def main():
top = 1000
#Routine for all n less than the top value
longest = 0
val = 0
t = time.time()
for num in range(2, top + 1):
r = findRecurrence(num)
if r > longest:
longest = r
val = num
t = time.time() - t
print('Including non-primes:')
print(val)
print('It took {0} seconds'.format(t))
#Routine for all primes less than the top value
longest = 0
val = 0
primes = pemaths.Eratosthenes(top)
t = time.time()
for num in primes:
r = findRecurrence(num)
if r > longest:
longest = r
val = num
t = time.time() - t
print('Using only primes:')
print(val)
print('It took {0} seconds'.format(t))
def elegant(interval):
t = time.time()
primes = pemaths.Eratosthenes(interval)
powers = [0] * (interval + 1)
for i in range(2, interval + 1):
pfacts = pemaths.primeFactorize(i, primes)
for prime, power in pfacts:
if power > powers[prime]:
powers[prime] = power
value = 1
for v in range(interval + 1):
if powers[v]:
value = value * (v ** powers[v])
print(value)
print('The elegant method took {0} seconds'.format(time.time() - t))
#28: 1,2,4,7,14,28
#We can see that 28 is the first triangle number to have over five divisors.
#
#What is the value of the first triangle number to have over five hundred
#divisors?
#
#Calculating triangle numbers will be simple, checking the number of divisors
#will be less so. If we work with prime factorization, this will be faster.
import time
import os.path, sys
# An OS-independent hack for importing from the parent directory
sys.path.append(os.path.abspath(os.path.split(os.getcwd())[0]))
import pemaths
primes = pemaths.Eratosthenes(10000) #Generate a list of primes
t = time.time()
tnum = 1 # Triangle number
n = 1
c = 0
while c <= 500: # While the count of divisors is <= 500
c = 1
n += 1
tnum += n # New generated triangle number
test = tnum
for prime in primes: # Check prime divisors
if prime > test ** 0.5:
c = 2 * c
break
e = 0
#Problem 7 from Project Euler
#Solution by Paul Barton
#
#Here is the text of the problem:
#By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see
#that the 6th prime is 13.
#
#What is the 10001st prime number?
#
#This is a relatively simple question. It is also easily within the scope of a
#prime sieve solution.
import time
import os.path, sys
# An OS-independent hack for importing from the parent directory
sys.path.append(os.path.abspath(os.path.split(os.getcwd())[0]))
import pemaths
t = time.time()
myprimes = pemaths.Eratosthenes(200000)
print('The 100001st prime is {0}'.format(myprimes[10000]))
print('This took {0} seconds'.format(time.time() - t))
for p in primes:
if val == p:
return True
elif val < p:
return False
raise ValueError('Primes list was too short! Test value exceeded highest \
prime value.')
t = time.time() # Start the timer
CONST = 1000
#The largest potential prime is determined by the highest n, a, and b
#values of b and a are constrained, n is unknown, so we guess an upper limit and
#checkPrime() will let us know if it is too small. The function is also written
#to remove the penalty of this value being too large.
primes = pemaths.Eratosthenes(20000)
#b must be prime, a requirement of the case n = 0
bvals = pemaths.Eratosthenes(CONST)
a = (CONST - 1) * (-1)
max_n = 0
max_a_b = (0, 0)
while a < CONST:
for b in bvals:
n = 0
while checkPrime(qForm(n, a, b), primes):
n += 1
if n > max_n:
max_n = n
max_a_b = a, b
#a must be odd: consider the quadratic formula as n(n + a) + b = Prime(odd)