Is this a new Fibonacci Identity? The Next CEO of Stack Overflowfibonacci identity using...



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 $n^{2m+1}=sumlimits_{0leq k leq m}(-1)^{m-k}U_m(n,k)cdot n^k$A question on the Faulhaber's formula












5












$begingroup$


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!






nt.number-theory co.combinatorics







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
    2 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    2 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
    2 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
    1 hour 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
    1 hour ago
















5












$begingroup$


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!






nt.number-theory co.combinatorics







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
    2 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    2 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
    2 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
    1 hour 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
    1 hour ago














5












5








5


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!






nt.number-theory co.combinatorics







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




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 2 hours ago









GrassiGrassi

10626




10626








  • 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
    2 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    2 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
    2 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
    1 hour 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
    1 hour 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
    2 hours ago










  • $begingroup$
    It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
    $endgroup$
    – Grassi
    2 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
    2 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
    1 hour 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
    1 hour 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
2 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
2 hours ago












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




$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
2 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
2 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
2 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

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









$endgroup$





















    0












    $begingroup$

    "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









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      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









      $endgroup$


















        2












        $begingroup$

        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









        $endgroup$
















          2












          2








          2





          $begingroup$

          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 1 hour ago









          Cherng-tiao PerngCherng-tiao Perng

          805148




          805148























              0












              $begingroup$

              "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









              $endgroup$


















                0












                $begingroup$

                "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









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  "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









                  $endgroup$



                  "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












                  share|cite



                  share|cite










                  answered 8 mins ago









                  Ira GesselIra Gessel

                  8,3722642




                  8,3722642






























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