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Suppose we are trying to avoid 3-term arithmetic progressions. There are two relevant sequences in the OEIS pertaining to this:

A003278: The sequence whose $n^{\text{th}}$ term is the smallest number $k$ for which a sequence of length $k$ avoiding any 3-term AP exists.

A065825: The sequence whose $n^{\text{th}}$ term is the smallest number $k$ for which the range $1\ldots k$ contains some sequence of $n$ numbers that avoids a 3-term AP.

A three-term AP is of course a set of numbers $a, b, c$ such that $$ a + c = 2b$$

My question is this:

Are there known sequences/results in either of the above formulations for a different "fibonacci-like" obstacle: namely, a 3-term sequence $a, b, c$ such that

$$a + b = c$$

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    $\begingroup$ I think if you search for "sum-free sets" you'll find a lot of work on these. Alternatively, figure out the first few terms, and look it up in the OEIS. $\endgroup$
    – Gerry Myerson
    Commented Apr 1, 2015 at 3:03
  • $\begingroup$ I would suggest this question would be more interesting if you sought a sequence that is Fibonacci free, alongwith being 3-AP free simultaneously. Also, to avoid trivial all-even or all-odd solutions, I'd enforce both the following variants of Fibonacci freeness: $f(a)+ f(b) \neq f(c)$ and $f(a) + f(b) \neq f(c) + 1$. $\endgroup$
    – user17348
    Commented Apr 1, 2015 at 17:29
  • $\begingroup$ @user17348 that's a good point, and is actually what I was really looking for. I thought I'd ask about this as an intermediate case. $\endgroup$
    – Suresh Venkat
    Commented Apr 1, 2015 at 20:22

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