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Jan 1, 2015 at 17:34 comment added Victor Kleptsyn Did you try using Proposition 1 of the paper you cite? For instance, its third conclusion says that $a(F_{n+1}-1)=a(F_{n-3}-1)$, and as your conjecture holds for first 4 values of $n$, it thus holds always (and moreover, you get an explicit formula -- signs are $-,-,+,+$ with a period of four).
Dec 31, 2014 at 7:04 comment added joro @DavidS.Newman I don't get numerical support for your claim.
Dec 31, 2014 at 4:07 comment added David S. Newman My guess about a(F_n-1) being odd was true for the first 21 values of n, but not true for larger values.
Dec 31, 2014 at 3:37 comment added Gjergji Zaimi I seem to remember having done this problem as an exercise in Stanley's EC1. If memory serves correct a(n) is odd iff all the strings of consecutive zeros in the Zeckendorff representation of n are odd.
Dec 31, 2014 at 2:25 history edited David S. Newman CC BY-SA 3.0
I changed a(F_n) to read a(F_n-1) to correct some confusion about the coefficient of the constant term
S Dec 30, 2014 at 23:09 history suggested Tadashi
Added relevant tag
Dec 30, 2014 at 22:56 review Suggested edits
S Dec 30, 2014 at 23:09
Dec 30, 2014 at 22:43 history edited David S. Newman CC BY-SA 3.0
Added missing parentheses
S Dec 30, 2014 at 22:40 history suggested barak manos CC BY-SA 3.0
Add some LaTex
Dec 30, 2014 at 22:32 review Suggested edits
S Dec 30, 2014 at 22:40
Dec 30, 2014 at 22:24 history asked David S. Newman CC BY-SA 3.0