Puzzle : Is there an ODD positive integer, such that $$n^{n+1}+(n+1)^{n+2}$$ is a prime number ?

Hey Peter, nice that you came in a visit, I can think a little about that problem, how far did you do the computations?

Integer $n$

I passes $n=8\ 000$

What are the reasons for not investigating the even n?

For even $n$ I found some primes

And for odd $n$ , there are several restrictions.

Do you think that this sequence has only a finite number of primes for all n?

The naive heuristic predicts infinite many primes, but the next above $n = 392$ could be huge.

Heuristic may very well be wrong.

If $n$ is odd and the expression prime, $n$ must be of the form $6k+5$ and $n+2$ cannot be prime.

It is usually impossible to decide whether a sequence produces infinite many primes. We have nothing better than a heuristic.

Apart from some special examples, like $an+b$ or $a^2+b^4$

Yes, those two are two famous theorems in number theory.

And $n^2+1$ is already too difficult , apparently.

@Grešnik You can post your order problem here with the question whether $32$ can be beaten !

Where, in this chatroom?

Why not, isn't this the room for such puzzles ?

Of course it is, but I was thinking that it could be better if I post that on-demand, that is, if some users come in this room and ask us what are we trying to settle?

Are factors of the above sequence often close to $\sqrt{n}^n$?

When the numbers are semiprimes.

I am not aware of a factorization with two prime factors of approximately the same magnitude.

So, there are not much semiprimes?

Semiprimes are numbers with two (possibly repeated) prime factors in the factorization (also the square of a prime is allowed). I did not look for which $n$ we have a semiprime, but probably MathPhile would be interested in that. I can search for some examples.

You can try in some small ranges if you want.

The first few $n$ giving a semiprime :

1 5 7 8 11 14 28

In the case of $n=1$, we have the square of a prime.

I think that can happen only when n=1.

I also think that, maybe it can even be proven.

For $n=54$ , we have another semiprime.

And $n = 114$ gives a semiprime. I think, no others are there until that limit.

So you know about 250 000 dollars award for finding a "large" prime?

@Grešnik hello

how are you?

Hey.