Infinite integer series where the next number is the sum of the two preceding it
Not to be confused with
Lucas sequences , the general class of sequences to which the Lucas numbers belong.
The Lucas spiral, made with quarter-arcs , is a good approximation of thegolden spiral when its terms are large. However, when its terms become very small, the arc's radius decreases rapidly from 3 to 1 then increases from 1 to 2. TheLucas sequence is aninteger sequence named after the mathematicianFrançois Édouard Anatole Lucas (1842–1891), who studied both thatsequence and the closely relatedFibonacci sequence . Individual numbers in the Lucas sequence are known asLucas numbers . Lucas numbers and Fibonacci numbers form complementary instances ofLucas sequences .
The Lucas sequence has the samerecursive relationship as the Fibonacci sequence, where each term is the sum of the two previous terms, but with different starting values.[ 1] golden ratio , and in fact the terms themselves areroundings ofinteger powers of the golden ratio.[ 2] [ 3]
The first few Lucas numbers are
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, ... . (sequenceA000032 OEIS ) which coincides for example with the number ofindependent vertex sets forcyclic graphs C n {\displaystyle C_{n}} n ≥ 2 {\displaystyle n\geq 2} [ 1]
As with the Fibonacci numbers, each Lucas number is defined to be the sum of its two immediately previous terms, thereby forming aFibonacci integer sequence . The first two Lucas numbers areL 0 = 2 {\displaystyle L_{0}=2} L 1 = 1 {\displaystyle L_{1}=1} F 0 = 0 {\displaystyle F_{0}=0} F 1 = 1 {\displaystyle F_{1}=1}
The Lucas numbers may thus be defined as follows:
L n := { 2 if n = 0 ; 1 if n = 1 ; L n − 1 + L n − 2 if n > 1. {\displaystyle L_{n}:={\begin{cases}2&{\text{if }}n=0;\\1&{\text{if }}n=1;\\L_{n-1}+L_{n-2}&{\text{if }}n>1.\end{cases}}} (wheren belongs to thenatural numbers )
All Fibonacci-like integer sequences appear in shifted form as a row of theWythoff array ; the Fibonacci sequence itself is the first row and the Lucas sequence is the second row. Also like all Fibonacci-like integer sequences, the ratio between two consecutive Lucas numbersconverges to thegolden ratio .
Extension to negative integers [ edit ] UsingL n − 2 = L n − L n − 1 {\displaystyle L_{n-2}=L_{n}-L_{n-1}}
..., −11, 7, −4, 3, −1, 2, 1, 3, 4, 7, 11, ... (termsL n {\displaystyle L_{n}} − 5 ≤ n ≤ 5 {\displaystyle -5\leq {}n\leq 5} The formula for terms with negative indices in this sequence is
L − n = ( − 1 ) n L n . {\displaystyle L_{-n}=(-1)^{n}L_{n}.\!} Relationship to Fibonacci numbers [ edit ] The first identity expressed visually The Lucas numbers are related to the Fibonacci numbers by manyidentities . Among these are the following:
Theirclosed formula is given as:
L n = φ n + ( 1 − φ ) n = φ n + ( − φ ) − n = ( 1 + 5 2 ) n + ( 1 − 5 2 ) n , {\displaystyle L_{n}=\varphi ^{n}+(1-\varphi )^{n}=\varphi ^{n}+(-\varphi )^{-n}=\left({1+{\sqrt {5}} \over 2}\right)^{n}+\left({1-{\sqrt {5}} \over 2}\right)^{n}\,,} whereφ {\displaystyle \varphi } golden ratio . Alternatively, as forn > 1 {\displaystyle n>1} ( − φ ) − n {\displaystyle (-\varphi )^{-n}} L n {\displaystyle L_{n}} φ n {\displaystyle \varphi ^{n}} φ n + 1 / 2 {\displaystyle \varphi ^{n}+1/2} ⌊ φ n + 1 / 2 ⌋ {\displaystyle \lfloor \varphi ^{n}+1/2\rfloor }
Combining the above withBinet's formula ,
F n = φ n − ( 1 − φ ) n 5 , {\displaystyle F_{n}={\frac {\varphi ^{n}-(1-\varphi )^{n}}{\sqrt {5}}}\,,} a formula forφ n {\displaystyle \varphi ^{n}}
φ n = L n + F n 5 2 . {\displaystyle \varphi ^{n}={{L_{n}+F_{n}{\sqrt {5}}} \over 2}\,.} For integersn ≥ 2, we also get:
φ n = L n − ( − φ ) − n = L n − ( − 1 ) n L n − 1 − L n − 3 + R {\displaystyle \varphi ^{n}=L_{n}-(-\varphi )^{-n}=L_{n}-(-1)^{n}L_{n}^{-1}-L_{n}^{-3}+R} with remainderR satisfying
| R | < 3 L n − 5 {\displaystyle \vert R\vert <3L_{n}^{-5}} Many of the Fibonacci identities have parallels in Lucas numbers. For example, theCassini identity becomes
L n 2 − L n − 1 L n + 1 = ( − 1 ) n 5 {\displaystyle L_{n}^{2}-L_{n-1}L_{n+1}=(-1)^{n}5} Also
∑ k = 0 n L k = L n + 2 − 1 {\displaystyle \sum _{k=0}^{n}L_{k}=L_{n+2}-1} ∑ k = 0 n L k 2 = L n L n + 1 + 2 {\displaystyle \sum _{k=0}^{n}L_{k}^{2}=L_{n}L_{n+1}+2} 2 L n − 1 2 + L n 2 = L 2 n + 1 + 5 F n − 2 2 {\displaystyle 2L_{n-1}^{2}+L_{n}^{2}=L_{2n+1}+5F_{n-2}^{2}} whereF n = L n − 1 + L n + 1 5 {\displaystyle \textstyle F_{n}={\frac {L_{n-1}+L_{n+1}}{5}}}
L n k = ∑ j = 0 ⌊ k 2 ⌋ ( − 1 ) n j ( k j ) L ( k − 2 j ) n ′ {\displaystyle L_{n}^{k}=\sum _{j=0}^{\lfloor {\frac {k}{2}}\rfloor }(-1)^{nj}{\binom {k}{j}}L'_{(k-2j)n}} whereL n ′ = L n {\displaystyle L'_{n}=L_{n}} L 0 ′ = 1 {\displaystyle L'_{0}=1}
For example ifn isodd ,L n 3 = L 3 n ′ − 3 L n ′ {\displaystyle L_{n}^{3}=L'_{3n}-3L'_{n}} L n 4 = L 4 n ′ − 4 L 2 n ′ + 6 L 0 ′ {\displaystyle L_{n}^{4}=L'_{4n}-4L'_{2n}+6L'_{0}}
Checking,L 3 = 4 , 4 3 = 64 = 76 − 3 ( 4 ) {\displaystyle L_{3}=4,4^{3}=64=76-3(4)} 256 = 322 − 4 ( 18 ) + 6 {\displaystyle 256=322-4(18)+6}
Generating function [ edit ] Let
Φ ( x ) = 2 + x + 3 x 2 + 4 x 3 + ⋯ = ∑ n = 0 ∞ L n x n {\displaystyle \Phi (x)=2+x+3x^{2}+4x^{3}+\cdots =\sum _{n=0}^{\infty }L_{n}x^{n}} be thegenerating function of the Lucas numbers. By a direct computation,
Φ ( x ) = L 0 + L 1 x + ∑ n = 2 ∞ L n x n = 2 + x + ∑ n = 2 ∞ ( L n − 1 + L n − 2 ) x n = 2 + x + ∑ n = 1 ∞ L n x n + 1 + ∑ n = 0 ∞ L n x n + 2 = 2 + x + x ( Φ ( x ) − 2 ) + x 2 Φ ( x ) {\displaystyle {\begin{aligned}\Phi (x)&=L_{0}+L_{1}x+\sum _{n=2}^{\infty }L_{n}x^{n}\\&=2+x+\sum _{n=2}^{\infty }(L_{n-1}+L_{n-2})x^{n}\\&=2+x+\sum _{n=1}^{\infty }L_{n}x^{n+1}+\sum _{n=0}^{\infty }L_{n}x^{n+2}\\&=2+x+x(\Phi (x)-2)+x^{2}\Phi (x)\end{aligned}}} which can be rearranged as
Φ ( x ) = 2 − x 1 − x − x 2 {\displaystyle \Phi (x)={\frac {2-x}{1-x-x^{2}}}} Φ ( − 1 x ) {\displaystyle \Phi \!\left(-{\frac {1}{x}}\right)} negative indexed Lucas numbers ,∑ n = 0 ∞ ( − 1 ) n L n x − n = ∑ n = 0 ∞ L − n x − n {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}L_{n}x^{-n}=\sum _{n=0}^{\infty }L_{-n}x^{-n}}
Φ ( − 1 x ) = x + 2 x 2 1 − x − x 2 {\displaystyle \Phi \!\left(-{\frac {1}{x}}\right)={\frac {x+2x^{2}}{1-x-x^{2}}}} Φ ( x ) {\displaystyle \Phi (x)} functional equation
Φ ( x ) − Φ ( − 1 x ) = 2 {\displaystyle \Phi (x)-\Phi \!\left(-{\frac {1}{x}}\right)=2} As thegenerating function for the Fibonacci numbers is given by
s ( x ) = x 1 − x − x 2 {\displaystyle s(x)={\frac {x}{1-x-x^{2}}}} we have
s ( x ) + Φ ( x ) = 2 1 − x − x 2 {\displaystyle s(x)+\Phi (x)={\frac {2}{1-x-x^{2}}}} whichproves that
F n + L n = 2 F n + 1 , {\displaystyle F_{n}+L_{n}=2F_{n+1},} and
5 s ( x ) + Φ ( x ) = 2 x Φ ( − 1 x ) = 2 1 1 − x − x 2 + 4 x 1 − x − x 2 {\displaystyle 5s(x)+\Phi (x)={\frac {2}{x}}\Phi (-{\frac {1}{x}})=2{\frac {1}{1-x-x^{2}}}+4{\frac {x}{1-x-x^{2}}}} proves that
5 F n + L n = 2 L n + 1 {\displaystyle 5F_{n}+L_{n}=2L_{n+1}} Thepartial fraction decomposition is given by
Φ ( x ) = 1 1 − ϕ x + 1 1 − ψ x {\displaystyle \Phi (x)={\frac {1}{1-\phi x}}+{\frac {1}{1-\psi x}}} whereϕ = 1 + 5 2 {\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}}} ψ = 1 − 5 2 {\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}} conjugate .
This can be used to prove the generating function, as
∑ n = 0 ∞ L n x n = ∑ n = 0 ∞ ( ϕ n + ψ n ) x n = ∑ n = 0 ∞ ϕ n x n + ∑ n = 0 ∞ ψ n x n = 1 1 − ϕ x + 1 1 − ψ x = Φ ( x ) {\displaystyle \sum _{n=0}^{\infty }L_{n}x^{n}=\sum _{n=0}^{\infty }(\phi ^{n}+\psi ^{n})x^{n}=\sum _{n=0}^{\infty }\phi ^{n}x^{n}+\sum _{n=0}^{\infty }\psi ^{n}x^{n}={\frac {1}{1-\phi x}}+{\frac {1}{1-\psi x}}=\Phi (x)} Congruence relations [ edit ] IfF n ≥ 5 {\displaystyle F_{n}\geq 5} F n {\displaystyle F_{n}}
The Lucas numbers satisfyGauss congruence . This implies thatL n {\displaystyle L_{n}} congruent to 1 modulon {\displaystyle n} n {\displaystyle n} prime . Thecomposite values ofn {\displaystyle n} Fibonacci pseudoprimes .
L n − L n − 4 {\displaystyle L_{n}-L_{n-4}}
ALucas prime is a Lucas number that isprime . The first few Lucas primes are
2, 3, 7, 11, 29, 47, 199, 521, 2207, 3571, 9349, 3010349, 54018521, 370248451, 6643838879, ... (sequenceA005479 OEIS ). The indices of these primes are (for example,L 4 = 7)
0, 2, 4, 5, 7, 8, 11, 13, 16, 17, 19, 31, 37, 41, 47, 53, 61, 71, 79, 113, 313, 353, 503, 613, 617, 863, 1097, 1361, 4787, 4793, 5851, 7741, 8467, ... (sequenceA001606 OEIS ). As of September 2015[update] L 148091 , which has 30950 decimal digits.[ 4] [update] probable prime isL 5466311 , with 1,142,392 decimal digits.[ 5]
IfLn is prime thenn is 0, prime, or apower of 2 .[ 6] L 2m is prime form = 1, 2, 3, and 4 and no other known values of m .
In the same way asFibonacci polynomials are derived from theFibonacci numbers , theLucas polynomials L n ( x ) {\displaystyle L_{n}(x)} polynomial sequence derived from the Lucas numbers.
Continued fractions for powers of the golden ratio [ edit ] Closerational approximations for powers of the golden ratio can be obtained from theircontinued fractions .
For positive integersn , the continued fractions are:
φ 2 n − 1 = [ L 2 n − 1 ; L 2 n − 1 , L 2 n − 1 , L 2 n − 1 , … ] {\displaystyle \varphi ^{2n-1}=[L_{2n-1};L_{2n-1},L_{2n-1},L_{2n-1},\ldots ]} φ 2 n = [ L 2 n − 1 ; 1 , L 2 n − 2 , 1 , L 2 n − 2 , 1 , L 2 n − 2 , 1 , … ] {\displaystyle \varphi ^{2n}=[L_{2n}-1;1,L_{2n}-2,1,L_{2n}-2,1,L_{2n}-2,1,\ldots ]} For example:
φ 5 = [ 11 ; 11 , 11 , 11 , … ] {\displaystyle \varphi ^{5}=[11;11,11,11,\ldots ]} is the limit of
11 1 , 122 11 , 1353 122 , 15005 1353 , … {\displaystyle {\frac {11}{1}},{\frac {122}{11}},{\frac {1353}{122}},{\frac {15005}{1353}},\ldots } with the error in each term being about 1% of the error in the previous term; and
φ 6 = [ 18 − 1 ; 1 , 18 − 2 , 1 , 18 − 2 , 1 , 18 − 2 , 1 , … ] = [ 17 ; 1 , 16 , 1 , 16 , 1 , 16 , 1 , … ] {\displaystyle \varphi ^{6}=[18-1;1,18-2,1,18-2,1,18-2,1,\ldots ]=[17;1,16,1,16,1,16,1,\ldots ]} is the limit of
17 1 , 18 1 , 305 17 , 323 18 , 5473 305 , 5796 323 , 98209 5473 , 104005 5796 , … {\displaystyle {\frac {17}{1}},{\frac {18}{1}},{\frac {305}{17}},{\frac {323}{18}},{\frac {5473}{305}},{\frac {5796}{323}},{\frac {98209}{5473}},{\frac {104005}{5796}},\ldots } with the error in each term being about 0.3% that of thesecond previous term.
Lucas numbers are the second most common pattern insunflowers after Fibonacci numbers, when clockwise and counter-clockwise spirals are counted, according to an analysis of 657 sunflowers in 2016.[ 7]
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