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Is there an easy way to prove that $|\operatorname{Re}(a_n)| \to \infty$ where $a_n=\left(\frac{1}{2}+i\frac{\sqrt{7}}{2}\right)^n$?

Of course $|a_n| \to \infty$, but we have $$ \operatorname{Re}(a_n)=2^{-n}\sum_{0 \leq k \leq [n/2]} (-1)^k\binom{n}{k}7^k $$ where $[n/2]$ stands for the integer part of $n/2$.

However I don't see how this could be computed differently, or estimated in order to prove the claim...

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    $\begingroup$ Asked and answered some years ago on Math Stack Exchange. The result is nontrivial but can be proved in several ways using known techniques. math.stackexchange.com/questions/705877/an-exotic-sequence $\endgroup$
    – Noam D. Elkies
    Commented May 26, 2023 at 12:56
  • $\begingroup$ Thank you, it is indeed hard! $\endgroup$
    – J.Mayol
    Commented May 26, 2023 at 13:04
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    $\begingroup$ Your definition of $a_n$ seems to be an absolute value, so it's already real. Did you mean to use parentheses? Also, use \left and \right to make them the right size, thus $$a_n = \left(\frac12+i\frac{\sqrt7}{2}\right)^n.$$ $\endgroup$
    – Joe Silverman
    Commented May 26, 2023 at 13:06
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    $\begingroup$ And did you mean $|\operatorname{Re}(a_n)|\to\infty$? $\endgroup$
    – Dave Benson
    Commented May 26, 2023 at 18:13
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    $\begingroup$ Yes, indeed. Sorry for this mess. The answer of @NoamD.Elkies is exactly answering this statement (and not the two previous iterations of my question, sorry). $\endgroup$
    – J.Mayol
    Commented May 29, 2023 at 11:14

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