicated fractions

(n=15)

$3+\frac{1}{2+15}+\frac{\frac{12}{7+\frac{10}{6\cdot8(14-\frac{9}{13})}}-\frac{4}{5}}{11}\approx\pi+8.60\text{e-}16$

3+((((12/(7+(((10/(14-(9/13)))/8)/6)))-(4/5))/11)+(1/(2+15))) = 85563208/27235615

3+(2/((9/(8-((7/(((5+10)*(14*15))-(11/6)))/(12-(1/4)))))+13))   err=  9.6984e-16  14204192/100317295  

How

Exhaustive search
Let $F(S)$ be the set of numbers you can "achieve" using the numbers in set $S$.

For example,
$F(\{3\}) = \{3\}$
$F(\{3,4\}) = \{3,4,7,-1,1,12,\frac{3}{4},\frac{4}{3}\}$

Then the problem is to find the element in $F(\{1,2,3,...,n\})$ nearest $\pi$.

How can you calculate $F(S)$? Let's consider this element of $F(\{1,2,3,4\})$: $$ \frac{1}{3}+4\cdot2 $$ The outer-most operation is addition. So it's an element of $F(\{1,3\})$ $+$ an element of $F(\{2,4\})$. Generalizing this gives you a way to compute $F(\{1,2,...,n\})$. Note that this involves computing $F$ for $2^{n}-1$ subsets.

Smart search

Consider the element in $F(\{1,2,...,10\})$ nearest $\pi$, (where $\epsilon$ is about $-2.33e-08$): $$ \pi +\epsilon =3+\frac{1}{7+\frac{10}{4\cdot5\cdot8-\frac{2}{6\cdot9}}} $$ Let's move some of the numbers to the left side of the equation: $$ \frac{10}{\frac{1}{(\pi +\epsilon)-3}-7} =4\cdot5\cdot8-\frac{2}{6\cdot9} $$ Hmm, so this version could have been discovered by calculating $F(\{\pi,1,3,7,10\})$ and $F(\{2,4,5,6,8,9\})$ and considering pairs of elements that are nearly equal (and then solving for $\epsilon$).

Better yet, for each element of $F(\{\pi,1,3,7,10\})$, replace $\pi$ with $\pi+0.00001$ and then $\pi-0.00001$ to get a range of values. For example, $$ \frac{10}{\frac{1}{(\pi \pm 0.00001)-3}-7} $$ gives the range $[158.699,161.252]$ and $4\cdot5\cdot8-\frac{2}{6\cdot9}$ is within this range. So, solving for $\pi$ guarantees an error less than $0.00001$.

Generalizing this observation gives the "Smart Search". Split $\{\pi,1,2,...,n\}$ into two pieces, where the side without $\pi$ has exactly $k$ elements (it's best for $k$ to be near $n/2$). In the example above $k$ was $6$. Notice that $k=7$ would uncover the same result, but $k=5$ would miss it, because $4\cdot5\cdot8$ and $\frac{2}{6\cdot9}$ use three numbers each.

Anyways, this technique allows my computer to search up to $n=14$ with $k=8$, instead of the exhaustive search which uses too much RAM at $n=11$ (and is much slower).

Answer 6

I found some new records using Simulated Annealing. My constructions use general expressions formed by binary expression trees.

$n=11$

error 2.74E-10
$3 + 1 / (7 + 5 / (8 * 10 - 9 / 11 / 2 / 4 / 6)) = 3.1415926533157648$

$n=12$

error 2.91E-11 $3 + 2 / (7 + 6 + 9 / (8 - 1 / (5 + 4 * 10 * 11 * 12))) = 3.1415926536189365$

$n=13$

error 1.61E-12 $3 + 1 / (7 + 5 / (2 + 6*13 + 11 / (8 + 9) / (10 - 4 * 12))) = 3.141592653591404$

$n=14$

error 3.82E-14 $ 3 + 2 / (14 + (1 - 6 / 5 / 11 / (7 / 12 - 10 * 4) / 13) / 8) = 3.141592653589755 $

$n=18$

error 4.1E-16 $726714064/231320271 = 3 + 2 / 14 - 10 / (15 * ((4 / (18 + (1 / (9 + (7 * (11 + (6 / 13))))))) - (5 - 16 - 12 - 8) * 17))))$

$n=20$

error 5.37E-18.
$644339207239/205099539720 = 3 + (11 + 6) / 12 / 10 + 18 / ((4 + 15) / 17 / 13 - ((20 * 5 * 16 * 19 - (1 / (2 / 7 - (9 + 14)))) * 8))$

Old results

$n=13$, error 1.45E-9.
$3 + (((9 / 11 / 13 / 12 - 5) / 6 - 2) / 8 / 4 / 10 + 1) / 7 = 3.141592652139527$.
$n=14$, error 1.56E-10.
$5-2 + (((7 / 13 / 3 + 10) / 12 / 8 + 14 + 4) / 6 / 11 + 1) / 9 = 3.1415926537454313$. $n=16$, error 4.95E-12.
$3 + ((((7 / 11 / 4 + 6) / 13 / 9 - 10) / 5 / 14 / 16 + 2) / 15 + 1) / 8 = 3.141592653584841$. $n=17$, error 8.68E-13.
$3 + (((((5 / 9 / 6 - 11) / 17 + 15) / 4 + 13) / 12 / 8 / 10 - 2) / 16 / 14 + 1) / 7$. $ = 3.141592653590661$

Answer 7

$n=10$:

Attempt 1.1: Expansion of arctan(1) [3 terms]

Yeah, this might not be a great idea. We know that

$$4\arctan1=\pi=4\sum_{n\ge0}\dfrac{(-1)^n}{2n+1}$$

So we can approximate this with three terms to get

$$4\arctan1\approx4\left(1-\dfrac13+\dfrac15\right)\\=4\times\dfrac{13}{15}=4\times\dfrac{7+6}{3\times5}$$which is not accurate at all. ($=3.4\overline6$)

Attempt 1.2: Expansion of arctan(1) [4 terms]

We can now try this with 4 terms with

$$4\arctan1\approx4\left(1-\dfrac13+\dfrac15-\dfrac17\right)\\=4\times\dfrac{76}{105}=4\times\dfrac{6(10+1+2)}{3\times5\times7}$$which is also not accurate.

Attempt 1.3: Expansion of arctan(1) [5 terms]

Using 5 terms, we can try to get to 315 in our denominator and 263 in our numerator. The only problem is I am unsure how to. Here is my attempt (since we need the 4 to multiply our expansion):

$$263:\text{ }2^8+7\text{ (numbers left: }1,3,5,6,9,10)\\315:\text{ (unable to do, closest was }316)\text{ }3^5+7\times9+10=316$$and now we do$$4\times\dfrac{263}{316}\approx3.3291$$which is actually more accurate than if I had managed to get 315^[1], and we can get this more accurate with using up 6 and 1, which I managed to not use to get$$4\times\dfrac{263}{323}=4\times\dfrac{2^8+7}{3^5+7\times9+10+1+6}\approx3.25696594$$which is even more accurate than before (and in my testing with expansions, is the most accurate I managed to get it) but still not close enough to even be considered probably. I'll come back to this in the future.

Attempt 2: Unoriginality (original method by 2012rcampion)

Yeah, we're using $\dfrac{355}{113}$ again. However, this time, we can use all of the numbers:

$$\dfrac{355}{113}=\left(3+\dfrac{2\times8}{6\times4\times5-7}\right)(10-9)\approx3.141592920\dots$$which yet again has an error of $2.67E-7$.

However, I will try to provide a fully original solution for $n=10$ in the future.

---

^[1]As I finish up writing this post, I just realized I'm stupid. I could have just subtracted 1 from 10 to get $7\times9+10-1=7\times9+9=9(7+1)=8\times9$ which would have been all that I would have needed to do to get the $315$ I wanted. But hey, I've already found a better solution for my expansion of $4\arctan1$ already, so it doesn't matter.

---

Edit 1: I have a solution for $n=17$:

$n=17$:

Attempt 1: Small digit approximation method.

We get this from using the small digit approximation for $\pi$: [source]

$$.1^{-2/3}+(4/.5)^{-6}-.7-.8$$

So what do we do?

First off, we can rewrite this as$$(1/10)^{-2/3}+8^{-6}-(7/10)-(8/10)$$Now, rewrite $-2/3$ as $(4-6)/3$ and $7/10$ as $7/(12-2)$ so now it is$$\left(\dfrac1{10}\right)^{(4-6)/3}+8^{-6}-\dfrac7{12-2}-\dfrac8{10}$$Now, to get our -6, we replace that with 5-11 (which is the only replacement that I found that I was able to make that would not require me to restart this) and finally, we use $\dfrac{17-13}{14-9}=\dfrac45$ to rewrite our approximation for $n=17$ as (without using 16 or 15)$$\left(\dfrac1{10}\right)^{(4-6)/3}+8^{5-11}-\dfrac7{12-2}-\dfrac{17-13}{14-9}$$which now the approximation is improved to just be $5.28E-9$ less than $\pi$, my most accurate solution so far.

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Edit 2: $n=11$ and $n=13$ solution

$n=11$

Attempt 1: Continued fraction for $\pi$

This is a method using the continued fraction expression of $\pi$. The first few terms for this are

$$3+\dfrac1{7+\dfrac1{16}}$$

which we can rewrite as

$$3+\dfrac1{7+\dfrac{5-4}{6+9+11-10}}=\dfrac{355}{113}$$

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$n=13$

Attempt 1: Continued fraction

This uses the same method as $n=11$, but this time we can write it as

$$3+\dfrac1{7+\dfrac{5-4}{6+9+\dfrac{11-10}{13-12}}}=\dfrac{355}{113}$$