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Oct 4, 2017 at 22:33 comment added Gerry Myerson The link I gave no longer works. The bibliographic information on the publication is G. Myerson and A. J. van der Poorten. Some problems concerning recurrence sequences. Amer. Math. Monthly 102(8):698–705, 1995. Those with jstor access will find it at jstor.org/stable/2974639?seq=1#page_scan_tab_contents
Oct 25, 2014 at 12:24 comment added Joseph O'Rourke @GerryMyerson: Great paper, Gerry! So many nice questions and results. E.g., "it is impossible for every rational number to occur in a recurrence sequence."
Oct 25, 2014 at 12:20 vote accept Joseph O'Rourke
Oct 25, 2014 at 3:31 comment added Gerry Myerson This sequence is the first thing in a paper I wrote with Alf van der Poorten some years ago. maths.mq.edu.au/~alf/www-centre/alfpapers/a106.pdf
Oct 25, 2014 at 0:24 comment added Richard Stanley To follow up on Qiaochu's comment, this is exactly how I found the recurrence.
Oct 25, 2014 at 0:13 comment added Qiaochu Yuan @Joseph: as opposed to polynomial recurrences, the theory of linear recurrences is well-understood, and in particular it's well-understood exactly what sequences can be described in this way: they are precisely the sequences of the form $f(n) = \sum_i p_i(n) r_i^n$ where the $p_i$ are polynomials. The sequence in question here is $(0, 1, 0, 2, 0, 3, \dots) + (0, 0, -1, 0, -2, 0, -3, \dots)$ which is $\frac{1 + (-1)^n}{2} \frac{n}{2} - \frac{1 - (-1)^n}{2} \frac{n-1}{2}$ and it's not hard to reverse-engineer the recurrence from here.
Oct 24, 2014 at 23:57 comment added Joseph O'Rourke Beautiful solution to Q1! : $\;0, 1, -1, 2, -2, 3, -3, 4, -4, 5, 5, \ldots$.
Oct 24, 2014 at 23:50 history answered Richard Stanley CC BY-SA 3.0