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Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$. Then, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

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  • $\begingroup$ I get a different left hand side. Rewrite the n+2 term as the sum of n+1 and n terms, and then compute the difference of cubes and divide by three. Algebraically you get the product of the n term and the n+1 term and (the sum of n+1 and n terms). This seems to have more to do with (a+b)^n - a^n - b^n than with Fibonacci. Gerhard "Unsure Of Any Combinatorial Interpretation" Paseman, 2019.03.26. $\endgroup$ – Gerhard Paseman Mar 26 at 18:33
  • $\begingroup$ Thanks, edited accordingly. $\endgroup$ – T. Amdeberhan Mar 26 at 18:40
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    $\begingroup$ $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next. $\endgroup$ – Noam D. Elkies Mar 26 at 19:58
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$F_n$ is the number of compositions (ordered partitions) of $n-1$ into parts equal to 1 or 2. The number of triples $(a,b,c)$ of such compositions is $F_n^3$. The number such that $a,b,c$ all begin with 1 is $F_{n-1}^3$. The number such that $a,b,c$ all begin with 2 is $F_{n-2}^3$. Otherwise either one of $a,b,c$ begins with 1 and the others begin with 2, or vice versa. There are $3F_{n-1}F_{n-2}^2$ such triples of the first type. Similarly there are $3F_{n-1}^2F_{n-2}$ of the second type, i.e., one of $a,b,c$ begins with 2 and the others begin with 1. Hence \begin{eqnarray*} F_n^3 & = & F_{n-1}^3 + F_{n-2}^3 +3(F_{n-1}^2F_{n-2}+F_{n-1}F_{n-2}^2)\\ & = & F_{n-1}^3 + F_{n-2}^3 +3F_{n-1}F_{n-2}(F_{n-1}+F_{n-2})\\ & = & F_{n-1}^3 + F_{n-2}^3 + 3F_nF_{n-1}F_{n-2}. \end{eqnarray*}

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    $\begingroup$ With the greatest respect, and mostly out of curiosity, would you really prefer such a bijective proof to the algebra in e.g. Elkies's comment? $\endgroup$ – Lucia Mar 27 at 0:01
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    $\begingroup$ If it's simply a matter of proving the identity, then I prefer Elkies. If you want to understand it combinatorially, then a bijective proof is better. $\endgroup$ – Richard Stanley Mar 27 at 0:13
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    $\begingroup$ The OP specified a desire for "combinatorial" or "conceptual" explanations. But the distinction between combinatorics and algebra is blurry here. You have to choose one ball from each of three urns, each containing $a$ amaranth and $b$ blue balls. How many choices don't have all three balls the same color? On the one hand, $(a+b)^3-a^3-b^3$. On the other, choose any cyclic permutation of (amaranth, blue, either) to get $3ab(a+b)$. $\endgroup$ – Noam D. Elkies Mar 27 at 0:14
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    $\begingroup$ +1 for use of "amaranth" as a color. $\endgroup$ – Michael Lugo Mar 27 at 14:16
  • $\begingroup$ This generalizes to give $F_{n-1} F_{n-2} = (F_n^2 - F_{n-1}^2 - F_{n-2}^2)/2$ (by considering pairs of compositions) but one doesn't get anything similarly nice for fourth powers as far as I can tell. $\endgroup$ – Michael Lugo Mar 27 at 14:22
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This is just the following identity: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$ Since $$F_n+(-F_{n-1})+(-F_{n-2})=0,$$ your formula follows.

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    $\begingroup$ Simpler yet: since $F_n = F_{n-1} + F_{n+2}$ it's enough to use the two-variable identity $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$ which is a quick consequence of the binomial expansion of $(a+b)^3$. $\endgroup$ – Noam D. Elkies Mar 26 at 20:00
  • $\begingroup$ @NoamD.Elkies: Thanks for this approach too. $\endgroup$ – T. Amdeberhan Mar 27 at 14:47

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