number theory - Are there infinitely many primes with digit sums of the form $n^2+1$?

The motivation for this question is from one of well known Landau Problems which asks for proof of the statement:

Are there infinitely many primes $p$ such that $p − 1$ is a perfect square? In other words: Are there infinitely many primes of the form $n^2 + 1$.

But this is not what I am asking. Before I ask my question let me explain the scenario.

A prime $p$ is called $\color{blue}{\text{Nice prime}} \ $if sum of its digits is of the form $n^2+1$ For example $37\rightarrow 3+7=3^2+1$ . The primes sum of whose digits is of the form $n^2$ (Eg $31\rightarrow 4=2^2$) or $n^2-1$ (Eg $71\rightarrow 8=3^2-1$) will be called Almost Nice primes.

The question is are there infinitely many Nice primes?

Now, I tried to find Nice and Almost Nice primes by hand till $400$ and here is what I've got:

Nice primes are:

$5,11,19,23,37,41,73,89,101,109,113,127,131,163,179,181,197,271,307,311,359,373$.

While Almost Nice primes are:

$n^2\rightarrow 1,3,31,79,97,103,211,277,367,349$

$n^2-1\rightarrow 3,17,53,71,107,233,251$

There is a reason why I called primes whose sum of digits is of the form $n^2$ and $n^2-1$ Almost nice primes.

If you have an Almost Nice Prime of the form $n^2-1$ then adding $2\times10^k$ to it (here $k$ is the highest power of $10$ in decimal expansion of $n^2-1$) will give you a Nice prime if it is a prime (by it I mean $n^2-1+2\times10^k$). In a similar manner, if a prime $p$ is an Almost Nice prime of the form $n^2$ then if $n^2+10^k$ is a prime then it will be a Nice prime.

But introducing Almost Nice prime is not much helpful as we have to make sure that $\text{Almost Nice prime (of the form $n^2$)}+10^k$ $\text{or Almost Nice prime (of the form $n^2-1$)+$2\times10^k$}$ is a prime before concluding that it met our Nice prime criteria.

Since the post is long, I again remind you the question. Are there infinitely many Nice primes?

Thanks.

Answer

Even the set of positive integers with digit sum $101$, only having the digits $0$ and $1$ and ending with a $1$, contains $$\binom{n-2}{99}$$ $n$-digit numbers.

This means, that we have , for example, about $\large \color\red {10^{438}}$ numbers with a million digits in the set. Plenty of them should be primes, since they share no common factor.

If $n$ increases, the binomial coefficient grows much faster than $n$ itself. So there is an overwhelming statistical evidence that infinite many nice primes exist.

Of course, such an evidence is no proof.