Totient and Sigma <--> Primes and Biprimes

Though it was obvious when I looked into it, I only recently realised that, for any prime number, the average of its totient and sum of divisors is equal to the number itself. This is because for a prime number $p$, it's only divisors are itself and $1$, thus the sum of divisors is $p+1$. For the totient, the only number that is coprime with $p$ is 1 and thus the totient is $p-1$. Adding the sum of divisors and the totient together gives $2p$ and the average is $p$. Thus formally stated:

In a similar vein, it was only today that I discovered that for a number that is biprime, sometimes called semiprime, the average of its sum of divisors and totient is always one more than the number. Again, this was obvious once I looked into it. The reason is that a biprime number $n$ has only two factors, let's say $a$ and $b$. Thus $n=ab$. The sum of divisors is thus $n+a+b+1$. For the totient, there are $b-1$ multiples of $a$ and $a-1$ multiples of $b$ that are coprime with $n$. Thus the totient is $n-a-b-1$, remembering that 1 is regarded as being coprime as well. Adding the sum of divisors and the totient together and averaging, we get:$$\frac{(n+a+b+1)-(n-a-b-1)}{2}=\frac{2n+2}{2}=n+1$$This is all very simple stuff but I'd never seen this connection between the totients and sums of divisors of primes and biprimes formally stated before. Thus formally stated: