Products and sum of cubes in FibonacciPrincipal term of the Dirichlet Divisor problem, from the work of A.F....
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 CatalanLarge cubes in sum/difference setsA link between hooks and contents: Part II
$begingroup$
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
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
$endgroup$
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?
$$F_nF_{n-1}F_{n-2}=frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$$
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
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
$endgroup$
$begingroup$
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
6 hours ago
$begingroup$
Thanks, edited accordingly.
$endgroup$
– T. Amdeberhan
6 hours ago
2
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
5 hours ago
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?
$$F_nF_{n-1}F_{n-2}=frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$$
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
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
$endgroup$
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
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
edited 6 hours ago
T. Amdeberhan
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
asked 7 hours ago
T. AmdeberhanT. Amdeberhan
18k229131
18k229131
$begingroup$
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
6 hours ago
$begingroup$
Thanks, edited accordingly.
$endgroup$
– T. Amdeberhan
6 hours ago
2
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
5 hours ago
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
6 hours ago
$begingroup$
Thanks, edited accordingly.
$endgroup$
– T. Amdeberhan
6 hours ago
2
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
5 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
Thanks, edited accordingly.
$endgroup$
– T. Amdeberhan
6 hours ago
$begingroup$
Thanks, edited accordingly.
$endgroup$
– T. Amdeberhan
6 hours ago
2
2
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
5 hours ago
$begingroup$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
$endgroup$
– Noam D. Elkies
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
$endgroup$
4
$begingroup$
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
5 hours ago
add a comment |
$begingroup$
$F_n$ 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 + F_nF_{n-1}F_{n-2}. end{eqnarray*}
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
$endgroup$
1
$begingroup$
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
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
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
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
$endgroup$
4
$begingroup$
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
5 hours ago
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
$endgroup$
4
$begingroup$
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
5 hours ago
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
$endgroup$
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
answered 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
answered 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
answered 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
660147
660147
4
$begingroup$
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
5 hours ago
add a comment |
4
$begingroup$
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
5 hours ago
4
4
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
add a comment |
$begingroup$
$F_n$ 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 + F_nF_{n-1}F_{n-2}. end{eqnarray*}
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
$endgroup$
1
$begingroup$
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
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
$F_n$ 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 + F_nF_{n-1}F_{n-2}. end{eqnarray*}
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
$endgroup$
1
$begingroup$
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
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
$F_n$ 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 + F_nF_{n-1}F_{n-2}. end{eqnarray*}
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
$endgroup$
$F_n$ 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 + F_nF_{n-1}F_{n-2}. end{eqnarray*}
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
answered 1 hour ago
Richard StanleyRichard Stanley
29k9115189
29k9115189
1
$begingroup$
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
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
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 $3ab(a+b)$.
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– Noam D. Elkies
1 hour ago
add a comment |
1
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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
1 hour ago
1
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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.
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– Richard Stanley
1 hour ago
1
$begingroup$
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
1 hour ago
1
1
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
1
1
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
1
1
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
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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.
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– Gerhard Paseman
6 hours ago
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Thanks, edited accordingly.
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– T. Amdeberhan
6 hours ago
2
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$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next.
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– Noam D. Elkies
5 hours ago