Is this a new Fibonacci Identity? The Next CEO of Stack Overflowfibonacci identity using generating functionPrimality criterion for generalized Fermat numbers similar to the LLT ?Solvable parametric $7$th and $13$th degree equations using $eta(q)/eta(q^p)$?To what extent is it possible to generalise a natural bijection between trees and $7$-tuples of trees, suggested by divergent series?Two spaces attached to mod 2 level 9 modular forms--a conjectural Hecke isomorphismCan someone explain this appearance of the Fibonacci series in the formula of the Fibonacci series?On the automorphism group of binary quadratic formsFor what (other) families of graphs does the clique-coclique bound hold?Coefficients $U_m(n,k)$ in the identity $n^2m+1=sumlimits_0leq k leq m(-1)^m-kU_m(n,k)cdot n^k$A question on the Faulhaber's formula
Is this a new Fibonacci Identity?
The Next CEO of Stack Overflowfibonacci identity using generating functionPrimality criterion for generalized Fermat numbers similar to the LLT ?Solvable parametric $7$th and $13$th degree equations using $eta(q)/eta(q^p)$?To what extent is it possible to generalise a natural bijection between trees and $7$-tuples of trees, suggested by divergent series?Two spaces attached to mod 2 level 9 modular forms--a conjectural Hecke isomorphismCan someone explain this appearance of the Fibonacci series in the formula of the Fibonacci series?On the automorphism group of binary quadratic formsFor what (other) families of graphs does the clique-coclique bound hold?Coefficients $U_m(n,k)$ in the identity $n^2m+1=sumlimits_0leq k leq m(-1)^m-kU_m(n,k)cdot n^k$A question on the Faulhaber's formula
$begingroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
asked 6 hours ago
GrassiGrassi
11626
$endgroup$
|
show 1 more comment
$begingroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
asked 6 hours ago
GrassiGrassi
11626
$endgroup$
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
6 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
6 hours ago
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
6 hours ago
1
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
5 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
4 hours ago
|
show 1 more comment
$begingroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
asked 6 hours ago
GrassiGrassi
11626
$endgroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
nt.number-theory co.combinatorics
asked 6 hours ago
GrassiGrassi
11626
asked 6 hours ago
GrassiGrassi
11626
asked 6 hours ago
GrassiGrassi
11626
asked 6 hours ago
GrassiGrassi
11626
asked 6 hours ago
GrassiGrassi
11626
11626
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
6 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
6 hours ago
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
6 hours ago
1
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
5 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
4 hours ago
|
show 1 more comment
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
6 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
6 hours ago
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
6 hours ago
1
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
5 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
4 hours ago
1
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
6 hours ago
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
6 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
6 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
6 hours ago
1
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
6 hours ago
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
6 hours ago
1
1
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
5 hours ago
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
5 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
4 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
4 hours ago
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ ($m,nin mathbbZ$) where $s_n$ is the Somos-4 seqence has rank $2$. For Somos-6 corresponding matrix has rank $4$ etc.
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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%2f326922%2fis-this-a-new-fibonacci-identity%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
add a comment |
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
add a comment |
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
answered 4 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
835148
add a comment |
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
answered 3 hours ago
Ira GesselIra Gessel
8,4122642
8,4122642
add a comment |
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ ($m,nin mathbbZ$) where $s_n$ is the Somos-4 seqence has rank $2$. For Somos-6 corresponding matrix has rank $4$ etc.
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
$endgroup$
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ ($m,nin mathbbZ$) where $s_n$ is the Somos-4 seqence has rank $2$. For Somos-6 corresponding matrix has rank $4$ etc.
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
$endgroup$
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ ($m,nin mathbbZ$) where $s_n$ is the Somos-4 seqence has rank $2$. For Somos-6 corresponding matrix has rank $4$ etc.
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
$endgroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ ($m,nin mathbbZ$) where $s_n$ is the Somos-4 seqence has rank $2$. For Somos-6 corresponding matrix has rank $4$ etc.
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
answered 46 mins ago
Alexey UstinovAlexey Ustinov
6,90445979
6,90445979
add a comment |
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%2f326922%2fis-this-a-new-fibonacci-identity%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
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
6 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
6 hours ago
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
6 hours ago
1
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
5 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
4 hours ago