igorvanloo

def A(k,n):
    total = 0
    f0 = 1
    modulo = 1000000007
    power = pow(2, n//k, modulo)
    for a in range(0, n//k + 1):
        total = (total + (f0 * pow(power, k-2*a, modulo))) % modulo
        f0 = (f0*(k-2*a)*(k-2*a-1)) % modulo
        f0 = (f0*pow((a+1)**2, -1, modulo)) % modulo
    return total

igorvanloo

def compute():
    eulercoins = [1504170715041707]
    current_eulercoin = 1504170715041707
    inv = pow(1504170715041707, -1, 4503599627370517)
    n = 2
    while True:
        number = 1504170715041707*n % 4503599627370517 #Search downwards
        if number < current_eulercoin:
            current_eulercoin = number
            eulercoins.append(number)

igorvanloo

def compute(limit):
    f = [int(4*(x**2)+1) for x in range(limit+1)] #Initialise the list f 
    max_prime_factor = [0]*(limit+1) #Initialise the list max_prime_factor
    for x in range(1,len(f)): #Go through f
        div = f[x] #Initialise divisor
        if div > 1: #Check if divisor > 1
            curr1 = x % div
            while curr1 <= limit: #while curr = x + k*f[x] < limit we continue
                if f[curr1] % div == 0: #Check if f[x+k*f[x]] is divisible by f[x]
                    max_prime_factor[curr1] = max(max_prime_factor[curr1], div)

igorvanloo

def compute():
    count = []
    for n in range(1,10):
        for i in range(1,22):
            if i == len(str(n**i)):
                count.append(n**i) 
    return len(count)
  
#Used the following function to find the bounds for n
#n = 9 bound is 22 because 9^22

igorvanloo

def compute(cipher):
    possiblecodes = []
    for x in range(97,123):
        for y in range(97, 123):
            for z in range(97,123):
                password = chr(x) + chr(y) + chr(z)
                count = 0
                decrypt = []
                for i in cipher:
                    temp = i^(ord(password[count % 3]))

igorvanloo

def compute():
    x = 100
    while True:
        count = 1
        for i in range(2, 7):
            if sorted(str(x)) != sorted(str(i*x)):
                break
            count += 1
        if count == 6:
            return x

igorvanloo

def compute():
    primes = sorted(list(set(list_primes(10000)) - set((list_primes(1000)))))
    
    supercandidates = []
    while len(primes) != 0:
        curr = primes.pop()
        candidates = [curr]
        for y in primes:
            if sorted(str(curr)) == sorted(str(y)):
                candidates.append(y)

igorvanloo

def list_primality_modified(n):
	result = [0] * (n + 1) #Initialize an array of length n+1
	for i in range(2,int((n)) + 1): #We want to go all the way to the length instead of sqrt(n) 
					#to include all prime factors
		if result[i] == 0: #If this value has not been marked yet, continue
			for j in range(2 * i, len(result), i):
				result[j] += 1 #Add all the prime factors
	return result
#We will obtain a list where array[x] represents the number of prime factors of x,
#if array[x] = 0 then x is a prime

igorvanloo

def goldbachs(number):
    for x in range(1, int(math.sqrt(number))+1):
        temp_var = number - 2*(x**2)
        if is_prime(temp_var) == True:
            return True
    return False

def compute():
    count = 33
    while True:

igorvanloo

def all_numbers(alist):
    nums = set()
    for x in alist:
        for y in x:
            nums.add(y)
    return list(nums)
    
def method1():
    possible_numbers = (all_numbers(keys))
    possible_passcode = list(itertools.permutations(possible_numbers))