#!/bin/python3

import sys
'''
Subproblem: OPT[i] = max sequence to finish i-length chocolate
Base Cases: OPT[0] = 0, OPT[1] = 1
Recursion: OPT[i] = OPT[j] + 1, where j is the factor of i with max sequence
No. of subproblems: length of chocolate(n)
Running Time: n * O(n) = O(n^2)
Solution: OPT(n)
'''
def get_factors(chocolate_length):
    for i in range(1, chocolate_length//2 + 1):
        if chocolate_length % i == 0:
            yield i

def longestSequence(a):
    biggest_chocolate = max(a)
    OPT = [0 for i in range(biggest_chocolate + 1)]
    OPT[0] = 0
    OPT[1] = 1
    for i in range(2, biggest_chocolate + 1):
        max_seq = 0
        for factor in get_factors(i):
            if max_seq < (i//factor)*OPT[factor]:
                max_seq = (i//factor)*OPT[factor]
                
        OPT[i] = max_seq + 1
        
    res = 0
    for chocolate in a:
        res += OPT[chocolate]
        
    return res
    #  Return the length of the longest possible sequence of moves.

if __name__ == "__main__":
    n = int(input().strip())
    a = list(map(int, input().strip().split(' ')))
    result = longestSequence(a)
    print(result)