Let $k$ be any positive integer then how to prove that the sequence $$Q_k=10^{\lceil k \log_{10}(n) \rceil }+n^{k-1}$$ Contains infinitely many primes?
It seems like because if you look at some small values of $k$ we can think of infinude of primes $$Q_2=\{11,12,13,104,105,106,107,108,109,1010,1011,1012,1013 \cdots \}$$ $$Q_3=\{11,14,109,116,1025,10251036,1049,1064,1081,10100 ,111331 \cdots\}$$
Is this true? If false then how many terms are prime in the general $Q_n$ sequence?
Where this came from: Consider the sequence of of numbers formed by concatenating (keeping side by side) $n$ and $n^k$ . For $k=2$. The numbers would be $11 ,24,39,416,525,636 ,749,864 \cdots$ Let they form a sequence $s(n,k)$ . Thus $$s(n,2)= \{ 11 ,24,39,416,525,636 ,749,864 \cdots \}$$
Similarly
$$s(n,3)=\{11,28,327,464,5125 \cdots \}$$
We can easily see that $$s(n,k)=n \cdot 10^{\lceil k \log(n) \rceil }+n^k$$
The sequence in question $Q_k$ is just
$$Q_k= \frac{s(n,k)}{n}$$
The exponent $\lceil k\log_{10}n\rceil$ is constant for large stretches of $n$, so the question can be rephrased:
For $k=1$ this is
which is probably false (these are generalized Fermat primes with base 10, so $e$ must be a power of 2 and the expected number of primes is just 2, 11 and 101). For $k=2$ this is
which is true under stronger-than-currently-known results on prime gaps.
For larger $k$ this is surely unknown, since it requires finding primes in quadratic or higher polynomials.