Find primes in the Continued Fraction approximation to Pi
Motivated by Matt Parker's Stand-up Maths video "What is the biggest tangent of a prime?" https://www.youtube.com/watch?v=A7eJb8n8zAw&t=1s. In that video, looking for integers such that tan(N) > N boils down to finding N which are close to (K+1/2)Pi. This equivalent to finding rational approximations to Pi of the form (2N)/(2*K+1). Instead of brute forcing this, it occurred to me that the Continued Fraction approximation technique already produces the sequence of progressively closer rational approximations to a value. No searching needed.
https://en.wikipedia.org/wiki/Approximations_of_%CF%80#Continued_fractions https://en.wikipedia.org/wiki/Continued_fraction#Generalized_continued_fraction
Unfortunately, the sequence in PI = [3; 7,15,1,292,1,1,1,2,...] has no obvious formula. But conveiniently, the first 1000 Convergents (fractional approximations) are available in https://oeis.org/A002485 (numerator) and https://oeis.org/A002486 (denominator).
Download the numerator sequence https://oeis.org/A002486/b002486.txt read them into some python code, and check for numerators of the form 2*N. Then check N to see if it isprime().
Interestingly, line 86 looks like this:
86 2339618734654425141409627264213705772778073822
And that is EXACTLY 2x the prime mentioned in the video: 1,169,809,367,327,212,570,704,813,632,106,852,886,389,036,911
See piprime1.py for details.
It uses the Python package sympy to get the isprime() function: https://www.geeksforgeeks.org/python-sympy-isprime-method/.
https://docs.sympy.org/latest/modules/ntheory.html?highlight=isprime#sympy.ntheory.primetest.isprime
It uses the Miller-Rabin test https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test.
And https://en.wikipedia.org/wiki/Baillie%E2%80%93PSW_primality_test
Standalone implimentations for different languages are available here: https://rosettacode.org/wiki/Miller%E2%80%93Rabin_primality_test#Python:_Proved_correct_up_to_large_N
To go further, you have to generate more convergents. Conveniently, the Continued Fraction terms are available here https://oeis.org/A001203. Downloading the first 20000 values from https://oeis.org/A001203/b001203.txt
Running piprime2.py checks the values and finds one big one at step 1984 (1017 digits):
230835870782558831561617186504559084198719501221763995608082253627620752053749345488376393822837250198036536001853828659466202612019525543362322174085744303421231446484541625047630462908919109308644634605051209877750956648014568322183373423523622941806761765245932401727973436579786298208782013178059220103271409347616696556052706562092799953175234183483071403726145726928572372071037042523626350312132351311366806233135093893271182587352730075523143635168510803804031460442796778933680674070124730971307185688425634077096234482442639666385695677866015904370207368846631450100939158029908242779848800640038255592227473300237596577845602369215568916732445980431078426390412264603773550384039765410088966381694110344811198325354315338629604946794192217817288101344643511450133142277670683067655250506551517767422160650566385017503208608678491109517443585115317845289832567015746473548492179557935154400719019569904865219030736244089287736334048402066257337090606092966121806567484954460809024219605952851728610326005069
But this process gets a bit slow for terms beyond about 10000. Running this out to 20000 takes hours and doesn't find anything. Out there the terms are over 10000 digits.
180 Million terms in the sequence are availble here: http://chesswanks.com/seq/cfpi/. But it would take a long time to check those all. Computing the Convergents is pretty fast, so you could run a bunch of searches in parallel, say by having K different computers generate convergents up to some starting point (say 20000*i for i in (1..K)) and only then check for primes for the next 20000 convergents.
Happy Hunting!