A companion page on Linear Recurrences and their generating Functions for Fibonacci Numbers, Continued Fraction convergents, Pythagorean triples and other series of numbers.
As used here  Vajda  Dunlap  Knuth  Definition  Description  

Phi Φ 
τ  τ  φ, α 
 Koshy uses α (page 78)  
phi φ  –σ  –φ  –β 
 Koshy uses –β (page 78)  
abs(x) x  x  x  x  abs(x) = x if x≥0; abs(x) = –x if x<0  the absolute value of a number, its magnitude; ignore the sign;  
floor(x) x  [x]  trunc(x), not used for x<0  x  the nearest integer ≤ x. 
When x>0, this is "the integer part of x" or "truncate x"
i.e. delete any fractional part after the decimal point. 3=floor(3)=floor(3.1)=floor(3.9), 4=floor(4)=floor(3.1)=floor(3.9)  
round(x) [x] 

trunc(x + 1/2)  the nearest integer to x; trunc(x+0.5)  3=round(3)=round(3.1), 4=round(3.9), 4=round(4)=round(3.9), 3=round(3.1) 4=round(3.5), 3=round(3.5)  
ceil(x) x      x  the nearest integer ≥ x.  3=ceil(3), 4=ceil(3.1)=ceil(3.9), 3=ceil(3)=ceil(3.1)=ceil(3.9)  
fract(x) frac(x)      x mod 1  x – floor(x)  the fractional part of x, i.e. the part of abs(x) after the decimal point  





_{n}C_{r} n choose r; the element in row n column r of Pascal's Triangle the coefficient of x^{r} in (1+x)^{n} the number of ways of choosing r objects from a set of n different objects. n≥0 and r≥0 (otherwise value is 0) 
Fibonaccitype series with the rule S(i)=S(i1)+S(i2) for all integers i:
i ... –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 ... Fibonacci
F(i)... –8 5 –3 2 –1 1 0 1 1 2 3 5 8 ... Lucas
L(i)... 18 –11 7 –4 3 –1 2 1 3 4 7 11 18 ... General Fib
G(a,b,i)... 13a–8b –8a+5b 5a–3b –3a+2b 2a–b –a+b a b a+b a+2b 2a+3b 3a+5b 5a+8b ...
Formula  Refs  Comments  
F(0) = 0, F(1) = 1, F(n+2) = F(n + 1) + F(n)    Definition of the Fibonacci series  
F(–n) = (–1)^{n + 1} F(n)  Vajda2, Dunlap5  Extending the Fibonacci series 'backwards'  
L(0) = 2, L(1) = 1, L(n + 2) = L(n + 1) + L(n)    Definition of the Lucas series  
L(–n) = (–1)^{n} L(n)  Vajda4, Dunlap6  Extending the Lucas series 'backwards'  
G(n + 2) = G(n + 1) + G(n)  Vajda3, Dunlap4  Definition of the Generalised Fibonacci series, G(0) and G(1) needed  

Dunlap63 
Phi and –phi are the roots of x^{2} = x + 1  

Dunlap65 
Beware! Dunlap occasionally uses φ to
represent our phi = 0.61803.., but more frequently he uses
φ to represent –0.61803.. !  
2 F(n + 1) = F(n) + √(5 F(n)^{2} + 4(–1)^{n})  F(n+1) from F(n): Problem B42, S Basin, FQ, 2 (1964) page 329 
F(n + 2) + F(n) + F(n – 2) = 4 F(n)  B&Q(2003)Identity 18 
F(n + 2 ) + F(n) = L(n + 1)  by Definition of L(n), Vajda6, HoggattI8, Dunlap14, Koshy5.14 
F(n + 2) – F(n) = F(n + 1)  by Definition of F(n) 
F(n + 3) + F(n) = 2 F(n + 2)  B&Q(2003)Identity 16 
F(n + 3) – F(n) = 2 F(n + 1)   
F(n + 4) + F(n) = 3 F(n + 2)  B&Q(2003)Identity 17 
F(n + 2) + F(n – 2) = 3 F(n)  B&Q(2003)Identity 7 
F(n + 2) – F(n – 2) = L(n)  HoggattI14 
F(n + 4) – F(n) = L(n + 2)   
F(n + 5) + F(n) = F(n + 2) + L(n + 3)   
F(n + 5) – F(n) = L(n + 2) + F(n + 3)   
F(n + 6) + F(n) = 2 L(n + 3)   
F(n + 6) – F(n) = 4 F(n + 3)   
F(n) + 2 F(n – 1) = L(n)  (Dunlap32) 
F(n + 2) – F(n – 2) = L(n)  Vajda7a, Dunlap15, Koshy5.15 
F(n + 3) – 2 F(n) = L(n)  possible correction for Dunlap31 
F(n + 2) – F(n) + F(n – 1) = L(n)  possible correction for Dunlap31 
F(n) + F(n + 1) + F(n + 2) + F(n + 3) = L(n + 3)  C Hyson(*) 
L(n – 1) + L(n + 1) = 5 F(n)  Vajda5, Dunlap13, Koshy5.16, B&Q(2003)Identity 34, HoggattI9 
L(n) + L(n + 3) = 2 L(n + 2)   
L(n) + L(n + 4) = 3 L(n + 2)   
2 L(n) + L(n + 1) = 5 F(n + 1)  B&Q(2003)Identity 52 
L(n + 2) – L(n – 2) = 5 F(n)   
L(n + 3) – 2 L(n) = 5 F(n)   
F(n) + L(n) = 2 F(n + 1)  Vajda7b, Dunlap16, B&QIdentity 51 
L(n) + 5 F(n) = 2 L(n + 1)   
3 F(n) + L(n) = 2 F(n + 2)  Vajda26, Dunlap28 
3 L(n) + 5 F(n) = 2 L(n + 2)  Vajda27, Dunlap29 
Phi = 
 ; phi = 

Phi phi = 1  Vajda page 51(3), Dunlap65 
Phi + phi = √5   
Phi / phi = Phi + 1   
phi / Phi = 1 – phi   
Phi – phi = 1   
Phi = phi + 1 = √5 – phi   
phi = Phi – 1 = √5 – Phi   
Phi^{2} = 1 + Phi  Vajda page 51(4), Dunlap64 
phi^{2} = 1 – phi  Vajda page 51(4), Dunlap64 
Phi^{n+2} = Phi^{n+1} + Phi^{n}   
(–phi)^{n+2} = (–phi)^{n+1} + (–phi)^{n}   
phi^{n} = phi^{n+1} + phi^{n+2}   
(–Phi)^{n} = (–Phi)^{n+1} + (–Phi)^{n+2}   

Vajda101  

Vajda101a  
 "Binet's" Formula De Moivre(1718), Binet(1843), Lamé(1844), Vajda58, Dunlap69, Hoggattpage 11, B&Q(2003)Identity 240  
L(n) = Phi^{n} + (–phi)^{n}  Vajda59, Dunlap70, B&Q(2003)Identity 241  
 Vajda62, Dunlap71 corrected, B&Q(2003)Identity 240 Corollary 30  
L(n) = round(Phi^{n}),if n≥2  Vajda63, Dunlap72, B&Q(2003)Corollary 35  
   
L(–n) = round( (–phi)^{–n} ), n≥2    
   
F(n + 1) = round(Phi F(n)),if n≥2  Vajda64, Dunlap73  
L(n + 1) = round(Phi L(n)),if n≥4  Vajda65, Dunlap74  
fract( F(2n) phi ) = 1 – phi^{2n}  Knuth vol 1, Ex 1.2.8 Qu 31 with ψ=phi  
fract( F(2n+1) phi ) = phi^{2n+1}  Knuth vol 1, Ex 1.2.8 Qu 31  

Rabinowitz25, B&Q(2003)Identity 242, Vajda page 125  
Phi^{n} = Phi F(n) + F(n–1)  Rabinowitz28, B&Q(2003)Corrolary 33  
Phi^{n} = F(n+1) + F(n) phi  Rabinowitz28, B&Q(2003)Corollary 33  

I Ruggles (1963) FQ 1.2 pg 80, Rabinowitz25, B&Q(2003)Identity 243, Vajda page 125  

I Ruggles (1963) FQ 1.2 pg 80  
(–phi)^{n} = –phi F(n) + F(n–1)  Rabinowitz28  
(–phi)^{n} = F(n+1) – Phi F(n)  Vajda103b, Dunlap75  
√5 Phi^{n} = Phi L(n) + L(n–1)    
√5 (–phi)^{n} = phi L(n) – L(n–1)   
F(nk) is a multiple of F(n) F(nk) ≡ 0 (mod F(k))  B&Q(2003)Theorem 1, Vajda Theorem I page 82  
gcd(F(m),F(n)) = F(gcd(m,n))  Lucas (1878) B&Q(2003)Theorem 6,Vajda Theorem II page 83  
F(mn+r) ≡ ± F(r) (mod F(n) )  Knuth Vol 1 Ex 1.2.8 Qu. 32, Vajda page 86  
gcd(L(m),L(n)) = L(gcd(m,n)), if both L(m)/gcd(L(m),L(n)) and L(n)/gcd(L(m),L(n)) are odd integers  Vajda page 86  
L(mn+r) ≡ ± L(r) (mod L(n) )  (Vajda page 87)  
 B&Q(2003)Theorem 2  
 Vajda85  
 Vajda86  
 Vajda87  
L(t) is not a factor of L(kt) for even k  
 Vajda88  
L(t) is not a factor of F(kt) for odd k and t≥3 
F(n)^{2} + 2 F(n – 1)F(n) = F(2n)   
F(n + 1)^{2} + F(n)^{2} = F(2n + 1)  Vajda11, Dunlap7, Lucas(1878), B&Q(2003)Identity 13, HoggattI11 
F(n + 1)^{2} – F(n – 1)^{2} = F(2n)  Lucas(1878), B&Q(2003)Identity 14, HoggattI10 
F(n + 1)^{2} – F(n)^{2} = F(n + 2) F(n – 1)  Vajda12, Dunlap8 
F(n + 2)^{2} = 3 F(n + 1)^{2} – F(n)^{2} – 2 (–1)^{n}  V E Hoggatt B208 FQ 9 (1971) pg 217. 
F(n+3)^{2} + F(n)^{2} = 2 ( F(n+1)^{2} + F(n+2)^{2} )  B&Q(2003)Identity 30 
F(n + k + 1)^{2} + F(n – k)^{2} = F(2k + 1)F(2n + 1)  Sharpe(1965), a generalization of Vajda11,Dunlap7 Melham(1999) 
F(n + k)^{2} + F(n – k)^{2} =F(n + k –2)F(n + k + 1) + F(2k – 1)F(2n – 1)  Sharpe (1965) 
F(n + k + 1)^{2} – F(n – k)^{2} = F(n – k – 1)F(n – k + 2) + F(2 k)F(2n + 2)  Sharpe (1965) 
F( n+p )^{2} – F( n–p )^{2} = F( 2n )F( 2p )  I Ruggles (1963) FQ 1.2 pg 77; HoggattI25, Sharpe (1965) 
F(n + 1) F(n – 1) – F(n)^{2} = (–1)^{n} 
Cassini's Formula(1680), Simson(1753), Vajda29, Dunlap9, HoggattI13 special case of Catalan's Identity with r=1 B&Q(2003)Identity 8 
F(n)^{2} – F(n + r)F(n – r) = (1)^{nr}F(r)^{2}  Catalan's Identity(1879) 
F(n)F(m + 1) – F(m)F(n + 1) = (1)^{m}F(n – m) 
d'Ocagne's Identity, special case of Vajda9 with G=F 
F(n + m) = F(n + 1)F(m + 1) – F(n – 1)F(m – 1)  B&Q(2003)Identity 231 
F(n + m) = F(m) F(n + 1) + F(m – 1) F(n)  alternative to Dunlap10, B&Q(2003)Identity 3;
variation of R T Hansen FQ (1972) "Generating Identities for Fibonacci and Lucas Triples" p 571578 
F(n) = F(m) F(n + 1 – m) + F(m – 1) F(n – m)  I Ruggles (1963) FQ 1.2 pg 79; Dunlap10, special case of Vajda8 
F(n) F(n + 1) = F(n – 1) F(n + 2) + (–1)^{n1}  Vajda20a special case: i:=1;k:=2;n:=n1; HoggattI19 
F(n + i) F(n + k) – F(n) F(n + i + k) = (–1)^{n} F(i) F(k)  Vajda20a=Vajda18 (corrected) with G:=H:=F 
F(a)F(b) – F(c)F(d) = (–1)^{r}( F(a – r)F(b – r) – F(c – r)F(d – r) ) a+b=c+d for any integers a,b,c,d,r 
Johnson FQ 42 (2004) B960 'A Fibonacci Iddentity', solution pg 90 also Johnson7 Cassini, Catalan and D'Ocagne's Identities are all special cases of this formula 
( F(n1)F(n+2) )^{2} + (2 F(n)F(n+1) )^{2} = (F(n+1)F(n+2) – F(n1)F(n))^{2} = F(2n+1)^{2} 
A F Horadam FQ 20 (1982) pgs 121122, B&Q(2003)Identity 19 (corrected) special case of Generalised Fibonacci Pythagorean Triples 
L(n + 2)^{2} = 3 L(n + 1)^{2} – L(n)^{2} + 10(–1)^{n}  V E Hoggatt B208 FQ 9 (1971) pg 217. 
L(n + 2) L(n – 1) = L(n + 1)^{2} – L(n)^{2}   
L(n + 1) L(n – 1) – L(n)^{2} = –5 (–1)^{n}  B&Q(2003)Identity 60 
L(2n) + 2 (–1)^{n} = L(n)^{2}  Vajda17c, Dunlap12, B&Q(2003)Identity 36 
L(n + m) + (–1)^{m} L(n – m) = L(m) L(n)  Vajda17a, Dunlap11 (special cases: HoggattI15,I18) 
L(4n) + 2 = L(2n)^{2}  HoggattI15, special case of Vajda17a 
2 L(n + 1) = L(n) + √5 √(L(n)^{2} – 4(–1)^{n})  L(n+1) from L(n): Problem B42, S Basin, FQ 2 (1964) page 329 
F(2n) = F(n) L(n)  Vajda13, HoggattI7, Koshy5.13, B&Q(2003)Identity 33 
5 F(n) = L(n + 1) + L(n – 1)  
L(n + 1)^{2} + L(n)^{2} = 5 F(2n + 1)  Vajda25a 
L(n + 1)^{2} – L(n – 1)^{2} = 5 F(2n)  
L(n + 1)^{2} – 5 F(n)^{2} = L(2n + 1)  
L(2n) – 2 (–1)^{n} = 5 F(n)^{2}  Vajda23, Dunlap25 
L(n)^{2} – 4(–1)^{n} = 5 F(n)^{2}  B&Q(2003)Identity 53, HoggattI12 
F(n+k) + F(n–k) = F(n)L(k), k even;  Bergum and Hoggatt (1975) equn (5) 
F(n+k) + F(n–k) = L(n)F(k), k odd;  Bergum and Hoggatt (1975) equn (6) 
F(n+k) – F(n–k) = F(n)L(k), k odd;  Bergum and Hoggatt (1975) equn (7) 
F(n+k) – F(n–k) = L(n)F(k), k even;  Bergum and Hoggatt (1975) equn (8) 
L(n+k) + L(n–k) = L(n)L(k), k even  Bergum and Hoggatt (1975) equn (9) 
L(n+k) + L(n–k) = 5F(n)F(k), k odd  Bergum and Hoggatt (1975) equn (10) 
L(n+k) – L(n–k) = L(n)L(k), k odd  Bergum and Hoggatt (1975) equn (11) 
L(n+k) – L(n–k) = 5F(n)F(k), k even  Bergum and Hoggatt (1975) equn (12) 
F(n + 1) L(n) = F(2n + 1) + (–1)^{n}  Vajda30, Vajda31, Dunlap27, Dunlap30 
L(n + 1) F(n) = F(2n + 1) – (–1)^{n}   
F(2n + 1) = F(n + 1) L(n + 1) – F(n) L(n)  Vajda14, Dunlap18 
L(2n + 1) = F(n + 1) L(n + 1) + F(n) L(n)   
L(m) L(n) + L(m – 1) L(n – 1) = 5 F(m + n – 1)  R T Hansen FQ (1972) "Generating Identities for Fibonacci and Lucas Triples" p 571578 
L(n)^{2} – 2 L(2n) = –5 F(n)^{2}  Vajda22, Dunlap24 
5 F(n)^{2} – L(n)^{2} = 4 (–1)^{n + 1}  Vajda24, Dunlap26 
F(n)^{2} + L(n)^{2} = 4 F(n + 1)^{2} – 2 F(2n)  FQ (2003)vol 41, B936, M A Rose, page 87 
5 (F(n)^{2} + F(n + 1)^{2}) = L(n)^{2} + L(n + 1)^{2}  Vajda25 
F(n) L(m) = F(n + m) + (–1)^{m} F(n – m)  a recurrence relation for F(n+km): Vajda15a, Dunlap19 
L(n) F(m) = F(n + m) – (–1)^{m} F(n – m)  Vajda15b, Dunlap20 
5 F(m) F(n) = L(n + m) – (–1)^{m} L(n – m)  Vajda17b, Dunlap23, (special cases:HoggattI16,I17) 
2 F(n + m) = L(m) F(n) + L(n) F(m)  Vajda16a, Dunlap2, FQ (1967) B106 H H Ferns pp 466467 
2 L(n + m) = L(m) L(n) + 5 F(n) F(m)  FQ (1967) B106 H H Ferns pp 466467 
F(m) L(n) + F(m – 1) L(n – 1) = L(m + n – 1)  R T Hansen FQ (1972) "Generating Identities for Fibonacci and Lucas Triples" p 571578 
(–1)^{m} 2 F(n – m) = L(m) F(n) – L(n) F(m)  Vajda16b, Dunlap22 
L(n + i) F(n + k) – L(n) F(n + i + k) = (–1)^{n + 1} F(i) L(k)  Vajda19a 
F(n + i) L(n + k) – F(n) L(n + i + k) = (–1)^{n} F(i) L(k)  Vajda19b 
L(n + k + 1)^{2} + L(n – k)^{2} = 5 F(2n + 1)F(2k + 1)  Melham (1999) Theorem 1 
L(n + i) L(n + k) – L(n) L(n + i + k) = (–1)^{n + 1} 5 F(i) F(k)  Vajda20b 
(–1)^{k}F(n)F(m–k) + (–1)^{m}F(k)F(n–m) + (–1)^{n}F(m)F(k–n) = 0  FQ 11 (1973) B228 page 108 
(–1)^{k}L(n)F(m–k) + (–1)^{m}L(k)F(n–m) + (–1)^{n}L(m)F(k–n) = 0  FQ 11 (1973) B229 page 108 
5 F(jk+r) F(ju+v) = L(j(k+u)+(r+v))  (1)^{ju+v}L(j(ku)+(rv))  FQ 16 General Identities For Linear Fibonacci And Lucas Summations, R T Hansen, 121128 
F(jk+r) L(ju+v) = F(j(k+u)+(r+v)) + (1)^{ju+v}F(j(ku)+(rv))  FQ 16 General Identities For Linear Fibonacci And Lucas Summations, R T Hansen, 121128 
L(jk+r) L(ju+v) = L(j(k+u)+(r+v)) + (1)^{ju+v}L(j(ku)+(rv))  FQ 16 General Identities For Linear Fibonacci And Lucas Summations, R T Hansen, 121128 
5F(a)F(b) – L(c)L(d) = (–1)^{r}( 5F(a – r)F(b – r) – L(c – r)L(d – r) )
a+b=c+d for any integers a,b,c,d,r 
Johnson 
F(a) L(b) – F(c) L(d) = (–1)^{r}( F(a–r) L(b–r) – F(c–r) L(d–r) with a+b=c+d 
Johnson32, special case of Johnson44 
F(3n) = F(n + 1)^{3} + F(n)^{3} – F(n – 1)^{3}  B&Q(2003)Identity 232  
F(n + 1)F(n + 2)F(n + 6) – F(n + 3)^{3} = (–1)^{n}F(n)
F(n)F(n + 4)F(n + 5) – F(n + 3)^{3} = (–1)^{n+1}F(n + 6) 
FQ 41 (2003) pg 142, Melham. The second is a variant with n for n and using Vajda2 

F(n–2)F(n–1)F(n+3) – F(n)^{3} = (–1)^{n1}F(n–3) F(n+2)F(n+1)F(n–3) – F(n)^{3} = (–1)^{n}F(n+3)  Fairgrieve and Gould (2005) versions of the above two formulae of Melham 

F(n–2)F(n+1)^{2} – F(n)^{3} = (–1)^{n1} F(n–1) F(n+2)F(n–1)^{2} – F(n)^{3} = (–1)^{n} F(n+1)  Fairgrieve and Gould (2005)  
F(n+a+b)F(n–a)F(n–b) – F(nab)F(n+a)F(n+b) = (–1)^{n+a+b}F(a)F(b)F(a+b)L(n)  Melham (2011) Theorem 1  
F(n+a+b–c)F(n–a+c)F(n–b+c) – F(n–a–b+c)F(n+a)F(n+b) = (–1)^{n+a+b+c}F(a+b–c)( F(c)F(n+a+b–c) + (–1)^{c}F(a–c)F(b–c)L(n) ) 
Melham (2011) Theorem 5  
F(i+j+k) = F(i+1)F(j+1)F(k+1) + F(i)F(j)F(k) – F(i–1)F(j–1)F(k–1) for any integers i,j,k  Johnson's (6)  
L(5n) = L(n) (L(2n) + 5F(n) + 3)( L(2n) – 5F(n) + 3), n odd  Aurifeuille's Identity (1879) FQ 42 (2004) R S Melham, pgs 155160 
F(n–1)^{2}F(n+1)^{2} – F(n–2)^{2}F(n+2)^{2} = 4(–1)^{n}F(n)^{2}  Melham (2011) 21 
F(n–3)F(n–1)F(n+1)F(n+3) – F(n)^{4} = (–1)^{n}L(n)^{2}  Melham (2011) 22 
F(n)^{2} F(m + 1) F(m – 1) – F(m)^{2} F(n + 1) F(n – 1) = (–1)^{n – 1} F(m + n) F(m – n)  Vajda32 
F(n – 2)F(n – 1)F(n + 1)F(n + 2) + 1 = F(n)^{4}  GelinCesàro Identity (1880) (see Dickson page 401) FQ 41 (2003) pg 142, B&Q(2003)Identity 31 HoggattI29, Simson(1753) 
L(n – 2)L(n – 1)L(n + 1)L(n + 2) + 25 = L(n)^{4}  B&Q(2003)Identity 56 
F(n+a+b+c)F(n–a)F(n–b)F(n–c) – F(nabc)F(n+a)F(n+b)F(n+c) = (–1)^{n+a+b+c}F(a+b)F(a+c)F(b+c)F(2n)  Melham (2011) Theorem 2 
F(n+a+b+c–d)F(n–a+d)F(n–b+d)F(n–c+d) –
F(n–a–b–c+2d)F(n+a)F(n+b)F(n+c) = (–1)^{n+a+b+c}F(a+b–d)F(a+c–d)F(b+c–d)F(2n+d) 
Melham (2011) Theorem 6 
(F(n)^{2} + F(n+1)^{2} + F(n+2)^{2} )^{2} = 2 ( F(n)^{4} + F(n+1)^{4} + F(n+2)^{4} )  Candido's Identity (1951) FQ 42 (2004) R S Melham, pgs 155160 
[ L(n1)L(n+2) ]^{2} + [ 2L(n)L(n+1) ]^{2} = [ 5F(2n+1) ] ^{2}  Wulczyn FQ 18 (1980) pg 188 special case of Generalised Fibonacci Pythagorean Triples 
F(n)F(n+1)F(n+2)F(n+4)F(n+5)F(n+6) + L(n+3)^{2} = [ F(n+3)( 2F(n+2)F(n+4) – F(n+3)^{2}) ]^{2} 
J Morgado Note on some results of A F Horadam and A G Shannon
concerning Catalan's Identity on Fibonacci Numbers Portugaliae Math. 44 (1987) pgs 243252  
 De Moivre Analogue, S Fisk (1963) FQ 1.2 Problem B10, pg 85. HoggattI44 
We define F!(n) = F(n)F(n1)...F(2)F(1), n>0; F!(0)=1
for which some authors use n!_{F}, to compare with
n! = n(n1)...3.2.1.
There is no universal notation for the Fibonomial. The fibonomial "Fibonacci n choose k" is defined as:
 = 
 = 
 if n ≥ k ≥ 0  
= 0, otherwise 
((  n  )) 
k 
[  n  ] 
k 
_{n} ^{k}  0  1  2  3  4  5  6  7 

0  1  
1  1  1  
2  1  1  1  
3  1  2  2  1  
4  1  3  6  3  1  
5  1  5  15  15  5  1  
6  1  8  40  60  40  8  1  
7  1  13  104  260  260  104  13  1 

Vajda page 74, "add the two numbers above" analogy from Pascal's triangle  
 Melham (1999)....  
 .... examples  
0 = F(n)  F(n1)  F(n2) 0 = F(n)^{2}  2 F(n1)^{2}  2 F(n2)^{2} + F(n3)^{2} 0 = F(n)^{3}  3 F(n1)^{3}  6 F(n2)^{3} + 3 F(n3)^{3} + F(n4)^{3} 0 = F(n)^{4}  5 F(n1)^{4}  15 F(n2)^{4} + 15 F(n3)^{4} + 5 F(n4)^{4}  F(n5)^{4} ...  Brousseau (1968)...but the general formula was not given. For this see next line:  
 Knuth AoCP Vol 1 section 1.2.8 Exercise 30, (1997)  

 

 

 


G(n) = G(0) F(n – 1) + G(1) F(n)  B&Q(2003)Identity 37  
G(–n) = (–1)^{n} (G(0) F(n + 1) – G(1) F(n))  ditto  applying Vajda2  
√5 G(n) = ( G(0) phi + G(1) ) Phi^{n} + (G(0) Phi – G(1)) ( –phi )^{n}  Vajda55/56, Dunlap77, B&Q(2003)Identity 244  
 Amer Math Monthly (2005) "Fibonacci, Chebyshev and Orthogonal Polynomials" D Aharonov, A Beardam, K Driver, p612630  
2 G(k) = ( 2 G(1) – G(0) ) F(k) + G(0) L(k)  Johnson46  
G(n + m) = F(m – 1) G(n) + F(m) G(n + 1)  Vajda8, Dunlap33, B&Q(2003)Identity 38, Johnson40  
G(n – m) = (–1)^{m} (F(m + 1) G(n) – F(m) G(n + 1))  Vajda9, Dunlap34, B&Q(2003)Identity 47  
G(n + m) + (–1)^{m} G(n – m) = L(m) G(n)  Vajda10a, Dunlap35, B&Q(2003)Identity 45, Bergum & Hoggatt (1975) (36) and (38)  
G(n + m) – (–1)^{m} G(n – m) = F(m) ( G(n–1) + G(n+1))  Vajda10b, Dunlap36, B&Q(2003)Identity 48, Bergum & Hoggatt (1975) (37) and (39)  
G(m) F(n) – G(n) F(m) = (–1)^{n+1} G(0) F(m – n)  Vajda21a  
G(m) F(n) – G(n) F(m) = (–1)^{m} G(0) F(n – m)  Vajda21b  
G(m+k) F(n+k) + (–1)^{k+1} G(m) F(n) = F(k) G(m + n + k)  Howard(2003) 
G(n + i) H(n + k) – G(n) H(n + i + k) = (–1)^{n} (G(i) H(k) – G(0) H(i + k)) 
Vajda18 (corrected), B&Q(2003)Identity 44 a special case of Johnson's: 
G(p)H(q) – G(r)H(s) = (1)^{n}[ G(pn)H(qn) – G(rn)H(sn) ] if p+q = r+s and p,q,r,s,n are integers 
Johnson44 
G(n + 1) G(n – 1) – G(n)^{2} = (–1)^{n} (G(1)^{2} – G(0) G(2))  Vajda28, B&Q(2003)Identity 46 
4 G(n–1)G(n) + G(n–2)^{2} = G(n+1)^{2}  B&Q(2003)Identity 65 
G(n + 3)^{2} + G(n)^{2} = 2( G(n+1)^{2} + G(n+2)^{2} )  B&Q(2003)Identity 70 
G(i+j+k) = F(i+1)F(j+1)G(k+1) + F(i)F(j)G(k) – F(i–1)F(j–1)G(k–1) for any integers i,j,k  Johnson (39a) 
4G(i)^{2}G(i+1)^{2} + G(i–1)^{2}G(i+2)^{2} = ( G(i)^{2} + G(i+1)^{2} )^{2}  Generalised Fibonacci Pythagorean Triples A F Horadam Special Properties of the Sequence w_{n}(a,b;p,q) FQ 5 (1967) pgs 424434 
G(n + 2)G(n + 1)G(n – 1)G(n – 2) + ( G(2)G(0) – G(1)^{2} )^{2} = G(n)^{4} 
B&Q(2003)Identity 59 

HoggattI1, Lucas(1878), B&Q 2003Identity 1  

B&Q 2003Identity 21  

HoggattI2  

  

  

HoggattI6, Lucas(1878), B&Q(2003)Identity 12  

HoggattI5, Lucas(1878), B&Q(2003)Identity 2  
   

Vajda37a(adapted), Dunlap42(adapted), B&Q(2003)Identity 10  

B&Q(2003)Identity 236  
 B&Q(2003)Identity 23  
 B&Q(2003)Identity 24 (corrected)  
 B&Q(2003)Identity 25 (corrected)  

B&Q 2003Identity 27  

B&Q 2003Identity 26  

B&Q 2003Identity 29  

B&Q 2003Identity 28  

Vajda97, Dunlap54  

B&Q(2003)Identity 55 
1/89  =  0.0 1 1 2 3 5 ... 
1/9899  =  0.00 01 01 02 03 05 08 13 21 ... 
 Hudson and Winans (1981)  
If P(n) = a P(n1) + b P(n2) for n≥2; P(0) = c; P(1) = d and m and N are defined by B^{2} = m + Ba + b, N = cm + dB + bc, then
provided that (a+√(a^{2}+4b))/(2B) < 1 and  (a–√(a^{2}+4b))/(2B)  < 1  Long (1981) 

Vajda60, Dunlap51  

  

  

  

Vajda61, Dunlap52  

  

Vajda77(corrected), Dunlap53(corrected)  

Vajda89 (corrected)  

R L Graham (1963) FQ 1.1, Problem B9, pg 85, FQ 1.4 page 79  

R L Graham (1963) FQ 1.1, Problem B9, pg 85  

Johnson11, Vajda102 

Vajda45, Dunlap5, HoggattI3, Lucas(1878), Koshy77, B&Q(2003)Identity 9 (Identity 233 variant)  

HoggattI4  

  

Vajda40, Dunlap45  

  

Vajda42, Dunlap47  

  

Vajda93  

Vajda94  

Vajda95, B&Q(2003)Identity 234  

Vajda page 70  

Vajda96, B&Q(2003)Identity 54  

Vajda page 70  

Vajda98, Dunlap55, B&Q(2003)Identity 58  

Vajda99, Dunlap56, B&Q(2003)Identity 57  

Vajda100, Dunlap57, B&Q(2003)Identity 35  

V Hoggatt (1965) Problem B53 FQ 3, pg 157 

adapted from Benjamin, Carnes, Cloitre (2009)  

see A005969  

Ohtsuka and Nakamura (2010) Theorem 1  

Ohtsuka and Nakamura (2010) Theorem 2 

L G Brökling (1964) FQ 2.1 Problem B20 solution, pg76; Vajda33; Dunlap38; B&Q(2003)Identity 39  

  

Vajda34, Dunlap37, B&Q(2003)Identity 61  

Vajda35, Dunlap39, B&Q(2003)Identity 62  

Vajda36, Dunlap40  

Vajda37, Dunlap41, B&Q(2003)Identity 69  

Vajda38, Dunlap43, B&Q(2003)Identity 49  

Vajda39, Dunlap44, B&Q(2003)Identity 41  

Vajda41, Dunlap46  

Vajda43, Dunlap48, B&Q(2003)Identity 64  

Fibonacci with a Golden Ring KungWei Yang Mathematics Magazine 70 (1997), pp. 131135.  

Vajda44, Dunlap49, B&Q(2003)Identity 67  

Stan Rabinowitz, "SecondOrder Linear Recurrences" card, Generating Function special case (x=1/r, P=1, Q=1)  

  
 B&Q(2003)Identity 42 

B&Q(2003) Identity4  

Vajda54(corrected), Dunlap84(corrected)  

B&Q(2003)Identity 165  

B&Q(2003)Identity 166  

S Basin & V Ivanoff (1963) Problem B4, FQ 1.1 pg 74, FQ1.2 pg 79; B&Q(2003)Identity 6  

I Ruggles (1963) FQ 1.2 pg 77  

I Ruggles (1963) FQ 1.2 pg 77  

B&Q(2003)Identity 20  

B&Q(2003)Identity 238, Vajda68  

Vajda50, Dunlap82  

HoggattI41 (special case p=0: Vajda69, Dunlap85)  

Vajda71, Dunlap87  

HoggattI42 (special case p=0: Vajda70, Dunlap86)  

Vajda72, Dunlap88  

Vajda73, Dunlap89,HoggattI45  

Vajda75, Dunlap91, HoggattI46  

Vajda74, Dunlap90, HoggattI47  

Vajda76, Dunlap92  

Vajda91, B&Q(2003)Identity 235, Catalan 1857  

Vajda92, B&Q(2003)Identity 237, Catalan (1857)see Vajda pg 69  

Rabinowitz17 (special case of Vajda66)  

Rabinowitz17 (special case of Vajda66)  

Vajda66  

B&Q(2003) Identity 5  
 Lucas (1878) equns 7476, this form due to Hoggatt and Lindt (1969), see Gould (1977) 

Vajda80  

Vajda81  

Vajda78  

Vajda79  

On a General Fibonacci Identity, J H Halton, Fib Q, 3 (1965), pp 3143 

I Ruggles (1963) FQ 1.2 pg 77; Vajda47; Dunlap80  

B&Q(2003)Identity 239  
 Vajda46, Dunlap79, B&Q(2003)Identity 40  

Vajda49, Dunlap81  

Vajda51, Dunlap83 

D Lind, Problem H93, FQ 4 (1966), page 332  

D Lind, Problem H93, FQ 4 (1966), page 252, corrected page 332 
 from Binet's formula  
 
 from Binet's formula  
 
L( 2n) = 2 cosh( ng )  from Binet's formula  
L( 2n+1 ) = 2 sinh( ng )  from Binet's formula 

D Lind, Problem H64, FQ 3 (1965), page 116  
 from Rabinowitz7 corrected  
 from Rabinowitz7 corrected  
L(n) = 2 i^{–n} cos(–i n ln( i Phi) )  from Rabinowitz7 corrected  
L(n) = 2 i^{–n} cosh( n ln( i Phi) )  from Rabinowitz7 corrected  
√i/√Phi = [1+i, ]  = √Phi+I J Good (1993)  
2 √i) √Phi + (1 – i) √phi  = (1 +I J Good (1993) 
G(x) =  ∞ Σ i=0  S(i) x^{i}  = S(0) + S(1) x + S(2) x^{2} + S(3) x^{3} + ... 
To shift to the right (insert a 0 at the start of the series so all other terms have an index increased by 1),
multiply the GF by x; to shift to the left, divide by x.
There is much more on GFs on my Fibonomials page.
Fibonacci(n) 0,1,1,2,3,... 

Lucas(n) 2,1,3,4,7,... 

G(a,b,n) a,b,a+b,a+2b,... 


Fibonacci(2n) 0,1,3,8,21,... 

Lucas(2n) 2,3,7,18,... 

G(a,b,2n) a,a+b,2a+3b,... 


Fibonacci(2n+1) 1,2,5,13,... 

Lucas(2n+1) 1,4,11,29,... 

G(a,b,2n+1) b,a+2b,3a+5b,... 


Fibonacci(3n) 2,8,34,144,... 

Lucas(3n) 2,4,18,76,... 

G(a,b,3n) a,a+2b,5a+8b,... 


Fibonacci(3n+1) 1,3,13,55,... 

Lucas(3n+1) 3,11,47,199,... 

G(a,b,3n+1) a+b,3a+5b,13a+21b,... 


Fibonacci(3n+2) 1,5,21,89,... 

Lucas(3n+2) 2,4,18,76,... 

G(a,b,3n+2) a,a+2b,5a+8b,... 


Fibonacci(k n) 

Lucas(k n) 

G(a,b,kn) 


Fibonacci(n)^{2} 0^{2},1^{2},1^{2},2^{2},3^{2},... 

Lucas(n)^{2} 2^{2},1^{2},3^{2},4^{2},... 

G(a,b,n)^{2} a^{2},b^{2},(a+b)^{2},... 


Fib(n)Fib(n+1) 1×1,1×2,2×3,3×5,... 

Lucas(n)Lucas(n+1) 2×1,1×3,3×4,4×7,... 

G(a,b,n)G(a,b,n+1) ab,b(a+b), (a+b)(a+2b),... 


Fibonacci(n)^{3} 0^{3},1^{3},1^{3},2^{3},3^{3},... 

Lucas(n)^{3} 2^{3},1^{3},3^{3},4^{3},... 

The GF of 1,2,5,13,... is that of
Fib(2n+1)
which is
(1 – x)/(x^{2} – 3 x + 1)
so 1,0,2,0,5,0,13,... has GF
(1 – x^{2})/(x^{4} – 3 x^{2} + 1)
To insert an extra 0 at the start, multiply the GF by x.
So the GF for the oddindexed Fibonacci numbers only in their correct positions in the Fibonacci series
with Fib(2n+1) is the coefficient of x^{2n+1}
is therefore x (1 – x^{2})/(x^{4}  3 x^{2} + 1)
for the series 0,1,0,2,0,5,0,13,... .
Adding these two GFs, that is, for Fib(2n) as the coefficient of x^{2n}
and Fib(2n+1) as the coefficient of x^{2n+1}
should then give the complete
Fibonacci series GF:
0,0,1,0,3,0,8, 0,21, ... +We can check that x^{2}/(x^{4}  3 x^{2} + 1) + x (1 – x^{2})/(x^{4}  3 x^{2} + 1) = x/(1 – x – x^{2})
0,1,0,2,0,5,0,13, 0, ...
0,1,1,2,3,5,8,13,21,...
Multiplying a GF by a constant k multiples all the members of the series by k.
A series formed by adding two series S and T elementwise to form the series {S(n)+T(n) for n=1,2,3,...},
has a GF which is the sum of the two separate GFs.
Check that a Fib[n1] + b Fib[n] gives the GF of G(a,b).
 See, e.g., Solving Linear Recurrences from Differential Equations in the Exponential Manner and Vice Versa W Oberschelp in Applications of Fibonacci Numbers Vol 6 (1996) pages 365380 
: a book;  
: an article (chapter) in a journal (book);  
: a web resource.  
FQ  : The Fibonacci Quarterly journal, all papers are available free in PDF (shown below) except for the last 5 years which are only available online for subscribers to the FQ. 
Fibonacci and Phi in the Arts 
Fibonacci Home Page
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