jakedobkin

from math import sqrt
from decimal import *
# you need more than 100 digits, b/c otherwise you get rounding errors
getcontext().prec = 105

squares=[]
for a in range (2,10):
	squares.append(a**2)

total=0

jakedobkin

# http://projecteuler.net/problem=78
from math import floor,sqrt

def pent(n):
	return ((3*n*n)-n) / 2

def sign(x):
	a = floor(x/2)
	if a%2 == 0:
		return 1

jakedobkin

# http://projecteuler.net/problem=77
# i couldn't find a recursive generating function like i used for #76
# one exists, http://mathworld.wolfram.com/EulerTransform.html - but i couldn't figure it out
# so i had to go back to the change counting problem, #31, and repurpose one of those functions

for x in range (11,80):
	coins=[]
	
	# generate primes for coin algorithm
	n=x+1

jakedobkin

# http://projecteuler.net/problem=76
# we'll use euler's formula for generating the partition series

def pent(n):
	return ((3*n*n)-n) / 2

def p(k):
	if k in dict:
		return dict[k]
	elif k == 0:

jakedobkin

#http://projecteuler.net/problem=75

def generatora(a,b,c):
	if a+b+c <= 1500000:
		global count
		k=1
		while k*(a+b+c) <= 1500000:
			count[k*(a+b+c)]+=1
			k+=1
		aa = a-2*b+2*c

jakedobkin

from math import factorial 

def fadd(n):
	sum = 0
	for i in range (0,len(str(n))):
		sum += factorial(int(str(n)[i:i+1]))
	return sum

def roundfadd(n,array,p):
	if n in sequences:

jakedobkin

# http://projecteuler.net/problem=73

from fractions import Fraction, gcd

count=0
# start at 5, b/c that's the first number that has a fraction in this range
for d in range (5,12001):
	for e in range (d/3,d/2):
		if gcd(e,d) == 1:
			count+=1

jakedobkin

# http://projecteuler.net/problem=72
# i got to thinking that this is really just asking to take a sum of euleur's totient from 2 to 1MM
# and that turns out to be true.

factors = [1]*1000001
for i in range (0,len(factors)):
	factors[i]=i


# set up prime sieve to n

jakedobkin

from fractions import Fraction, gcd
from math import floor

# n /d < 3/7
# so n < d*(3/7)
# so for each denominator, we'll look for the closest n that satisfies this equation
# if n and d have gcd==1, then they're a reduced fraction
# and the last one should be the closest fraction to 3/7 under 1MM

for d in range (1,1000000):

jakedobkin

factors = [1]*10000001
for i in range (0,len(factors)):
	factors[i]=i


# set up prime sieve to n
n=10000000
myPrimes = [True]*(n+1)
last = 0
for i in range (2,n):