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Does there exist an integer sequence $\{a_n\}_{n = 1}^\infty$ that satisfies the following properties:

  1. $\forall t > 1, n^t = o(a_n)$
  2. $\forall p > 1, q > 0, a_n = o(p^{n^q})$ ?

The only thing I managed to determine, was, that the convergence radius of its generating function is $1$ and thus, by Pringsheim theorem it has a singularity in $1$, which is essential, because if it were a pole of degree $k$, then it would have meant, that $a_n = O(n^k)$, which is certainly not true.

However, that does not seem to be much helpful.

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    $\begingroup$ $\lfloor n^{\log n}\rfloor$ should work. $\endgroup$ – Wojowu Apr 17 at 14:31
  • $\begingroup$ @Wojowu, yes, this one indeed seems to work. You might probably want to post this as an answer, so I could accept it... $\endgroup$ – Yanior Weg Apr 17 at 15:46

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