Excessive Integers

Question

For a positive integer n with the prime factorization n = p1^e1 * p2^e2 * ... pk^ek where p1,...,pk are primes and e1,...,ek are positive integers, we can define two functions:

• Ω(n) = e1+e2+...+ek the number of prime divisors (counted with multiplicity) (A001222)
  • ω(n) = k the number of distinct prime divisors. (A001221)

With those two functions we define the excess e(n) = Ω(n) - ω(n) (A046660). This can be considered as a measure of how close a number is to being squarefree.

Challenge

For a given positive integer n return e(n).

Examples

For n = 12 = 2^2 * 3 we have Ω(12) = 2+1 and ω(12) = 2 and therefore e(12) = Ω(12) - ω(12) = 1. For any squarefree number n we obivously have e(n) = 0. The first few terms are

1       0
2       0
3       0
4       1
5       0
6       0
7       0
8       2
9       1
10      0
11      0
12      1
13      0
14      0
15      0

Some more details in the OEIS wiki.

Answer 1

Vyxal s, 3 bytes

ǐ¯†

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Ports 05AB1E. Vyncode makes the byte count worse.

Explained

ǐ¯†
ǐ    # Prime factors with duplicates
 ¯   # deltas
  †  # vectorised negation
# summed by the s flag