User:Alonso del Arte/Labos function

I don't know if the name will stick, but I'm calling this the Labos function, defined as $\Lambda (n)=\phi (\sigma (n)-\phi (n))$ for $n\in \mathbb {Z} ^{+}$ (see A077088). The interesting thing happens when you iterate the function. The trajectory might hit 4 or 24 and get stuck there. Others get into more interesting "orbits." The table below summarizes what I've found from a survey of a few small starting values:

• $p$, 1, 0 ($p$ is any positive, purely real prime)
• (9 or 14), 6, 4, 4, 4, ...
• (15 or 21 or 35 or 49), 8, 10, 6, 4, 4, 4, ...
• (25 or 26 or 27), 10, 6, 4, 4, 4, ...
• (28 or 57 or 62), 20, 16, 22, 12, 8, 10, 6, 4, 4, 4, ...
• (33 or 38 or 65 or 77), 12, 8, 10, 6, 4, 4, 4, ...
• (34 or 45), 18, 20, 16, 22, 12, 8, 10, 6, 4, 4, 4, ...
• (39 or 51 or 55), 16, 22, 12, 8, 10, 6, 4, 4, 4, ...
• (40 or 52), 36, 78, 48, 36, 78, 48, 36, 78, 48, ...
• (42 or 69 or 74 or 75), 24, 24, 24, ...
• (44 or 54 or 56 or 63 or 70), 32, 46, 20, 16, 22, 12, 8, 10, 6, 4, 4, 4, ...
• (50 or 64), 72, 108, 120, 160, 156, 168, 144, 280, 192, 144, 280, 192, 144, 280, 192, ...
• 58, 30, 32, 46, 20, 16, 22, 12, 8, 10, 6, 4, 4, 4, ...
• (66 or 80), 60, 72, 108, 120, 160, 156, 168, 144, 280, 192, 144, 280, 192, 144, 280, 192, ...
• 68, 46, 20, 16, 22, 12, 8, 10, 6, 4, 4, 4, ...
• 76, 48, 36, 78, 48, 36, 78, 48, 36, 78, ...

We will continue to use $p$ to mean any prime in $\mathbb {Z} ^{+}$.

Theorem. (Labos) $\Lambda (p)=1$, otherwise $\Lambda (n)$ is even.

Lemma 1. $\Lambda (p)=1$.

Proof. Since $p$ is prime, it has only two divisors, 1 and $p$ itself. Therefore $\sigma (p)=p+1$. On the other hand, $\phi (p)=p-1$. Therefore $\Lambda (p)=\phi ((p+1)-(p-1))=\phi (2)=1$.

Lemma 2. $\Lambda (p^{2})=\phi (2p+1)$.

Proof. $\sigma (p^{2})=p^{2}+p+1$. Recall that, given $\alpha >0$, $\phi (p^{\alpha })=(p-1)^{\alpha -1}$. Therefore $\phi (p^{2})=(p-1)p=p^{2}-p$. Then $\Lambda (p^{2})=\phi ((p^{2}+p+1)-(p^{2}-p))=\phi (2p+1)$.

The reason to focus specifically on the case of squares of primes rather than composite numbers in general will soon be apparent.

Proof. Lemma 1 takes care of the first clause of the theorem, so then it remains to show that if $n=1$ or if $n$ is composite, then $\Lambda (n)$ is even. The special case of $\Lambda (1)$ is disposed of if you agree that $\phi (0)=0$, in which case then $\Lambda (1)=0$. There is no such philosophical difficulty for composite $n$, for which we can assert that $\sigma (n)\geq n^{2}+n+1$—this is an equality if and only if $n$ is the square of a prime. The fewest number of divisors a composite number can have is 3, in which case it is the square of a prime. The more divisors a number has, the larger $\sigma (n)$ is. Concomittantly, as $\tau (n)$ increases, $\phi (n)$ decreases: $\phi (n)\leq n-{\sqrt {n}}$. This means that the gulf between $\sigma (n)$ and $\phi (n)$ increases, and thus $\sigma (n)-\phi (n)\geq 2n+1\geq 2$. Recall that $\phi (n)$ is even for all $n>2$ (see Euler's totient function#Properties). So, if $\Lambda (n)\geq \phi (2n+1)$, this means that $\Lambda (n)$ is even. □