正则表达式（ECMAScript的），276个 205 201 193 189字节
比较不同主要因子的多重性（指数）是使用ECMAScript正则表达式解决的一个有趣问题-缺少在循环迭代过程中持续存在的反向引用，这对计算任何事物都是一个挑战。即使可以对所讨论的数字特征进行计数，通常也可以采用更间接的方法进行更好的高尔夫运动。
与我在ECMA regex上发表的其他文章一样，我会发出扰人的警告：我强烈建议学习如何在ECMAScript regex中解决一元数学问题。对于我而言，这是一次引人入胜的旅程，对于那些可能想要自己尝试的人，尤其是对数字理论感兴趣的人，我都不想破坏它。请参阅此较早的帖子，以获取一系列连续扰流器标记的建议问题的列表，以逐一解决。
因此，如果您不想破坏一些高级一元正则表达式魔术，请不要再继续阅读。如果您确实想自己弄清楚该魔术，我强烈建议您先解决ECMAScript regex中的一些问题，如上面链接的那部分所述。
我以前开发的正则表达式的主要有效载荷非常适用于此挑战。这是找到最高多重性的质数的正则表达式。我的第一个解决方案很长，后来我分阶段进行了深入研究，首先将其重写以使用分子先行，然后使用高级技术将其移植回普通ECMAScript，以解决缺少分子先行的问题，随后打高尔夫球，使其比原始的普通ECMAScript解决方案小得多。
正则表达式中适用于此问题的部分是第一步，它找到Q，即共享所有素数的最小N因子。一旦有了这个数字，我们所要做的就是证明N是一个“恒定指数数”，就是将N除以Q，直到不再可用为止。如果结果为1，则所有素数都具有相同的多重性。
在使用我先前开发的算法找到Q并提交答案后，我意识到可以用一种完全不同的方式来计算它：找到N的最大无平方因子（使用与我的Carmichael number regex相同的算法）。事实证明，就避免缺乏分子前行和后向可变长后行（无需引入以前使用的高级技术）而言，这一点都不困难*，并且要短64字节！此外，它消除了将无平方N和素数N视为不同特殊情况的复杂性，从该解决方案中又删除了7个字节。
（仍然存在其他问题，这些问题需要以前在这里使用的先进技术来简化Q的计算，但目前我的PPCG帖子中都没有这些问题。）
我将多重性检验放在连续素数检验之前，因为后者要慢得多。首先进行可能会更快失败的测试，可使正则表达式更快地获得均匀分布的输入。首先使用高尔夫也是更好的选择，因为它使用更多的反向引用（如果使用两位数，则将花费更多）。
我可以使用Grimy发现的技巧从此正则表达式（193→189）中删除4个字节，该技巧可以在商被确保大于或等于除数的情况下进一步缩短除法。
^(?=(|(x+)\2*(?=\2$))((?=(xx+?)\4*$)(?=(x+)(\5+$))\6(?!\4*$))*x$)(?=.*$\2|((?=((x*)(?=\2\9+$)x)(\8*$))\10)*x$)(?!(((x+)(?=\13+$)(x+))(?!\12+$)(x+))\11*(?=\11$)(?!(\15\14?)?((xx+)\18+|x?)$))
在线尝试！
# For the purposes of these comments, the input number = N.
^
# Assert that all of N's prime factors are of equal multiplicity
# Step 1: Find Q, the largest square-free factor of N (which will also be the smallest
# factor of N that has all the same prime factors as N) and put it in \2.
# If N is square-free, \2 will be unset.
(?=
# Search through all factors of N, from largest to smallest, searching for one that
# satisfies the desired property. The first factor tried will be N itself, for which
# \2 will be unset.
(|(x+)\2*(?=\2$)) # for factors < N: \2 = factor of N; tail = \2
# Assert that tail is square-free (its prime factors all have single multiplicity)
(
(?=(xx+?)\4*$) # \4 = smallest prime factor of tail
(?=(x+)(\5+$)) # \5 = tail / \4 (implicitly); \6 = tool to make tail = \5
\6 # tail = \5
(?!\4*$) # Assert that tail is no longer divisible by \4, i.e. that that
# prime factor was of exactly single multiplicity.
)*x$
)
# Step 2: Require that either \2 is unset, or that the result of repeatedly
# dividing tail by \2 is 1.
(?=
.*$\2
|
(
# In the following division calculation, we can skip the test for divisibility
# by \2-1 because it's guaranteed that \2 <= \8. As a result, we did not need to
# capture \2-1 above, and can use a better-golfed form of the division.
(?=
( # \8 = tail / \2
(x*) # \9 = \8-1
(?=\2\9+$)
x
)
(\8*$) # \10 = tool to make tail = \8
)
\10 # tail = \8
)*
x$ # Require that the end result is 1
)
# Assert that there exists no trio of prime numbers such that N is divisible by the
# smallest and largest prime but not the middle prime.
(?!
( # \11 = a factor of N
( # \12 = a non-factor of N between \11 and \13
(x+)(?=\13+$) # \13 = a factor of N smaller than \11
(x+) # \14 = tool (with \15) to make tail = \13
)
(?!\12+$)
(x+) # \15 = tool to make tail = \12
)
\11*(?=\11$) # tail = \11
# Assert that \11, \12, and \13 are all prime
(?!
(\15\14?)? # tail = either \11, \12, or \13
((xx+)\18+|x?)$
)
)
*分子前瞻性仍然更清洁，N的特殊情况是无平方的。此滴6个字节，从而产生195 187 183字节溶液：
^(?=(?*(x+))\1*(?=\1$)((?=(xx+?)\3*$)(?=(x+)(\4+$))\5(?!\3*$))*x$)(?=((?=((x*)(?=\1\8+$)x)(\7*$))\9)*x$)(?!(((x+)(?=\12+$)(x+))(?!\11+$)(x+))\10*(?=\10$)(?!(\14\13?)?((xx+)\17+|x?)$))
# For the purposes of these comments, the input number = N.
^
# Assert that all of N's prime factors are of equal multiplicity
# Step 1: Find Q, the largest square-free factor of N (which will also be the smallest
# factor of N that has all the same prime factors as N) and put it in \1.
(?=
(?*(x+)) # \1 = proposed factor of N
\1*(?=\1$) # Assert that \1 is a factor of N; tail = \1
# Assert that tail is square-free (its prime factors all have single multiplicity)
(
(?=(xx+?)\3*$) # \3 = smallest prime factor of tail
(?=(x+)(\4+$)) # \4 = tail / \3 (implicitly); \5 = tool to make tail = \4
\5 # tail = \4
(?!\3*$) # Assert that tail is no longer divisible by \3, i.e. that that
# prime factor was of exactly single multiplicity.
)*x$
)
# Step 2: Require that the result of repeatedly dividing tail by \1 is 1.
(?=
(
# In the following division calculation, we can skip the test for divisibility
# by \1-1 because it's guaranteed that \2 <= \8. As a result, we did not need to
# capture \1-1 above, and can use a better-golfed form of the division.
(?=
( # \7 = tail / \1
(x*) # \8 = \7-1
(?=\1\8+$)
x
)
(\7*$) # \9 = tool to make tail = \7
)
\9 # tail = \7
)*
x$ # Require that the end result is 1
)
# Assert that there exists no trio of prime numbers such that N is divisible by the
# smallest and largest prime but not the middle prime.
(?!
( # \10 = a factor of N
( # \11 = a non-factor of N between \10 and \12
(x+)(?=\12+$) # \12 = a factor of N smaller than \10
(x+) # \13 = tool (with \14) to make tail = \12
)
(?!\11+$)
(x+) # \14 = tool to make tail = \11
)
\10*(?=\10$) # tail = \10
# Assert that \10, \11, and \12 are all prime
(?!
(\14\13?)? # tail = either \10, \11, or \12
((xx+)\17+|x?)$
)
)
在这里它被移植到可变长度的后面：
正则表达式（ECMAScript的2018），198个 195 194 186 182字节
^(?=(x+)(?=\1*$)(?<=^x((?<!^\5*)\3(?<=(^\4+)(x+))(?<=^\5*(x+?x)))*))((?=((x*)(?=\1\8+$)x)(\7*$))\9)*x$(?<!(?!(\14\16?)?((xx+)\12+|x?)$)(?<=^\13+)((x+)(?<!^\15+)((x+)(?<=^\17+)(x+))))
在线尝试！
# For the purposes of these comments, the input number = N.
^
# Assert that all of N's prime factors are of equal multiplicity
# Step 1: Find Q, the largest square-free factor of N (which will also be the smallest
# factor of N that has all the same prime factors as N) and put it in \1.
(?=
(x+)(?=\1*$) # \1 = factor of N; head = \1
(?<= # This is evaluated right-to-left, so please read bottom to top.
^x
(
(?<!^\5*) # Assert that head is no longer divisible by \6, i.e. that
# that prime factor was of exactly single multiplicity.
\3 # head = \4
(?<=(^\4+)(x+)) # \4 = head / \5 (implicitly); \3 = tool to make head = \4
(?<=^\5*(x+?x)) # \5 = smallest prime factor of head
)*
)
)
# Step 2: Require that the result of repeatedly dividing tail by \1 is 1.
(
# In the following division calculation, we can skip the test for divisibility
# by \1-1 because it's guaranteed that \2 <= \8. As a result, we did not need to
# capture \1-1 above, and can use a better-golfed form of the division.
(?=
( # \7 = tail / \1
(x*) # \8 = \7-1
(?=\1\8+$)
x
)
(\7*$) # \9 = tool to make tail = \7
)
\9 # tail = \7
)*
x$ # Require that the end result is 1
# Assert that there exists no trio of prime numbers such that N is divisible by the
# smallest and largest prime but not the middle prime.
# This is evaluated right-to-left, so please read bottom to top, but switch back to
# reading top to bottom at the negative lookahead.
(?<!
# Assert that \13, \15, and \17 are all prime.
(?!
(\14\16?)? # tail = either \13, \15, or \17
((xx+)\12+|x?)$
)
(?<=^\13+)
( # tail = \13
(x+) # \14 = tool to make tail = \15
(?<!^\15+)
(
(x+) # \16 = tool (with \14) to make tail = \17
(?<=^\17+)(x+) # \17 = a factor of N smaller than \13
) # \15 = a non-factor of N between \13 and \17
) # \13 = a factor of N
)