Products and sum of cubes in Fibonacci












6












$begingroup$


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 $F_n!=F_1cdots F_n$ 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
$$binom{n}3_F=frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$






nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities







share|cite|improve this question











$endgroup$












  • $begingroup$
    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






  • 13




    $begingroup$
    $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
    $endgroup$
    – Noam D. Elkies
    Mar 26 at 19:58
















6












$begingroup$


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 $F_n!=F_1cdots F_n$ 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
$$binom{n}3_F=frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$






nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities







share|cite|improve this question











$endgroup$












  • $begingroup$
    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






  • 13




    $begingroup$
    $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
    $endgroup$
    – Noam D. Elkies
    Mar 26 at 19:58














6












6








6


3



$begingroup$


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 $F_n!=F_1cdots F_n$ 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
$$binom{n}3_F=frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$






nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities







share|cite|improve this question











$endgroup$




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 $F_n!=F_1cdots F_n$ 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
$$binom{n}3_F=frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$






nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities




nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 26 at 18:40







T. Amdeberhan

















asked Mar 26 at 18:15









T. AmdeberhanT. Amdeberhan

18.1k229132




18.1k229132












  • $begingroup$
    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






  • 13




    $begingroup$
    $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
    $endgroup$
    – Noam D. Elkies
    Mar 26 at 19:58


















  • $begingroup$
    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






  • 13




    $begingroup$
    $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
    $endgroup$
    – Noam D. Elkies
    Mar 26 at 19:58
















$begingroup$
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






$begingroup$
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




$begingroup$
Thanks, edited accordingly.
$endgroup$
– T. Amdeberhan
Mar 26 at 18:40




13




13




$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
Mar 26 at 19:58




$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
Mar 26 at 19:58










2 Answers
2






active

oldest

votes


















15












$begingroup$

$F_n$ 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*}






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    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






  • 4




    $begingroup$
    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






  • 8




    $begingroup$
    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






  • 2




    $begingroup$
    +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



















14












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 8




    $begingroup$
    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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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









15












$begingroup$

$F_n$ 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*}






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    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






  • 4




    $begingroup$
    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






  • 8




    $begingroup$
    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






  • 2




    $begingroup$
    +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
















15












$begingroup$

$F_n$ 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*}






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    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






  • 4




    $begingroup$
    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






  • 8




    $begingroup$
    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






  • 2




    $begingroup$
    +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














15












15








15





$begingroup$

$F_n$ 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*}






share|cite|improve this answer











$endgroup$



$F_n$ 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*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 29 at 6:04









Aaron Meyerowitz

24.4k13388




24.4k13388










answered Mar 26 at 23:54









Richard StanleyRichard Stanley

29.2k9116191




29.2k9116191








  • 3




    $begingroup$
    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






  • 4




    $begingroup$
    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






  • 8




    $begingroup$
    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






  • 2




    $begingroup$
    +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














  • 3




    $begingroup$
    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






  • 4




    $begingroup$
    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






  • 8




    $begingroup$
    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






  • 2




    $begingroup$
    +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








3




3




$begingroup$
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




$begingroup$
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




4




4




$begingroup$
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




$begingroup$
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




8




8




$begingroup$
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




$begingroup$
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




2




2




$begingroup$
+1 for use of "amaranth" as a color.
$endgroup$
– Michael Lugo
Mar 27 at 14:16




$begingroup$
+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




$begingroup$
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











14












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 8




    $begingroup$
    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
















14












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 8




    $begingroup$
    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














14












14








14





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 26 at 19:52









Cherng-tiao PerngCherng-tiao Perng

770148




770148








  • 8




    $begingroup$
    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














  • 8




    $begingroup$
    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








8




8




$begingroup$
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




$begingroup$
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




$begingroup$
@NoamD.Elkies: Thanks for this approach too.
$endgroup$
– T. Amdeberhan
Mar 27 at 14:47


















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