Regex 🐇 (ECMAScript+(?*)x*+^RME / PCRE2 _^v10.35+), (削除) 60 (削除ここまで) (削除) 59 (削除ここまで) (削除) 54 (削除ここまで) 52 bytes

Attempt This Online! - PCRE2 v10.40+
RegexMathEngine : call with regex -X -xml,pq -nx -t0..17 --line-buffered -f regex.txt where regex.txt contains the regex with no newline

Regex 🐇 (PCRE2 _^v10.35+), (削除) 60 (削除ここまで) (削除) 59 (削除ここまで) (削除) 54 (削除ここまで) 52 bytes

Attempt This Online! - PCRE2 v10.40+

Regex 🐇 (ECMAScript+(?*)x*+^RME / PCRE2 _^v10.35+), (削除) 60 (削除ここまで) (削除) 59 (削除ここまで) 54 bytes

^((?*(?=(xx+?)2円*$|)((?=x2円)(?=(x+)(4円+$))5円)*+x+)x)*$

Attempt This Online! - PCRE2 v10.40+
Try it on replit.com - ECMAScript+(?*)+x*+ (RegexMathEngine -xml,pq)

It takes the base prime \$p\$ of every prime power \$p^k\le n\$, and calculates the product of that list of numbers. And because each \$p\$ will occur \$k\$ times in that list, this product is the same as the product of the prime powers themselves would be, if only the largest from each base prime were included.

^ # tail = N = input number
( # Loop the following:
 (?* # Non-atomic lookahead:
 # M = tail, which cycles from N down to 1
 (?=(xx+?)2円*$|) # 2円 = smallest prime factor of M if M ≥ 2;
 # otherwise, leave 2円 unchanged in PCRE, or
 # unset in ECMAScript
 (
 (?=x2円) # Keep iterating until tail ≤ 2,円 and because of
 # what 2円 is, this means at the end of the loop,
 # either tail == 2円 (if M is a prime power) or
 # tail == 1 (if M is not a prime power)
 (?=(x+)(4円+$)) # 4円 = {largest proper divisor of tail}
 # = tail / {smallest prime factor of tail};
  # 5円 = tail - 4円
 5円 # tail = tail - (tail - 4円) = 4円
 )*+ # Iterate the above as many times as possible,
 # minimum zero, and lock in the result using a
 # possessive quantifier.
 x+ # Multiply number of possible matches by tail
 )
 x # tail -= 1
)* # Loop the above a minimum of 0 times
$ # Assert that at the end of the above loop, tail == 0

Regex 🐇 (ECMAScript+(?*)x*+^RME / PCRE2 _^v10.35+), (削除) 60 (削除ここまで) (削除) 59 (削除ここまで) (削除) 54 (削除ここまで) 52 bytes

^((?*(?=(xx+?)2円*$|)((?=x2円)(x+)4円*(?=4円$))*+x+)x)*$

Attempt This Online! - PCRE2 v10.40+
RegexMathEngine : call with regex -X -xml,pq -nx -t0..17 --line-buffered -f regex.txt where regex.txt contains the regex with no newline

It takes the base prime \$p\$ of every prime power \$p^k\le n\$, and calculates the product of that list of numbers. And because each \$p\$ will occur \$\max \left\{ k ,円 \middle| ,円 p^k \le n \right\}\$ times in that list, this product is the same as the product of the prime powers themselves would be, if only the largest from each base prime were included.

^ # tail = N = input number
( # Loop the following:
 (?* # Non-atomic lookahead:
 # M = tail, which cycles from N down to 1
 (?=(xx+?)2円*$|) # 2円 = smallest prime factor of M if M ≥ 2;
 # otherwise, leave 2円 unchanged in PCRE, or
 # unset in ECMAScript
 (
 (?=x2円) # Keep iterating until tail ≤ 2,円 and because of
 # what 2円 is, this means at the end of the loop,
 # either tail == 2円 (if M is a prime power) or
 # tail == 1 (if M is not a prime power)
 (x+)4円*(?=4円$) # tail = 4円 = {largest proper divisor of tail}
 # = tail / {smallest prime factor of tail}
 )*+ # Iterate the above as many times as possible,
 # minimum zero, and lock in the result using a
 # possessive quantifier.
 x+ # Multiply number of possible matches by tail
 )
 x # tail -= 1
)* # Loop the above a minimum of 0 times
$ # Assert that at the end of the above loop, tail == 0

Regex 🐇 (ECMAScript+(?*)+()++^RME / PCRE2 _^v10.35+), (削除) 60 (削除ここまで) (削除) 59 (削除ここまで) 54 bytes

Attempt This Online! - PCRE2 v10.40+
Try it on replit.com - ECMAScript+(?*)+()++ (RegexMathEngine -xml,pq)

Regex 🐇 (ECMAScript+(?*)x*+^RME / PCRE2 _^v10.35+), (削除) 60 (削除ここまで) (削除) 59 (削除ここまで) 54 bytes

Attempt This Online! - PCRE2 v10.40+
Try it on replit.com - ECMAScript+(?*)+x*+ (RegexMathEngine -xml,pq)