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Suppose that $$c_0+c_1x+c_2x^2+ \cdots = \frac{1} {\lfloor{r}\rfloor+\lfloor{2r}\rfloor x+\lfloor{3r}\rfloor x^2+ \cdots},$$ where $r = (1+ \sqrt{5})/2$. Let $i_0 < i_1 < i_2 < \cdots $ be the numbers $i$ for which $c_i \ge 0.$ Is $i_j = 2j$ for every $j$?

For what other choices of $r$ is $i_j = 2j$ for every $j$?

(Added on Jan. 4, 2017) Here are some initial terms: $$ (c_0,c_1,c_2, \dots) = (1, -3, 5, -9, 17, -30, 52, -90, 154, -262, 446, -758, 1285, \dots ).$$ The first question is this: do the signs alternate for the whole sequence? The second question may lead to some surprises.

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  • $\begingroup$ This should work for $r=F_{t+1}/F_t$ for $t\geq5$; where $F_t$'s are the Fibonacci numbers. $\endgroup$
    – T. Amdeberhan
    Commented Dec 27, 2016 at 22:24
  • $\begingroup$ Sequence $\lfloor nr\rfloor$ in OEIS: oeis.org/A000201 $\endgroup$
    – Pietro Majer
    Commented Dec 28, 2016 at 20:26
  • $\begingroup$ Could you please write out the first few terms in this expansion? $\endgroup$
    – Pat Devlin
    Commented Jan 3, 2017 at 15:54
  • $\begingroup$ My guess would be: this is true for any real number $r$ between $3/2$ inclusive and $5/3$ exclusive. (This is an experimental observation, I don't have a proof.) $\endgroup$
    – Gro-Tsen
    Commented Jan 4, 2017 at 19:45

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