def eightDivisorCount(n):
ps = Primes.MakePrimeList(int(n**0.5)+2)
primePi = Primes.Prime_Pi()
f = 0
for p in ps:
if p**7 > n:
break
#print p
f += 1
for p in ps:
if 2*p**3 > n:
break
f += primePi(n//(p**3), ps)
print p,f,len(primePi.restricted_memo)
if (n//(p**3)) >= p:
f -= 1
#print p, primePi(n//(p**3), ps)
for n1,p1 in enumerate(ps):
if p1**3 >= n:
break
for p2 in ps[n1+1:]:
if p1*p2*p2 >= n:
break
f += primePi(n//(p1*p2),ps) - primePi(p2,ps)
#print p1,p2,primePi(n//(p1*p2),ps),primePi(p2,ps)
if p2 < 10000:
print p1,p2,f,len(primePi.restricted_memo)
print p1,f,len(primePi.restricted_memo)
return f
def new_432(m):
#primes = Primes.MakePrimeList(int(1.2*m**(1./2))+100)
primes = Primes.MakePrimeList(int(1.2 * m**(1. / 2)) + 120)
prime_pi = Primes.Prime_Pi()
ans = sum_range(1, 18) + func(m, 7, primes, prime_pi)
ans *= 510510
for ix in range(7):
ans *= (primes[ix] - 1)
ans /= primes[ix]
return ans
# Euler 484 - Arithmetic Derivative
# projecteuler.net/problem=484
# Kelvin Blaser 2014.11.23
import Primes
reload(Primes)
import scipy as sp
import time
prime_pi = Primes.Prime_Pi()
'''Q_mem = {}
def Q(n,primes):
#print 'Calculating Q[%d]'%(n,)
if n >= 10**7:
PRINT = True
else:
PRINT = False
try:
return Q_mem[n]
except KeyError:
pass
ans = n
r = 1
while r <= len(primes) and sp.prod(primes[:r]) <= n:
sign = (-1)**r
count = 0
for c in Primes.combProdLessThan(primes, r, n):
ans += sign * (n//int(sp.prod(c)))
count += 1
if PRINT:
print '%d from list of len %d: %d times; \t%d : %d'%(r,len(primes),
