Develop Reference

Why does this factorization program fail for numbers greated than 102030405060708001?

Following is a factorization program in Python. It works fine till 102030405060708001 but produces incorrect results for greater integers. Why is this so?
import math
a = []
def prime(n):
isPrime = True
for r in range(2,int(n**(1/2)+1)):
if n%r == 0:
isPrime = False
return isPrime
def factor(number):
if prime(number):
a.append(int(number))
for r in range(2, 1 + int(number**(1/2))):
if number%r == 0:
a.append(r)
number = number/r
factor(number)
break
return a
number = int(input("Enter a number whose factors you want:"))
factor(number)
print(f"Factor of {number} are ", end = ": ")
print(*a, sep = ' X ')
print(math.prod(a))
Example
Factor of 102030405060708001 are : 1867601 X 54631800401.
And the product of these factors is 102030405060708001
Factor of 102030405060708002 are : 2 X 2 X 2 X 2 X 2 X 3 X 3 X 5 X 5 X 5 X 6133 X 462119341.
But the product of these factors is 102030405060708000
Factor of 102030405060708003 are : 3 X 2 X 2 X 2 X 2 X 2 X 3 X 5 X 5 X 5 X 6133 X 462119341.
But the product of these factors is 102030405060708000

Is there a way to list super primes between 1 and n in NumPy

Is there a way to list super primes (primes in a prime position wikipedia article) between 1 and n using NumPy.

I got this without Numpy, is that okay?
here is the code based on
Sieve of Atkin
import math
is_prime = list()
limit = 100
for i in range(5, limit):
is_prime.append(False)
for x in range(1, int(math.sqrt(limit)) + 1):
for y in range(1, int(math.sqrt(limit)) + 1):
n = 4 * x ** 2 + y ** 2
if n <= limit and (n % 12 == 1 or n % 12 == 5) and n <= len(is_prime):
# print "1st if"
is_prime[n] = not is_prime[n]
n = 3 * x ** 2 + y ** 2
if n <= limit and n % 12 == 7:
# print "Second if"
is_prime[n] = not is_prime[n]
n = 3 * x ** 2 - y ** 2
if x > y and n <= limit and n % 12 == 11:
# print "third if"
is_prime[n] = not is_prime[n]
for n in range(5, int(math.sqrt(limit))):
if is_prime[n]:
for k in range(n ** 2, limit + 1, n ** 2):
if k <= len(is_prime):
is_prime[k] = False
for n in range(5, limit):
if n < len(is_prime) and is_prime[n]:
print(n)

Numpy Code Version
Sieve modification of Numpy Sieve by user2357112 supports Monica
import numpy as np
def sieve(n):
'''
Sieve of Erastosenes using numpy
'''
flags = np.ones(n, dtype=bool) # Set all values to True
# Set 0, 1, and even numbers > 2 to False
flags[0] = flags[1] = False
flags[4:n:2] = False
for i in range(3, n, 2):
# Check for odd primes
if flags[i]:
flags[i*i::i] = False
return np.flatnonzero(flags) # Indexes of True are prime
def superPrimes(n):
'''
Find superprimes
'''
# Find primes up to n using sieve
p = sieve(n)
# p(i) denotes the ith prime number, the numbers in this sequence are those of the form p(p(i))
# Want to find numbers we can index in the list (i.e. has to be less than len(p))
indexes = p[p < len(p)] # base 1 indexes within range
return p[indexes-1] # Subtract 1 since primes array is base 0 indexed
Test
print(superPrimes(200))
# Output: [ 3 5 11 17 31 41 59 67 83 109 127 157 179 191]

The sum of the digits is incorrect. Why?

i can't understand why the output of my code concatnates the digits instead of showing their sum:
#Get a number, and show the number of digits and the sum of the digits.
num = int(input('Enter a number: '))
j = 0
i = 1
k = 0
while i < num:
i = i*10
j += 1
k += (num - k)%i
print (f' The number has {j} digit(s), and the sum is: {k}')

Follow the code. Let's say num = 432:
i = 1 * 10 = 10
j = 0 + 1 = 1
k = 0 + (432 - 0)%10 = 2
---
i = 10 * 10 = 100
j = 1 + 1 = 2
k = 2 + (432 - 2)%100 = 2 + 32 = 34
---
i = 100 * 10 = 1000
j = 2 + 1 = 3
k = 34 + (432 - 34)%1000 = 34 + 398 = 432
This algorithm is most definitely not adding every digit. There are several ways to do what you intend in python. One way is inputting the number as a string and summing every digit casting them as integers inside a generator:
num = input('Enter a number: ')
total = sum(int(digit) for digit in num)
print(total)
If you want the number to be an integer since the beginning, you can also do this:
num = int(input('Enter a number: '))
total = 0
while num > 0:
digit = num%10
total += digit
num /= 10 # num //= 10 in python 3
print(total)

python: How to sum a series using an increment number

n = int(input('Enter n: '))
count = 1
sum = 0
number = 1
while (count <= n):
sum = sum + number
count = count + 1
number = number + 2
print('Sum =', sum)
Is it possible to use the same concept for 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 .... + n

You could use a list comprehension to make this more elegant and pythonic:
def sum_series(start, end):
return sum([i**2 for i in range(start, end+1)])
print(sum_series(1,10))
Output:
385
Or using higher order functions:
>>> sum(map(lambda x: x**2, range(1,11)))
385

Something like this should do
n = int(input('Enter n: '))
count = 1
sum = 0
while (count <= n):
sum = sum + count*count
print("{s}+".format(s=sum)
count = count + 1
print('Sum =', sum)

How can I put these numbers in list?

So I have this Collatz conjecture assignment. Basically I have to write a program to which I give number and it will calculate Collatz conjecture for it. Here is my problem though: the number that will come out will be written like this :
12
6
3
10
5
16
8
4
2
1
When they should be in list like this [12, 6, 3, 10, 5, 16, 8, 4, 2, 1].
And here is my code:
n = int(input("The number is: "))
while n != 1:
print(n)
if n % 2 == 0:
n //= 2
else:
n = n * 3 + 1
print(1)

You have to store the numbers in a list
result = []
while n != 1:
result.append(n)
if n % 2 == 0:
n //= 2
else:
n = n * 3 + 1
result.append(n)
print result

This is also an option. A silly one, but still:
n = int(input("The number is: "))
print('[', end='')
while n != 1:
print(n, end=', ')
if n % 2 == 0:
n //= 2
else:
n = n * 3 + 1
print('1]')

A recursive version, just for fun:
number = int(input("the number is: "))
def collatz(n):
if n == 1:
return [n]
elif n % 2 == 0:
return [n] + collatz(n/2)
else:
return [n] + collatz((3*n)+1)
print collatz(number)