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Here we are considering integers $n\geq 1$, and $\lfloor x\rfloor$ denotes the floor function. On the other hand one knows the so-called Bernoulli inequality, see for example this MathWorld. This morning I wondered about two puzzles that I've created involving previous inequality and two well known sequences of integers (in this post I was interested in prime numbers). These puzzles do not have a context in this inequality,
my creation or definition of the puzzles was artificious and has no a special mathematical meaning, but I am curious about if it is possible to deduce a solution.

Definition. For integers $n\geq 1$, I've considered when $$\lfloor n\log(1+n)-\log(1+n^2) \rfloor=\lfloor\log\frac{(1+n)^n}{1+n^2}\rfloor\tag{1}$$ is a prime number.

Computational fact (there was a mistake, see comments). Our sequence of prime numbers (those prime numbers arising from the sequence $(1)$) starts with $$3, 5, 13, 19, 29, 43, 47, 79, 83,\ldots$$ and seems that there isn't in The On-Line Encyclopedia of Integer Sequences.

Example. The prime number $3$ belongs of our sequence because $3$ is equals to $$\lfloor\log\frac{(1+4)^4}{1+4^2}\rfloor.$$

Question. Is it possible to deduce if there are infinitely many primes of the form $(1)$, or can you provide me a deduction about the corresponding counting function of our sequence of primes? What is your reasoning or heuristic? Many thanks.

I think that it is very difficult, but maybe some user knows how to count these primes.

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  • $\begingroup$ Yes many thanks since 3 divides $15$ @Peter $\endgroup$ – user243301 Jan 8 '18 at 11:36
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    $\begingroup$ The sequence grows very slowly and numerical analysis indicates that there are plenty of primes. $\endgroup$ – Peter Jan 8 '18 at 11:39
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    $\begingroup$ Yes @Peter , other reasoning is that 3 divides the su of the digits $8+7=15$ :) $\endgroup$ – user243301 Jan 8 '18 at 11:39
  • $\begingroup$ For easier analysis, it is useful that the numbers are equal $\large \lfloor n\cdot \ln(n+1)-\ln(n^2+1) \rfloor$ $\endgroup$ – Peter Jan 8 '18 at 11:47
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    $\begingroup$ Many thanks @Peter your examples are unbeatable. $\endgroup$ – user243301 Jan 8 '18 at 13:31

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