Prime factorization of a factorial

Question 1

Every number can be represented by a unique (up to re-ordering) multiplication of prime numbers, called the prime factorization of the number, as you are finding the prime factors that can uniquely create that number.

2^3=8

3^1=3

5^1=5

and 8*3*5=120

But this also means that: (2^3)*(3^1)*(5^1) = 120

It's not saying that 2 occurs 3 times as a digit in the number 120, which it obviously does not, but rather to multiply 2 by 2 by 2, for a total of 3 twos. Likewise for the 3 and 5, which occur once in the prime factorization of 120. The expression which you mention is showing you this unique prime factorization of the number 120. This is one way of getting the prime factorization of a number in Python:

def pf(number):
    factors=[]
    d=2
    while(number>1):
        while(number%d==0):
            factors.append(d)
            number=number/d
        d+=1
    return factors

Running it you get:

>>> pf(120)
[2, 2, 2, 3, 5]

Which multiplied together give you 120, as explained above. Here's a little diagram to illustrate this more clearly:

enter image description here

Question 2

Consider e.g. 33!. It's a product of:

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

the factors are:

2   2   2   2    2     2     2     2     2     2     2     2     2     2     2     2
    2       2          2           2           2           2           2           2
            2                      2                       2                       2
                                   2                                               2
                                                                                   2
  3     3     3        3        3        3        3        3        3        3        3
              3                          3                          3
                                                                    3
      5          5              5              5              5              5
                                                              5
          7                  7                    7                    7
                   11                               11                               11
                         13                                     13
                                     17
                                           19
                                                       23
                                                                         29    31

Do you see the pattern?

33! = 2^( 33 div 2 + 33 div 4 + 33 div 8 + 33 div 16 + 33 div 32) *
      3^( 33 div 3 + 33 div 9 + 33 div 27) *
      5^( 33 div 5 + 33 div 25) *
      ----
      7^( 33 div 7) * 11^( 33 div 11) * 13^( 33 div 13) *
      ----
      17 * 19 * 23 * 29 * 31

Thus, to find prime factorization of n! without doing any multiplications or factorizations, we just need to have the ordered list of primes not greater than n, which we process (with a repeated integer division and a possible summation) in three stages - primes that are smaller or equal to the square root of n; such that are smaller or equal to n/2; and the rest.

Actually with lazy evaluation it's even simpler than that. Assuming primes is already implemented returning a stream of prime numbers in order, in Haskell, factorial factorization is found as

ff n = [(p, sum . takeWhile (> 0) . tail . iterate (`div` p) $ n) 
         | p <- takeWhile (<= n) primes]

-- Prelude> ff 33
-- [(2,31),(3,15),(5,7),(7,4),(11,3),(13,2),(17,1),(19,1),(23,1),(29,1),(31,1)]

because 33 div 4 is (33 div 2) div 2, etc..

Question 3

2^3 is another way of writing 2^³, or two to the third power. (2^3)(3^1)(5^1) = 2^³ × 3 × 5 = 120.

(2^3)(3^1)(5^1) is just the prime factorization of 120 expressed in plain ASCII text rather than with pretty mathematical formatting. Your assignment requires output in this form simply because it's easier for you to output than it would be for you to figure out how to output formatted equations (and probably because it's easier to process for grading).

The conventions used here for expressing equations in plain text are standard enough that you can directly type this text into google.com or wolframalpha.com and it will calculate the result as 120 for you: (2^3)(3^1)(5^1) on wolframalpha.com / (2^3)(3^1)(5^1) on google.com

WolframAlpha can also compute prime factorizations, which you can use to get test results to compare your program with. For example: prime factorization of 1000!

A naïve solution that actually calculates the factorial will only handle numbers up to 12 (if using 32 bit ints). This is because 13! is ~6.2 billion, larger than the largest number that can be represented in a 32 bit int.

However it's possible to handle much larger inputs if you avoid calculating the factorial first. I'm not going to tell you exactly how to do that because either figuring it out is part of your assignment or you can ask your prof/TAs. But below are some hints.

a^b × a^c = a^b+c

equation (a)      10 = 2¹ × 5¹
equation (b)      15 = 3¹ × 5¹
10 × 15 = ?      Answer using the right hand sides of equations (a) and (b), not with the number 150.

10 × 15 = (2¹ × 5¹) × (3¹ × 5¹) = 2¹ × 3¹ × (5¹ × 5¹) = 2¹ × 3¹ × 5²

As you can see, computing the prime factorization of 10 × 15 can be done without multiplying 10 by 15; You can instead compute the prime factorization of the individual terms and then combine those factorizations.

Question 4

If you write out the factorial 5!:
1 * 2 * 3 * 4 * 5,
you will notice that there is one non-prime number: 4. 4 can be written as 2 * 2 or 2^2, which is where the extra 2s come from. Add up all of the occurrences (exponential forms are in parentheses; add exponents for like bases):
2 (2^1) * 3 (3^1) * 4 (2^2) * 5 (5^1), you get the proper answer.

Question 5

You can use O(n/2 log log n) algorithm using only sums (no need precalc primes).

This is a sieve using relation

f = a * b  ~>  f^k = a^k * b^k

then, we reduce all initial factors 1 * 2 * 3 * ... * n moving k from big numbers to small numbers.

Using Sieve of Atkin the Will Ness algorithm could be better for very big n if not, I think it could be better

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv) {
  int n = atoi(argv[1]);
  int *p = (int *) malloc(sizeof(int) * (n + 1));
  int i, j, d;
  for(i = 0; i <= n; i++)
    p[i] = 1;
  for(i = n >> 1; i > 1; i--)
    if(p[i]) {
      for(j = i + i, d = 2; j <= n; j += i, d++) {
        if(p[j]) {
          p[i] += p[j];
          p[d] += p[j];
          p[j] = 0;
        }
      }
    }
  printf("1");
  for(i = 2; i <= n; i++)
    if(p[i])
      printf(" * %i^%i", i, p[i]);
  printf("\n");
  return 0;
}