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 CatalanLarge cubes in sum/difference setsA link between hooks and contents: Part II
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?
$$F_nF_n-1F_n-2=fracF_n^3-F_n-1^3-F_n-2^33.$$
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
18.5k230132
$endgroup$
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
18.5k230132
$endgroup$
$begingroup$
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
18.5k230132
$endgroup$
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
18.5k230132
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.5k230132
edited Mar 26 at 18:40
T. Amdeberhan
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.5k230132
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.5k230132
asked Mar 26 at 18:15
T. AmdeberhanT. Amdeberhan
18.5k230132
18.5k230132
$begingroup$
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
24.5k13389
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
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
895159
$endgroup$
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
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326414%2fproducts-and-sum-of-cubes-in-fibonacci%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
24.5k13389
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
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
24.5k13389
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
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
24.5k13389
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
$endgroup$
$F_n$ 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
24.5k13389
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
edited Mar 29 at 6:04
Aaron Meyerowitz
24.5k13389
edited Mar 29 at 6:04
Aaron Meyerowitz
24.5k13389
edited Mar 29 at 6:04
Aaron Meyerowitz
24.5k13389
24.5k13389
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
answered Mar 26 at 23:54
Richard StanleyRichard Stanley
29.4k9118191
29.4k9118191
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
$begingroup$
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
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
895159
$endgroup$
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
895159
$endgroup$
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
895159
$endgroup$
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
895159
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
895159
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
895159
answered Mar 26 at 19:52
Cherng-tiao PerngCherng-tiao Perng
895159
895159
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 |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326414%2fproducts-and-sum-of-cubes-in-fibonacci%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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