Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
definition $$\small{a(p)-a(p-1)=1}\text{ for all primes }p\text{ and }a(km)=a(k)+a(m)\text{ for all }k,m\in \mathbb{N}$$
Euler's product formula $$\small{\varphi(n)=n\prod_{p|n}\left(1-{1\over p}\right)=n\prod_{p|n}{{p-1}\over p}}$$
where φ\$\varphi\$ denotes Euler's totient function and p\$p\$ varies only over prime numbers.
Combining both, we deduce the property
first property $$\small{a(\varphi(n))=a\left(n\prod_{p|n}{{p-1}\over p}\right)=a(n)+\sum_{p|n}\left(a(p-1)-a(p)\right)=a(n)-\sum_{p|n}{1=a(n)-\omega(n)}}$$
where ω\$\omega\$ denotes the number of distinct prime factors of n\$n\$.
Applying the resulting formula k + 1\$k + 1\$ times, where k\$k\$ is large enough so that φk+1(n) = 1\$\varphi^{k+1}(n)=1\$, we get
second property $$\small{ \begin{aligned} 0=a\left(\varphi^{k+1}(n)\right)=a\left(\varphi^k(n)\right)-\omega\left(\varphi^k(n)\right)&=a\left(\varphi^{k-1}(n)\right)-\omega\left(\varphi^{k-1}(n)\right)-\omega\left(\varphi^k(n)\right) \\ &= \cdots = a(n)-\sum_{j=0}^k\omega\left(\varphi^j(n)\right) \end{aligned} }$$
From this property, we obtain the formula
formula $$\small{a(n)=\sum_{j=0}^k\omega\left(\varphi^j(n)\right)=\sum_{j=0}^{+\infty}\omega\left(\varphi^j(n)\right)}$$
where the last equality holds because ω(1) = 0\$\omega(1)=0\$.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
where φ denotes Euler's totient function and p varies only over prime numbers.
Combining both, we deduce the property
where ω denotes the number of distinct prime factors of n.
Applying the resulting formula k + 1 times, where k is large enough so that φk+1(n) = 1, we get
From this property, we obtain the formula
where the last equality holds because ω(1) = 0.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
$$\small{a(p)-a(p-1)=1}\text{ for all primes }p\text{ and }a(km)=a(k)+a(m)\text{ for all }k,m\in \mathbb{N}$$
$$\small{\varphi(n)=n\prod_{p|n}\left(1-{1\over p}\right)=n\prod_{p|n}{{p-1}\over p}}$$
where \$\varphi\$ denotes Euler's totient function and \$p\$ varies only over prime numbers.
Combining both, we deduce the property
$$\small{a(\varphi(n))=a\left(n\prod_{p|n}{{p-1}\over p}\right)=a(n)+\sum_{p|n}\left(a(p-1)-a(p)\right)=a(n)-\sum_{p|n}{1=a(n)-\omega(n)}}$$
where \$\omega\$ denotes the number of distinct prime factors of \$n\$.
Applying the resulting formula \$k + 1\$ times, where \$k\$ is large enough so that \$\varphi^{k+1}(n)=1\$, we get
$$\small{ \begin{aligned} 0=a\left(\varphi^{k+1}(n)\right)=a\left(\varphi^k(n)\right)-\omega\left(\varphi^k(n)\right)&=a\left(\varphi^{k-1}(n)\right)-\omega\left(\varphi^{k-1}(n)\right)-\omega\left(\varphi^k(n)\right) \\ &= \cdots = a(n)-\sum_{j=0}^k\omega\left(\varphi^j(n)\right) \end{aligned} }$$
From this property, we obtain the formula
$$\small{a(n)=\sum_{j=0}^k\omega\left(\varphi^j(n)\right)=\sum_{j=0}^{+\infty}\omega\left(\varphi^j(n)\right)}$$
where the last equality holds because \$\omega(1)=0\$.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
where φ denotes Euler's totient function and p varies only over prime numbers.
Combining both, we deduce the property
where ω denotes the number of distinct prime factors of n.
Applying the resulting formula k + 1 times, where k is large enough so that φk+1(n) = 1, we get
From this property, we obtain the formula
where the last equality holds because ω(1) = 0.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
where φ denotes Euler's totient function and p varies only over prime numbers.
Combining both, we deduce the property
where ω denotes the number of distinct prime factors of n.
Applying the resulting formula k + 1 times, where k is large enough so that φk+1(n) = 1, we get
From this property, we obtain the formula
where the last equality holds because ω(1) = 0.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
where φ denotes Euler's totient function and p varies only over prime numbers.
Combining both, we deduce the property
where ω denotes the number of distinct prime factors of n.
Applying the resulting formula k + 1 times, where k is large enough so that φk+1(n) = 1, we get
From this property, we obtain the formula
where the last equality holds because ω(1) = 0.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
where φ denotes Euler's totient function and p varies only over prime numbers.
Combining both, we deduce the property
where ω denotes the number of distinct prime factors of n.
Applying the resulting formula k + 1 times, where k is large enough so that φk+1(n) = φk(n) = 1, we get
From this property, we obtain the formula
where the last equality holds because ω(1) = 0.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
where φ denotes Euler's totient function and p varies only over prime numbers.
Combining both, we deduce the property
where ω denotes the number of distinct prime factors of n.
Applying the formula k times, where k is large enough so that φk+1(n) = φk(n) = 1, we get
From this property, we obtain the formula
where the last equality holds because ω(1) = 0.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.
Jelly, 9 bytes
ÆṪÐL·ÆFL€S
Try it online! or verify all test cases.
Background
The definition of sequence A064097 implies that
where φ denotes Euler's totient function and p varies only over prime numbers.
Combining both, we deduce the property
where ω denotes the number of distinct prime factors of n.
Applying the resulting formula k + 1 times, where k is large enough so that φk+1(n) = 1, we get
From this property, we obtain the formula
where the last equality holds because ω(1) = 0.
How it works
ÆṪÐL·ÆFL€S Main link. Argument: n
ÐL· Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF Break each resulting integer into [prime, exponent] pairs.
L€ Compute the length of each list.
This counts the number of distinct prime factors.
S Add the results.