:: Lucas Numbers and Generalized {F}ibonacci Numbers
:: by Piotr Wojtecki and Adam Grabowski
::
:: Received May 24, 2004
:: Copyright (c) 2004-2021 Association of Mizar Users

theorem :: FIB_NUM3:1

for a being Real
for n being Element of NAT st a to_power n = 0 holds
a = 0

proof end;

theorem Th2: :: FIB_NUM3:2

for a being non negative Real holds (sqrt a) * (sqrt a) = a

proof end;

theorem Th3: :: FIB_NUM3:3

for a being Real holds a to_power 2 = (- a) to_power 2

proof end;

theorem Th4: :: FIB_NUM3:4

for k being non zero Nat holds (k -' 1) + 2 = (k + 2) -' 1

proof end;

theorem Th5: :: FIB_NUM3:5

for a, b being Element of NAT holds (a + b) |^ 2 = (((a * a) + (a * b)) + (a * b)) + (b * b)

proof end;

theorem Th6: :: FIB_NUM3:6

for n being Element of NAT
for a being non zero Real holds (a to_power n) to_power 2 = a to_power (2 * n)

proof end;

theorem Th7: :: FIB_NUM3:7

for a, b being Real holds (a + b) * (a - b) = (a to_power 2) - (b to_power 2)

proof end;

theorem Th8: :: FIB_NUM3:8

for n being Element of NAT
for a, b being non zero Real holds (a * b) to_power n = (a to_power n) * (b to_power n)

proof end;

registration

cluster tau -> positive ;

coherence
tau is positive by FIB_NUM2:33;

cluster tau_bar -> negative ;

coherence
tau_bar is negative by FIB_NUM2:36;

end;

theorem Th9: :: FIB_NUM3:9

for n being Nat holds (tau to_power n) + (tau to_power (n + 1)) = tau to_power (n + 2)

proof end;

theorem Th10: :: FIB_NUM3:10

for n being Nat holds (tau_bar to_power n) + (tau_bar to_power (n + 1)) = tau_bar to_power (n + 2)

proof end;

deffunc H₁( set , Element of [:NAT,NAT:]) -> Element of [:NAT,NAT:] = [($2 `2),(($2 `1) + ($2 `2))];

definition

func Lucas -> sequence of [:NAT,NAT:] means :Def1: :: FIB_NUM3:def 1
( it . 0 = [2,1] & ( for n being Nat holds it . (n + 1) = [((it . n) `2),(((it . n) `1) + ((it . n) `2))] ) );

existence
ex b₁ being sequence of [:NAT,NAT:] st
( b₁ . 0 = [2,1] & ( for n being Nat holds b₁ . (n + 1) = [((b₁ . n) `2),(((b<sub>1</sub> . n) `1) + ((b₁ . n) `2))] ) )

proof end;

uniqueness
for b₁, b₂ being sequence of [:NAT,NAT:] st b₁ . 0 = [2,1] & ( for n being Nat holds b₁ . (n + 1) = [((b₁ . n) `2),(((b<sub>1</sub> . n) `1) + ((b₁ . n) `2))] ) & b<sub>2</sub> . 0 = [2,1] & ( for n being Nat holds b<sub>2</sub> . (n + 1) = [((b<sub>2</sub> . n) `2),(((b₂ . n) `1) + ((b<sub>2</sub> . n) `2))] ) holds
b₁ = b₂

proof end;

end;

:: deftheorem Def1 defines Lucas FIB_NUM3:def 1 :
for b₁ being sequence of [:NAT,NAT:] holds
( b₁ = Lucas iff ( b₁ . 0 = [2,1] & ( for n being Nat holds b₁ . (n + 1) = [((b₁ . n) `2),(((b<sub>1</sub> . n) `1) + ((b₁ . n) `2))] ) ) );

definition

let n be Nat;

:: Lucas numbers

func Lucas n -> Element of NAT equals :: FIB_NUM3:def 2
(Lucas . n) `1 ;

coherence
(Lucas . n) `1 is Element of NAT ;

end;

:: deftheorem defines Lucas FIB_NUM3:def 2 :
for n being Nat holds Lucas n = (Lucas . n) `1 ;

theorem Th11: :: FIB_NUM3:11

( Lucas 0 = 2 & Lucas 1 = 1 & ( for n being Nat holds Lucas ((n + 1) + 1) = (Lucas n) + (Lucas (n + 1)) ) )

proof end;

theorem Th12: :: FIB_NUM3:12

for n being Nat holds Lucas (n + 2) = (Lucas n) + (Lucas (n + 1))

proof end;

theorem Th13: :: FIB_NUM3:13

for n being Nat holds (Lucas (n + 1)) + (Lucas (n + 2)) = Lucas (n + 3)

proof end;

theorem Th14: :: FIB_NUM3:14

Lucas 2 = 3

proof end;

theorem Th15: :: FIB_NUM3:15

Lucas 3 = 4

proof end;

theorem Th16: :: FIB_NUM3:16

Lucas 4 = 7

proof end;

theorem Th17: :: FIB_NUM3:17

for k being Nat holds Lucas k >= k

proof end;

theorem :: FIB_NUM3:18

for m being non zero Element of NAT holds Lucas (m + 1) >= Lucas m

proof end;

registration

let n be Element of NAT ;

cluster Lucas n -> positive ;

coherence
Lucas n is positive

proof end;

end;

theorem :: FIB_NUM3:19

for n being Element of NAT holds 2 * (Lucas (n + 2)) = (Lucas n) + (Lucas (n + 3))

proof end;

theorem :: FIB_NUM3:20

for n being Nat holds Lucas (n + 1) = (Fib n) + (Fib (n + 2))

proof end;

theorem Th21: :: FIB_NUM3:21

for n being Nat holds Lucas n = (tau to_power n) + (tau_bar to_power n)

proof end;

theorem :: FIB_NUM3:22

for n being Nat holds (2 * (Lucas n)) + (Lucas (n + 1)) = 5 * (Fib (n + 1))

proof end;

theorem Th23: :: FIB_NUM3:23

for n being Nat holds (Lucas (n + 3)) - (2 * (Lucas n)) = 5 * (Fib n)

proof end;

theorem :: FIB_NUM3:24

for n being Nat holds (Lucas n) + (Fib n) = 2 * (Fib (n + 1))

proof end;

theorem :: FIB_NUM3:25

for n being Nat holds (3 * (Fib n)) + (Lucas n) = 2 * (Fib (n + 2))

proof end;

theorem :: FIB_NUM3:26

for n, m being Element of NAT holds 2 * (Lucas (n + m)) = ((Lucas n) * (Lucas m)) + ((5 * (Fib n)) * (Fib m))

proof end;

theorem :: FIB_NUM3:27

for n being Element of NAT holds (Lucas (n + 3)) * (Lucas n) = ((Lucas (n + 2)) |^ 2) - ((Lucas (n + 1)) |^ 2)

proof end;

theorem Th28: :: FIB_NUM3:28

for n being Nat holds Fib (2 * n) = (Fib n) * (Lucas n)

proof end;

theorem :: FIB_NUM3:29

for n being Element of NAT holds 2 * (Fib ((2 * n) + 1)) = ((Lucas (n + 1)) * (Fib n)) + ((Lucas n) * (Fib (n + 1)))

proof end;

theorem :: FIB_NUM3:30

for n being Element of NAT holds (5 * ((Fib n) |^ 2)) - ((Lucas n) |^ 2) = 4 * ((- 1) to_power (n + 1))

proof end;

theorem :: FIB_NUM3:31

for n being Element of NAT holds Fib ((2 * n) + 1) = ((Fib (n + 1)) * (Lucas (n + 1))) - ((Fib n) * (Lucas n))

proof end;

definition

let a, b be Nat;
:: original: [
redefine func [a,b] -> Element of [:NAT,NAT:];
coherence
[a,b] is Element of [:NAT,NAT:]

proof end;

end;

definition

let a, b be Nat;
deffunc H₂( set , Element of [:NAT,NAT:]) -> Element of [:NAT,NAT:] = [($2 `2),(($2 `1) + ($2 `2))];

func GenFib (a,b) -> sequence of [:NAT,NAT:] means :Def2: :: FIB_NUM3:def 3
( it . 0 = [a,b] & ( for n being Nat holds it . (n + 1) = [((it . n) `2),(((it . n) `1) + ((it . n) `2))] ) );

existence
ex b₁ being sequence of [:NAT,NAT:] st
( b₁ . 0 = [a,b] & ( for n being Nat holds b₁ . (n + 1) = [((b₁ . n) `2),(((b<sub>1</sub> . n) `1) + ((b₁ . n) `2))] ) )

proof end;

uniqueness
for b₁, b₂ being sequence of [:NAT,NAT:] st b₁ . 0 = [a,b] & ( for n being Nat holds b₁ . (n + 1) = [((b₁ . n) `2),(((b<sub>1</sub> . n) `1) + ((b₁ . n) `2))] ) & b<sub>2</sub> . 0 = [a,b] & ( for n being Nat holds b<sub>2</sub> . (n + 1) = [((b<sub>2</sub> . n) `2),(((b₂ . n) `1) + ((b<sub>2</sub> . n) `2))] ) holds
b₁ = b₂

proof end;

end;

:: deftheorem Def2 defines GenFib FIB_NUM3:def 3 :
for a, b being Nat
for b₃ being sequence of [:NAT,NAT:] holds
( b₃ = GenFib (a,b) iff ( b₃ . 0 = [a,b] & ( for n being Nat holds b₃ . (n + 1) = [((b₃ . n) `2),(((b<sub>3</sub> . n) `1) + ((b₃ . n) `2))] ) ) );

definition

let a, b, n be Nat;

func GenFib (a,b,n) -> Element of NAT equals :: FIB_NUM3:def 4
((GenFib (a,b)) . n) `1 ;

coherence
((GenFib (a,b)) . n) `1 is Element of NAT ;

end;

:: deftheorem defines GenFib FIB_NUM3:def 4 :
for a, b, n being Nat holds GenFib (a,b,n) = ((GenFib (a,b)) . n) `1 ;

theorem Th32: :: FIB_NUM3:32

for a, b being Nat holds
( GenFib (a,b,0) = a & GenFib (a,b,1) = b & ( for n being Nat holds GenFib (a,b,((n + 1) + 1)) = (GenFib (a,b,n)) + (GenFib (a,b,(n + 1))) ) )

proof end;

theorem Th33: :: FIB_NUM3:33

for a, b being Element of NAT
for k being Nat holds ((GenFib (a,b,(k + 1))) + (GenFib (a,b,((k + 1) + 1)))) |^ 2 = (((GenFib (a,b,(k + 1))) |^ 2) + ((2 * (GenFib (a,b,(k + 1)))) * (GenFib (a,b,((k + 1) + 1))))) + ((GenFib (a,b,((k + 1) + 1))) |^ 2)

proof end;

theorem Th34: :: FIB_NUM3:34

for a, b, n being Nat holds (GenFib (a,b,n)) + (GenFib (a,b,(n + 1))) = GenFib (a,b,(n + 2))

proof end;

theorem Th35: :: FIB_NUM3:35

for a, b, n being Nat holds (GenFib (a,b,(n + 1))) + (GenFib (a,b,(n + 2))) = GenFib (a,b,(n + 3))

proof end;

theorem Th36: :: FIB_NUM3:36

for a, b, n being Element of NAT holds (GenFib (a,b,(n + 2))) + (GenFib (a,b,(n + 3))) = GenFib (a,b,(n + 4))

proof end;

theorem :: FIB_NUM3:37

for n being Element of NAT holds GenFib (0,1,n) = Fib n

proof end;

theorem Th38: :: FIB_NUM3:38

for n being Element of NAT holds GenFib (2,1,n) = Lucas n

proof end;

theorem Th39: :: FIB_NUM3:39

for a, b, n being Element of NAT holds (GenFib (a,b,n)) + (GenFib (a,b,(n + 3))) = 2 * (GenFib (a,b,(n + 2)))

proof end;

theorem :: FIB_NUM3:40

for a, b, n being Element of NAT holds (GenFib (a,b,n)) + (GenFib (a,b,(n + 4))) = 3 * (GenFib (a,b,(n + 2)))

proof end;

theorem :: FIB_NUM3:41

for a, b, n being Element of NAT holds (GenFib (a,b,(n + 3))) - (GenFib (a,b,n)) = 2 * (GenFib (a,b,(n + 1)))

proof end;

theorem :: FIB_NUM3:42

for a, b, n being non zero Element of NAT holds GenFib (a,b,n) = ((GenFib (a,b,0)) * (Fib (n -' 1))) + ((GenFib (a,b,1)) * (Fib n))

proof end;

theorem :: FIB_NUM3:43

for n, m being Nat holds ((Fib m) * (Lucas n)) + ((Lucas m) * (Fib n)) = GenFib ((Fib 0),(Lucas 0),(n + m))

proof end;

theorem :: FIB_NUM3:44

for n being Element of NAT holds (Lucas n) + (Lucas (n + 3)) = 2 * (Lucas (n + 2))

proof end;

theorem :: FIB_NUM3:45

for a, n being Element of NAT holds GenFib (a,a,n) = ((GenFib (a,a,0)) / 2) * ((Fib n) + (Lucas n))

proof end;

theorem :: FIB_NUM3:46

for a, b, n being Element of NAT holds GenFib (b,(a + b),n) = GenFib (a,b,(n + 1))

proof end;

theorem :: FIB_NUM3:47

for a, b, n being Element of NAT holds ((GenFib (a,b,(n + 2))) * (GenFib (a,b,n))) - ((GenFib (a,b,(n + 1))) |^ 2) = ((- 1) to_power n) * (((GenFib (a,b,2)) |^ 2) - ((GenFib (a,b,1)) * (GenFib (a,b,3))))

proof end;

theorem :: FIB_NUM3:48

for a, b, k, n being Element of NAT holds GenFib ((GenFib (a,b,k)),(GenFib (a,b,(k + 1))),n) = GenFib (a,b,(n + k))

proof end;

theorem Th49: :: FIB_NUM3:49

for a, b, n being Element of NAT holds GenFib (a,b,(n + 1)) = (a * (Fib n)) + (b * (Fib (n + 1)))

proof end;

theorem :: FIB_NUM3:50

for b, n being Element of NAT holds GenFib (0,b,n) = b * (Fib n)

proof end;

theorem :: FIB_NUM3:51

for a, n being Element of NAT holds GenFib (a,0,(n + 1)) = a * (Fib n)

proof end;

theorem :: FIB_NUM3:52

for a, b, c, d, n being Element of NAT holds (GenFib (a,b,n)) + (GenFib (c,d,n)) = GenFib ((a + c),(b + d),n)

proof end;

theorem :: FIB_NUM3:53

for a, b, k, n being Element of NAT holds GenFib ((k * a),(k * b),n) = k * (GenFib (a,b,n))

proof end;

theorem :: FIB_NUM3:54

for a, b, n being Element of NAT holds GenFib (a,b,n) = ((((a * (- tau_bar)) + b) * (tau to_power n)) + (((a * tau) - b) * (tau_bar to_power n))) / (sqrt 5)

proof end;

theorem :: FIB_NUM3:55

for a, n being Element of NAT holds GenFib (((2 * a) + 1),((2 * a) + 1),(n + 1)) = ((2 * a) + 1) * (Fib (n + 2))

proof end;