Is this a new Fibonacci Identity?












6












$begingroup$


I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}

where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



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!










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    8 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    8 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
    8 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
    7 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
    6 hours ago
















6












$begingroup$


I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}

where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



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!










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    8 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    8 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
    8 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
    7 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
    6 hours ago














6












6








6


1



$begingroup$


I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}

where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



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!










share|cite|improve this question









$endgroup$




I have found the following Fibonacci Identity (and proved it).



If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}

where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 8 hours ago









GrassiGrassi

11426




11426








  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    8 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    8 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
    8 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
    7 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
    6 hours ago














  • 1




    $begingroup$
    This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
    $endgroup$
    – user44191
    8 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    8 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
    8 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
    7 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
    6 hours ago








1




1




$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
8 hours ago




$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
8 hours ago












$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
8 hours ago




$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
8 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
8 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
8 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
7 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
7 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
6 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
6 hours ago










3 Answers
3






active

oldest

votes


















7












$begingroup$

Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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 $2$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    "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.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
      Concrete mathematics gives the following reference:



      enter image description here



      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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.






      share|cite|improve this answer











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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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 $2$.






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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 $2$.






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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 $2$.






            share|cite|improve this answer









            $endgroup$



            Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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 $2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 6 hours ago









            Cherng-tiao PerngCherng-tiao Perng

            855148




            855148























                4












                $begingroup$

                "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.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  "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.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    "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.






                    share|cite|improve this answer









                    $endgroup$



                    "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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 5 hours ago









                    Ira GesselIra Gessel

                    8,4222642




                    8,4222642























                        2












                        $begingroup$

                        This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
                        Concrete mathematics gives the following reference:



                        enter image description here



                        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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.






                        share|cite|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
                          Concrete mathematics gives the following reference:



                          enter image description here



                          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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.






                          share|cite|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
                            Concrete mathematics gives the following reference:



                            enter image description here



                            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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.






                            share|cite|improve this answer











                            $endgroup$



                            This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
                            Concrete mathematics gives the following reference:



                            enter image description here



                            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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 1 hour ago

























                            answered 2 hours ago









                            Alexey UstinovAlexey Ustinov

                            6,91445979




                            6,91445979






























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