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: