Products and sum of cubes in FibonacciPrincipal term of the Dirichlet Divisor problem, from the work of A.F. Lavrik?Mean number of $n$-simplices per $(n-2)$-simplex in a triangulated $n$-manifold determinant of fibonacci-sum graphssum of three cubes and parametric solutionsSum of consecutive cubesDistinctness of products of Fibonacci numbersFive cubes, Hadamard and ShklyarskiyGcd of Fibonacci and CatalanA link between hooks and contents: Part II“Oddity” of Fibonacci-Catalan numbers
Products and sum of cubes in Fibonacci
Principal term of the Dirichlet Divisor problem, from the work of A.F. Lavrik?Mean number of $n$-simplices per $(n-2)$-simplex in a triangulated $n$-manifold determinant of fibonacci-sum graphssum of three cubes and parametric solutionsSum of consecutive cubesDistinctness of products of Fibonacci numbersFive cubes, Hadamard and ShklyarskiyGcd of Fibonacci and CatalanA link between hooks and contents: Part II“Oddity” of Fibonacci-Catalan numbers
$begingroup$
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?
$$F_nF_n-1F_n-2=fracF_n^3-F_n-1^3-F_n-2^33.$$
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, $binomnk_F:=fracF_n!F_k!cdot F_n-k!$. Then, I was studying these coefficients and was lead to
$$binomn3_F=fracF_n^3-F_n-1^3-F_n-2^33!.$$
nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
$endgroup$
add a comment |
$begingroup$
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?
$$F_nF_n-1F_n-2=fracF_n^3-F_n-1^3-F_n-2^33.$$
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, $binomnk_F:=fracF_n!F_k!cdot F_n-k!$. Then, I was studying these coefficients and was lead to
$$binomn3_F=fracF_n^3-F_n-1^3-F_n-2^33!.$$
nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
$endgroup$
$begingroup$
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
16
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
Mar 26 at 19:58
add a comment |
$begingroup$
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?
$$F_nF_n-1F_n-2=fracF_n^3-F_n-1^3-F_n-2^33.$$
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, $binomnk_F:=fracF_n!F_k!cdot F_n-k!$. Then, I was studying these coefficients and was lead to
$$binomn3_F=fracF_n^3-F_n-1^3-F_n-2^33!.$$
nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
$endgroup$
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?
$$F_nF_n-1F_n-2=fracF_n^3-F_n-1^3-F_n-2^33.$$
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, $binomnk_F:=fracF_n!F_k!cdot F_n-k!$. Then, I was studying these coefficients and was lead to
$$binomn3_F=fracF_n^3-F_n-1^3-F_n-2^33!.$$
nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities
nt.number-theory reference-request co.combinatorics elementary-proofs combinatorial-identities
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
edited Mar 26 at 18:40
T. Amdeberhan
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.7k2 gold badges32 silver badges136 bronze badges
18.7k2 gold badges32 silver badges136 bronze badges
$begingroup$
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
16
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
Mar 26 at 19:58
add a comment |
$begingroup$
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
16
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. 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
16
16
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. 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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$F_n$ 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 $3F_n-1F_n-2^2$ 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
begineqnarray* F_n^3 & = & F_n-1^3 + F_n-2^3
+3(F_n-1^2F_n-2+F_n-1F_n-2^2)\ & = &
F_n-1^3 + F_n-2^3 +3F_n-1F_n-2(F_n-1+F_n-2)\ & = &
F_n-1^3 + F_n-2^3 + 3F_nF_n-1F_n-2. endeqnarray*
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
$endgroup$
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
5
$begingroup$
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
9
$begingroup$
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
3
$begingroup$
+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
|
show 2 more comments
$begingroup$
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.
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
$endgroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$F_n$ 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 $3F_n-1F_n-2^2$ 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
begineqnarray* F_n^3 & = & F_n-1^3 + F_n-2^3
+3(F_n-1^2F_n-2+F_n-1F_n-2^2)\ & = &
F_n-1^3 + F_n-2^3 +3F_n-1F_n-2(F_n-1+F_n-2)\ & = &
F_n-1^3 + F_n-2^3 + 3F_nF_n-1F_n-2. endeqnarray*
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
$endgroup$
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
5
$begingroup$
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
9
$begingroup$
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
3
$begingroup$
+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
|
show 2 more comments
$begingroup$
$F_n$ 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 $3F_n-1F_n-2^2$ 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
begineqnarray* F_n^3 & = & F_n-1^3 + F_n-2^3
+3(F_n-1^2F_n-2+F_n-1F_n-2^2)\ & = &
F_n-1^3 + F_n-2^3 +3F_n-1F_n-2(F_n-1+F_n-2)\ & = &
F_n-1^3 + F_n-2^3 + 3F_nF_n-1F_n-2. endeqnarray*
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
$endgroup$
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
5
$begingroup$
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
9
$begingroup$
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
3
$begingroup$
+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
|
show 2 more comments
$begingroup$
$F_n$ 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 $3F_n-1F_n-2^2$ 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
begineqnarray* F_n^3 & = & F_n-1^3 + F_n-2^3
+3(F_n-1^2F_n-2+F_n-1F_n-2^2)\ & = &
F_n-1^3 + F_n-2^3 +3F_n-1F_n-2(F_n-1+F_n-2)\ & = &
F_n-1^3 + F_n-2^3 + 3F_nF_n-1F_n-2. endeqnarray*
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
$endgroup$
$F_n$ 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 $3F_n-1F_n-2^2$ 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
begineqnarray* F_n^3 & = & F_n-1^3 + F_n-2^3
+3(F_n-1^2F_n-2+F_n-1F_n-2^2)\ & = &
F_n-1^3 + F_n-2^3 +3F_n-1F_n-2(F_n-1+F_n-2)\ & = &
F_n-1^3 + F_n-2^3 + 3F_nF_n-1F_n-2. endeqnarray*
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
edited Mar 29 at 6:04
Aaron Meyerowitz
24.9k1 gold badge34 silver badges90 bronze badges
24.9k1 gold badge34 silver badges90 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
30k9 gold badges120 silver badges195 bronze badges
30k9 gold badges120 silver badges195 bronze badges
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
5
$begingroup$
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
9
$begingroup$
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
3
$begingroup$
+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
|
show 2 more comments
3
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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?
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– Lucia
Mar 27 at 0:01
5
$begingroup$
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
9
$begingroup$
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
3
$begingroup$
+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
5
5
$begingroup$
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
9
9
$begingroup$
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
3
3
$begingroup$
+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
|
show 2 more comments
$begingroup$
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.
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
$endgroup$
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
add a comment |
$begingroup$
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.
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
$endgroup$
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
add a comment |
$begingroup$
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.
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
$endgroup$
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.
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
9151 gold badge5 silver badges9 bronze badges
9151 gold badge5 silver badges9 bronze badges
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
add a comment |
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
add a comment |
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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
16
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
Mar 26 at 19:58