I was seeking a binary operator on natural numbers that is intermediate between the sum and the product, and explored this natural candidate:
$$x \star y = \lceil (x y + x + y)/2 \rceil \;.$$
Then I wondered which numbers are prime with respect to $\star$, i.e., only have one factoring. For example, $11 = 1 \star 10$ is prime but $13 = 1 \star 12 = 2 \star 8$ is not. Computing these $\star$-primes, I found they begin:
$$ 2,3,5,11,23,29,41,53,83,89,113,131,173,179,191, \ldots $$
I tried to prove there were an infinite number of $\star$-primes, but then discovered my primes are precisely the Sophie Germain primes (primes $p$ such that $2p+1$ is also prime), and it is unknown if there are an infinite number of them.
Q1. Why are the $\star$-primes as I defined them precisely the Sophie Germain primes?
I see that factoring $xy + x + y$ to $(x+1)(y+1)-1$ reveals the connection, but my argument for iff is not precise. (Incidentally, the ceiling cannot be ignored: replacing ceiling by floor results in different "primes.")
Q2. Is it possible to express the number of $\star$-divisors of $n$ in terms of a mixture of the number of divisors $\tau(n)$ and the number of partitions $p(n)$?
For example, here are the factors of $n=40$: $$(1 \star 39), (2 \star 26), (3 \star 19), (4 \star 15), (7 \star 9), (8 \star 8) \;,$$ And so 40 has 11 $\star$-divisors: $1, 2, 3, 4, 7, 8, 9, 15, 19, 26, 39 \;.$
I label this recreational because I'm sure this is like eating candy for many of you! Enjoy the snack!
Addendum. Incidentally, if $\star$ is defined using floor rather than ceiling, then the $\lfloor \star \rfloor$-primes $>3$ are even numbers $n$ such that $n+1$ and $2n+1$ are (conventionally) prime. I don't know if these primes have been named, or if it is known whether there are an infinite supply.