def pgen():
yield 2
p = 3
while True:
if lib.prime(p):
yield p
# print(': prime',p)
p += 2
def recurse(digits, nums, prevdiv): # digits numbers made, last divider
global uniquesets, count
if prevdiv == len(digits): count += 1 # found a set with increasing primes
else: # make prime number with next digits and place divider
num = 0
for d in range(prevdiv, len(digits)): # try each digit
num = num * 10 + digits[d]
# only make sets with primes in increasing order for uniqueness
if lib.prime(num) and (len(nums) == 0 or num > nums[-1]):
recurse(digits, nums + [num], d + 1)
def recurse(num,dsum): # assume arguments satisfy num % dsum == 0
global hsum, limit
if num*10 >= limit: return
if num % dsum == 0: # harshad number
if lib.prime(num//dsum): # also strong harshad number
num *= 10
if lib.miller_rabin_verified(num+1): hsum += num+1; print(':',num+1)
if lib.miller_rabin_verified(num+3): hsum += num+3; print(':',num+3)
if lib.miller_rabin_verified(num+7): hsum += num+7; print(':',num+7)
if lib.miller_rabin_verified(num+9): hsum += num+9; print(':',num+9)
else: num *= 10
for d in range(10): # recursively make right trunc harshad numbers
if num % dsum == 0: recurse(num,dsum)
num += 1
dsum += 1
def is_prime(n):
if n <= pclim: return n in pcset
return lib.prime(n)
import libtkoz as lib
import math
N = 600851475143
# brute force basically, factoring by numbers up to square root
largest = 0
for d in range(1, int(math.sqrt(N)) + 1):
if N % d == 0:
if lib.prime(N // d): # n/d decreases so if prime, its the largest
largest = N // d
break
if lib.prime(d): # d is increasing
largest = d
print(largest)
# faster algorithm, dividing out smaller prime factors
n = N
while n % 2 == 0:
n //= 2
print(': factor', 2)
d = 3
while d <= int(math.sqrt(n)):
while n % d == 0:
n //= d
print(': factor', d)
d += 2
print(n)
import libtkoz as lib
# brute force, generate lexicographic permutations decreasing
# by considering digit sums we can show that only
# 7*8/2=28 and 4*5/2=10 are not divisible by 3, then there is 1*2/2=1
# 1 is not prime so the largest must have 4 or 7 digits
for l in [7, 4]:
digits = list(range(l, 0, -1))
print(': permuting', digits)
found_prime = False
while True: # go through all permutations
# test primality
if lib.prime(int(''.join(str(d) for d in digits))):
print(''.join(str(d) for d in digits))
found_prime = True
break
i1 = len(
digits) - 1 # find position of digit breaking right->left decrease
while i1 != 0 and digits[i1 - 1] <= digits[i1]:
i1 -= 1
i1 -= 1
if i1 == -1:
print(': finished all', l, 'length permutations')
break
i2 = len(digits) - 1 # find largest smaller digit to swap with it
while digits[i2] >= digits[i1]:
i2 -= 1
digits[i1], digits[i2] = digits[i2], digits[i1] # swap
i1 += 1 # sort starting from here
i2 = len(digits) - 1
while i1 < i2:
from functools import reduce
solutions = 1000
# 1/x+1/y=1/n, we must have x>n and y>n so use x=n+x0 and y=n+y0
# n(x+y)=xy --> n(2n+x0+y0)=n^2+nx0+ny0+x0y0 --> n^2=x0y0
# we need the smallest perfect square with enough distinct ways to factor it
# into x0*y0 such that x0<=y0. every factor x0<=n corresponds to a y0=n/x0 so
# n^2 must have 1000 divisors <= n, or 1999 overall (exclude counting n twice)
# any perfect square will have an odd number of divisors (counting divisors with
# the product of 1+ prime exponents) so we need a n^2 with >2*solutions divisors
primes = [] # find enough primes so that 3^(prime count)>2*solutions
n = 2
while 3**len(primes) <= 2*solutions:
if lib.prime(n): primes.append(n)
n += 1
print(': generated primes',primes)
smallest = reduce(lambda x,y:x*y, (p*p for p in primes))
print(': initial product of the squares is',smallest)
print(':',smallest,lib.divisors2(smallest),lib.prime_factorization(smallest))
# recurse on the prime list to find a smaller number with over 2000 divisors
# order exponents in decreasing order, 2^a*3^b has (a+1)(b+1) divisors so order
# a >= b (put the greater exponent on the smaller number
def recurse(number,primeindex,divisors,lastexp):
global primes, smallest, solutions
if divisors > 2*solutions:
print(':',number,divisors,lib.prime_factorization(number),
lib.divisors2(number))
smallest = min(smallest,number)
import libtkoz as lib
limit = 2000000
psum = 0
pcount = 0
# could use sum(lib.list_primes(...))
# brute force approach will use loop to prevent using memory to store primes
# took ~6 sec (i7-7600u)
for n in range(limit):
if lib.prime(n):
psum += n
pcount += 1
print(': prime count', pcount)
print(psum)
import libtkoz as lib
nval = 10 #10000
modulus = 10**100 #10**9+7
# generate the Pn and Cn sequences
pn = []
cn = []
i = 2
while len(pn) < nval or len(cn) < nval:
ip = lib.prime(i)
if ip and len(pn) < nval: pn.append(i)
if not ip and len(cn) < nval: cn.append(i)
i += 1
pn = list(lib.digital_root(i) for i in pn)
cn = list(lib.digital_root(i) for i in cn)
print(': P_n', pn)
print(': C_n', cn)
superint = []
pi = 0 # indexes in pn an cn
ci = 0
while pi < len(pn):
# start by picking smaller digit unless we can maximize intersecting digits
# by picking the other
if ci < len(cn):
if pn[pi] < cn[ci]:
superint += [pn[pi]]
pi += 1
elif cn[ci] < cn[pi]:
superint += [cn[ci]]
def is_prime(n):
global primelim, primecache
if n > primelim: return lib.prime(n)
return n in primecache
def is_prime(n):
if n <= cachemax: return n in cache
return lib.prime(n)
def prime(n):
global ptable
global ptablemax
return (n <= ptablemax and n in ptable) or lib.prime(n)
def is_prime(p):
global primecachesize
global primecache
if p < primecachesize: return p in primecache
else: return lib.prime(p)
import libtkoz as lib
ratio = 0.10 # prime ratio must get smaller than this
# brute force, test all primes in spiral, takes ~7.5 sec (i5-2540m)
step = 2 # difference between numbers in next loop
# side length is step+1 and number visited is 2*step+1 (+1 for center 1)
spiralnum = 1 # number to test for primality
primes = 0
while True:
# try 4 corner numbers in next loop
for i in range(4):
spiralnum += step
if lib.prime(spiralnum): primes += 1
if primes / (2 * step + 1) < ratio: # prime ratio small enough
print(step + 1)
break
step += 2
import libtkoz as lib
# phi(n) is n * product of every (1-1/p) so n/phi(n) is 1/(product of (1-1/p))
# to maximize the ratio, we need as many distinct primes as possible
# multiply distinct primes in increasing order until multiplying the next one
# would exceed the limit, the solution is that (and all multiples of it within
# the limit have the same ratio)
maxn = 1000000
prod = 1
p = 2
while True:
if lib.prime(p):
prod *= p
if prod > maxn:
prod //= p
break
p += 1
if prod * 2 <= maxn:
print(': multiple solutions, choosing the smallest')
print(':', prod, '/ phi(n) =', prod / lib.totient(prod))
print(prod)
