Is $\frac{3^n+5^n+7^n}{15}$ only prime if $n$ is prime? Let $f(n)=3^n+5^n+7^n$
It is easy to show that $\ 15\mid f(n)\ $ if and only if $\ n\ $ is odd.
I searched for prime numbers of the form $g(n):=\frac{3^n+5^n+7^n}{15}$ with odd $n$ and found the following $n$ leading to a prime number : $$7,17,61,71,457,8111$$ All those numbers are prime numbers.

Can we show that this is a necessary condition ?

The cases $3\mid n$ and $5\mid n$ are clear because $\frac{3^n+5^n+7^n}{15}$ is divisble by $\ 3\ $ and $\ 5\ $ respectively, but what about $\ g(77)\ $ which smallest prime factor is $\ 463\ $ and other cases ?
 A: If n is even then $3^n + 5^n + 7^n$ is not divisible by 15.
If n is divisible by 3 or 5 then $(3^n + 5^n + 7^n) / 15$ is divisible by 3 or 5.
So you are asking: Is there an n, not prime, not divisible by 2, 3 or 5, such that $(3^n + 5^n + 7^n) / 15$ is prime?
For given M, the number of candidates for n is about M * (4/15 - log M). The probability that $(3^n + 5^n + 7^n)/15$ is prime is about $1 / \log (7^n / 15)$ ≈ $1 /(1.95 \cdot n)$.
The sum of 1/n is about ln M, we multiply this by $4 / (15 * 1.95)$ giving about $\ln M / 7.3$. According to this very inaccurate calculation, we can expect one prime for n ≤ 1,500 and two primes for n ≤ 2,200,000. But an expected value of 1 or 2 means it is absolutely possible and not particularly unlikely that there are no primes in that range.
Unless you have a mathematical proof otherwise, I would assume that there is such an n, actually an infinite number of such n's, but an exhaustive search for n up to 2,000,000 or even 4 billion without finding one doesn't mean much. Obviously the heuristics can be improved a lot.
