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In this MSE question, I asked whether one can prove that a generating function has infinitely many coefficients equal to zero. The answer given (and accepted) to that rather broad question was “No”.

The respondent raised an interesting refinement to the question, which seemed MO-hard to me, so I'm asking it here.

Question. Let $G$ be a generating function (ordinary, exponential, or otherwise) such as $$ G(a_n;\, x) := \sum_{n=0}^\infty a_nx^n, $$ where the coefficients $a_n$ satisfy some known recurrence relation R (linear or otherwise). For which R is it decidable whether $G$ has a maximal zero coefficient $a_m=0$? Put another way, for which R can it be shown that $G$ has infinitely many coefficients equal to zero?

Some possibly related posts:
Is it decidable whether the support of a rational $\mathbb{Z}$-series is a regular language?
Given a generating function with "zeros", can one derive the function for ONLY the "zeros"?

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  • $\begingroup$ The answer by Qiaochu to the second question that you linked answers that it is an open problem for linear recurrence relations. $\endgroup$
    – Boris Bukh
    Commented Oct 11, 2014 at 13:54
  • $\begingroup$ But it doesn't seem to me that his answer is dependent on whether the coefficients satisfy a recurrence relation — just a “rational generating function”. Am I missing something? $\endgroup$ Commented Oct 11, 2014 at 14:08
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    $\begingroup$ A sequence satisfies a linear recurrence iff the generating function is a rational function. $\endgroup$
    – Boris Bukh
    Commented Oct 11, 2014 at 14:39
  • $\begingroup$ Ah… Well, if you write your comment up as an answer, I'll accept it. $\endgroup$ Commented Oct 11, 2014 at 14:49
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Commented Oct 11, 2014 at 15:32

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