Problems & Puzzles:
Puzzles
Puzzle 50. “The
best approximation to pi with primes”
As usual, the subject was coming to me after visiting
the Patrick De Geest beautiful pages. There was
the following one: “Approximating
pi with Palindromes”.
In that and the same place we are invited to see the Eric
Weisstein's pipages
where from the Castellano’s result was taken:
1.09999901 x
1.19999911 x
1.39999931 x
1.69999961 =
3.141592573
Then I modified a little bit the idea, and asked my
self for the best approximations
to pi with primes (not
necessarily palprimes). I devised also a procedure (maybe
a kind of arbitrary) to rank the different known
approximations.
I got some of the “best”
approximations from the Weisstein pipages
and made my own initial contribution. This is the result:
Pi and Primes
Pi =
3.1415926535898793238…
Author 
pi
approximation (1) 
Digits used 
Result 
Digits OK 
Points (2) 
Ramanujan 
17*sqrt(7)/(2*2^3) 
6 
3.141829682 
3 
50.00% 
Egyptians 
2*11/7 
4 
3.142857143 
2 
50.00% 
Stoschek 
2^(3^2)/163 
6 
3.141104294 
3 
50.00% 
Carlos Rivera 
211891/67447 
11 
3.141592658 
8 
72.73% 
Pi & Golden Ratio 
3(3+sqrt(5))/5 
4 
3.141640786 
3 
75.00% 
Jud McCranie 
1833616417 / 583658233 
19

3.14159265358979353 
15

78.95% 
Plouffe 
43^(7/23) 
5 
3.141539853 
4 
80.00% 
Jud McCranie 
See Below 
18

See Below 
15

83.33% 
Jud McCranie 
See Below 
20

See Below 
17

85.00% 
Gardner 
3+((2^2)^2)/113 
7 
3.14159292 
6 
85.71% 
Carlos Rivera 
See Below 
553

See Below 
547

98.91% 
Felice Russo 
sqrt(ln(33487/(sqrt(3)))) 
6

3.14159297 
6

100% 
Ramanujan & Olds 
5*71/113 
6 
3.14159292 
6 
100.00% 
Carlos Rivera 
ln(2^3*829453)/5 
9

3.141592653609 
9

100% 
Jud
McCranie 
See
below 
23 
See
below 
30 
130.4% 






Next ??? 




??? 
Paul
Leyland 
See
below in the "Other approaches" section 
See
below 
Arbitrary
large 
Arbitrary
large 
Arbitrary
large 
(1)You are able to use any
mathematical operation.
(2) Points = (Digits O.K.) / (Digits used) x 100
Would you like to improve
these results?
Solutions
See them in red color in the
table:
Carlos Rivera (23/04/99):
Inside the limit of the current Ubasic
(2032 digits) and working in Point 421 mode, you can
probe that: int(pi*10^547)  3*10^547
is the larger prime contained in
the decimal
digits of
pi (disregarding
the initial "3"). So the following number of
547 digits is prime (strongpseudoprime by now):
1415926535897932384626433832795028841971693993751058209749445
9230781640628620899862803482534211706798214808651328230664709
3844609550582231725359408128481117450284102701938521105559644
6229489549303819644288109756659334461284756482337867831652712
0190914564856692346034861045432664821339360726024914127372458
7006606315588174881520920962829254091715364367892590360011330
5305488204665213841469519415116094330572703657595919530921861
1738193261179310511854807446237996274956735188575272489122793
81830119491298336733624406566430860213949463952247371907021
Let's call it P(547).
Obviously 3 +
P(547)/(2*5)^547 produce pi with 547
correct digits, using a total of 553 digits with all the
primes (it happened that 547 is prime also!) for a rank
of 98.91%
This constitutes a method for
improving the approximation of pi, supposed you can find and probe the
primality of larger and larger primes in pi.
p.s. I would like to add
that some months ago (?) my friend G.L.Honaker
invited to me to look for primes in pi starting with the "3" and
disregarding the "decimal period". When I
started defining and making some contributions to this
puzzle I didn't related that search with it. Just this
morning I awake with the idea of rescuing that primes but
it happens that the largest of that kind of primes is 27
digits large, and I wanted something larger (specially
after the Jud's result...). Then suddenly I made the
final step: to throw away the initial "3" (why
not?) and started looking for primes in the decimal
portion of pi, with the result that there are primes
with 5, 12, 281 and 547 digits....!.So, thanks Honaker...
Jud McCranie
(23/04/99) has sent other two approximations to pi using only primes:
3+(5*2216117)/(3*53*577*853)
3.14159265358979316
Rank = 15/18 =83.33
3+(3^3*5*851159)/(2*7*71*816427)
3.1415926535897932390
Rank = 17/20 = 85.00
Felice Russo
sent (28/04/99):
Formula, value, digits employed,
digits OK, Rank
sqrt(ln(33487/(sqrt(3)))) = 3.141592977847856
, 6, 6, 100 %
sqrt(ln(5351^2/1481)) = 3.141592695171481, 9, 7, 77.8%
e^sqrt(10993/8389) = 3.141592616712708, 9, 7, 77.8%
ln(19739/853) = 3.141592105651308, 8, 6, 75%
e^75709/66137 = 3.141592631719165, 10, 7, 70%
Jud McCranie
has found  at last, 03/05/99 the most wanted >100%
approximation to pi:
ln(2^3*3*10939058860032031)/sqrt(163)
=
3.141592653589793238462643383279726619
23 digits used and 30 digits OK.... for a 130.4%
rank!!!!!!...
Have my most sincere
congratulations, friend Jud....
Jean Brette has
improved the Jud's result analyzing
& changing the argument of the ln
function. This is his amazing result:
= 3.14159265358979323846264338327973
which has the same 30 correct digits using only 15 digits of prime numbers for a pretty
Rank of 200.00%.
Other relations
between pi and prime numbers
1) Exact Relations
between pi and prime numbers
Jud McCranie
and Paul Leyland sent some exact
relations between pi and prime numbers, based in known
mathematical identities
Jud uses the
known identity 4*arctan(1) = pi and produces the relation
2*2*arctan(32) = pi. Others are 2*arcsin(32) = pi and 2*arccos(22) = pi.
Paul uses the
known identity e^(pi*i) = 1 to produce the relation pi=ln(23)/sqrt(23)
Of course, none of these
relations can be used to calculate pi in any extent.
2) pi of arbitrary accuracy using only
prime numbers to describe a Turing machine devoted to calculate pi
Paul Leyland devised the following
representation of pi with arbitrary accuracy using nothing
more than prime numbers:
"...However, if you don't want trig, exploit
Godel numbering and Turing
machines. Define a Turing machine and write a program for
it to calculate
pi to arbitrary precision. This is
tedious but clearly possible. The
Turing machine, by definition, is a finite state machine
and so can be
encoded in a finite number of symbols  use a distinct
prime for each
distinct symbol. Likewise, it is straightforward to write
a finite program
for such a machine which will output pi to an arbitrary
number of places.
The program can itself be encoded in a finite number of
different primes,
one for each distinct symbol. So far, it's been easy, but
we need to encode
the order in which the symbols appear. Suppose there are
N symbols in
total, not necessarily distinct. We start labeling the
distinct symbols
with successive primes starting from the (N+1)th prime.
We then write the
Turing machine and program as a product of powers of
these primes.
Everywhere symbol {i} appears, we raise it to the power
of its position in
the list of small primes. For all the symbols, we then
accumulate the
running product. The resultant product consists of
nothing but primes, multiplicationsymbols and exponentiationsymbols. On one
interpretation,
it's just a very large but finite integer. In another
interpretation, it's
an arbitrarily precise representation of pi. Interpretation is everything,
even in the examples you give on your page"
3) pi of arbitrary accuracy using only
2 prime numbers and two distinct and nested functions
Paul Leyland has also devised a 'formula' that
produces the pi value with arbitrary accuracy  as large
as the accuracy of pi obtained with the ratio A/B of two
previously known integers A & B  executing certain
"mathematical operations" explicitly over only
two prime numbers.
For example let's use the
approximation to pi 22/7 = 2*11/7 = 3.142857 that
has a Rank = 2/4*100 = 50%
The Leyland's
formula for this specific approximation to pi is:
lg(lg(SSSSSSSSSSSSSSSSSSSSSS(2)))/lg(lg(SSSSSSS(2)))
where "lg" means "log
base 2" and "SSS...SSS(2)"
means "square root of 2, as many times as large
is the Sstring".
Well, this representation of pi,
that uses only two digits in primes explicitly used, has
the same accuracy referred to pi
than 22/7, but a Rank = 2/2*100 = 100%. According to this
example it is possible to increase the Rank up to an
arbitrarily large quantity.
Subquestion:
How large can be the numerator of the quotient A/B in
order to execute correctly the computation of the Leyland's formula inside the limits of the
current highprecision arithmetic packages freely
available like Ubasic or PARI?
***
Laurent de Jerphanion From Paris, France sent (26/10/2000) the
following nice & new solutions:
"...it is relatively easy to get solutions at 100% :
3+sin sin (3) = 3.1406520...
7*e^cos(5/2) = 3.14170778...
2 * 7/(5e/5) = 3.14158896...
Solution at 125% :
e + e^e^e^(17/(3*3)) = 3.1415956...
Solution at 133% :
3+e^e^e^(2/5) = 3.14158389...
Solutions at 150% :
e + tan(2/5) = 3.14107504...
3/sin(3*e) = 3.14109534...
(ln ln (97+7/5))^e = 3.141592497...
Solutions at 200% :
e^e^e^2 = 3.1421918...
ln (e . ln (7!  61)) = 3.1415925681...
Although the last one would be downgraded to 120% in the form:
2/2 + ln ln (7!  61)
...not to mention ‘infinite’ solutions with no digits at all
(so they contain no nonprimes), such as e + sin(e) = 3.12906311... But
that is stretching things a bit..."
***
From time in time this puzzle brings new puzzlers!
Now is the turn for Polly T. Wang who sent (October 30, 2002) the
following solutions:
solution at 50%: E^((2^11+2*(2^2)!)/1831)=3.1415919.....
solution at 62.5%: ln(3*5*307/199)=3.14159312.....
solution at 70%:ln(2*463*929/3)/(2^2)=3.14159267....
solution at 75%: 3+(17/5!)=3.14166666.....
solution at 125%:ln(5*10691/11#)=3.1415926539317.....
***
One reader just named 'Jonathan' wrote:
' ...the solution "ln(2^3*3*10939058860032031)/sqrt(163) = ..." is
simply an inversion of: e^(pi*(sqrt(163))) =? that massive integer. and
was known to Ramanujan. widely referenced in American literature, and
appearing most notably in the April fools section on 1974 scientific
American (which claimed that it had been proven to be an exact equality)
so... it should be attributed to Ramanujan if anyone...'
Jud McCranie replied on my request, about this comment the following
short not:
Yes, I knew it wasn't original with me. I didn't mean to imply that I
originated it... I knew that fact about the close approximation to pi, I
didn't discover it.
***
10 years later this puzzle is still interesting to new puzzlers.
Arkadiusz Wesołowski wrote (July 09):
(sqrt(2+e^e)/e)^e = 3.141599009...
Rank = 500%
sqrt(2+e^e)^e/log(e^e^e) = 3.141599009...
Rank = 500%
***
