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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 2 hours ago
GrassiGrassi
10626
$endgroup$
add a 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 2 hours ago
GrassiGrassi
10626
$endgroup$
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
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
2 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
1 hour ago
add a 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 2 hours ago
GrassiGrassi
10626
$endgroup$
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 2 hours ago
GrassiGrassi
10626
asked 2 hours ago
GrassiGrassi
10626
asked 2 hours ago
GrassiGrassi
10626
asked 2 hours ago
GrassiGrassi
10626
asked 2 hours ago
GrassiGrassi
10626
10626
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
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
2 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
1 hour ago
add a 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
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
2 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
1 hour 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
2 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
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
2 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
2 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
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
add a comment |
2 Answers
2
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 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
$endgroup$
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 11 mins ago
Ira GesselIra Gessel
8,3822642
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
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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 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
$endgroup$
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 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
$endgroup$
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 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
$endgroup$
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 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
answered 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
answered 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
answered 1 hour ago
Cherng-tiao PerngCherng-tiao Perng
805148
805148
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 11 mins ago
Ira GesselIra Gessel
8,3822642
$endgroup$
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 11 mins ago
Ira GesselIra Gessel
8,3822642
$endgroup$
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 11 mins ago
Ira GesselIra Gessel
8,3822642
$endgroup$
"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 11 mins ago
Ira GesselIra Gessel
8,3822642
answered 11 mins ago
Ira GesselIra Gessel
8,3822642
answered 11 mins ago
Ira GesselIra Gessel
8,3822642
answered 11 mins ago
Ira GesselIra Gessel
8,3822642
8,3822642
add a comment |
add a comment |
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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
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
2 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
1 hour ago