Lucas number

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The Lucas sequence is an integer sequence named after the mathematician François Édouard Anatole Lucas (1842–1891), who studied both that sequence and the closely related Fibonacci sequence. Individual numbers in the Lucas sequence are known as Lucas numbers. Lucas numbers and Fibonacci numbers form complementary instances of Lucas sequences.

The Lucas sequence has the same recursive relationship as the Fibonacci sequence, where each term is the sum of the two previous terms, but with different starting values.^[1] This produces a sequence where the ratios of successive terms approach the golden ratio, and in fact the terms themselves are roundings of integer powers of the golden ratio.^[2] The sequence also has a variety of relationships with the Fibonacci numbers, like the fact that adding any two Fibonacci numbers two terms apart in the Fibonacci sequence results in the Lucas number in between.^[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, ... . (sequence A000032 in the OEIS)

which coincides for example with the number of independent vertex sets for cyclic graphs ${\displaystyle C_{n}}${\displaystyle C_{n}} of length ${\displaystyle n\geq 2}${\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 a Fibonacci integer sequence. The first two Lucas numbers are ${\displaystyle L_{0}=2}${\displaystyle L_{0}=2} and ${\displaystyle L_{1}=1}${\displaystyle L_{1}=1}, which differs from the first two Fibonacci numbers ${\displaystyle F_{0}=0}${\displaystyle F_{0}=0} and ${\displaystyle F_{1}=1}${\displaystyle F_{1}=1}. Though closely related in definition, Lucas and Fibonacci numbers exhibit distinct properties.

The Lucas numbers may thus be defined as follows:

${\displaystyle L_{n}:={\begin{cases}2&{\text{if }}n=0;\1円&{\text{if }}n=1;\\L_{n-1}+L_{n-2}&{\text{if }}n>1.\end{cases}}}${\displaystyle L_{n}:={\begin{cases}2&{\text{if }}n=0;\1円&{\text{if }}n=1;\\L_{n-1}+L_{n-2}&{\text{if }}n>1.\end{cases}}}

(where n belongs to the natural numbers)

All Fibonacci-like integer sequences appear in shifted form as a row of the Wythoff 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 numbers converges to the golden ratio.

Using ${\displaystyle L_{n-2}=L_{n}-L_{n-1}}${\displaystyle L_{n-2}=L_{n}-L_{n-1}}, one can extend the Lucas numbers to negative integers to obtain a doubly infinite sequence:

..., −11, 7, −4, 3, −1, 2, 1, 3, 4, 7, 11, ... (terms ${\displaystyle L_{n}}${\displaystyle L_{n}} for ${\displaystyle -5\leq {}n\leq 5}${\displaystyle -5\leq {}n\leq 5} are shown).

The formula for terms with negative indices in this sequence is

${\displaystyle L_{-n}=(-1)^{n}L_{n}.\!}${\displaystyle L_{-n}=(-1)^{n}L_{n}.\!}

The Lucas numbers are related to the Fibonacci numbers by many identities. Among these are the following:

• ${\displaystyle L_{n}=F_{n-1}+F_{n+1}=2F_{n+1}-F_{n}}${\displaystyle L_{n}=F_{n-1}+F_{n+1}=2F_{n+1}-F_{n}}
• ${\displaystyle L_{m+n}=L_{m+1}F_{n}+L_{m}F_{n-1}}${\displaystyle L_{m+n}=L_{m+1}F_{n}+L_{m}F_{n-1}}
• ${\displaystyle F_{2n}=L_{n}F_{n}}${\displaystyle F_{2n}=L_{n}F_{n}}
• ${\displaystyle F_{n+k}+(-1)^{k}F_{n-k}=L_{k}F_{n}}${\displaystyle F_{n+k}+(-1)^{k}F_{n-k}=L_{k}F_{n}}
• ${\displaystyle 2F_{2n+k}=L_{n}F_{n+k}+L_{n+k}F_{n}}${\displaystyle 2F_{2n+k}=L_{n}F_{n+k}+L_{n+k}F_{n}}
• ${\displaystyle L_{2n}=5F_{n}^{2}+2(-1)^{n}=L_{n}^{2}-2(-1)^{n}}${\displaystyle L_{2n}=5F_{n}^{2}+2(-1)^{n}=L_{n}^{2}-2(-1)^{n}}, so ${\displaystyle \lim _{n\to \infty }{\frac {L_{n}}{F_{n}}}={\sqrt {5}}}${\displaystyle \lim _{n\to \infty }{\frac {L_{n}}{F_{n}}}={\sqrt {5}}}.
• ${\displaystyle \vert L_{n}-{\sqrt {5}}F_{n}\vert ={\frac {2}{\varphi ^{n}}}\to 0}${\displaystyle \vert L_{n}-{\sqrt {5}}F_{n}\vert ={\frac {2}{\varphi ^{n}}}\to 0}
• ${\displaystyle L_{n+k}-(-1)^{k}L_{n-k}=5F_{n}F_{k}}${\displaystyle L_{n+k}-(-1)^{k}L_{n-k}=5F_{n}F_{k}}; in particular, ${\displaystyle F_{n}={L_{n-1}+L_{n+1} \over 5}}${\displaystyle F_{n}={L_{n-1}+L_{n+1} \over 5}}, so ${\displaystyle 5F_{n}+L_{n}=2L_{n+1}}${\displaystyle 5F_{n}+L_{n}=2L_{n+1}}.

Their closed formula is given as:

${\displaystyle L_{n}=\varphi ^{n}+(1-\varphi )^{n}=\varphi ^{n}+{\bigl (}{-\varphi ^{-1}}{\bigr )}^{n}=\left({1+{\sqrt {5}} \over 2}\right)^{n}+\left({1-{\sqrt {5}} \over 2}\right)^{n},}${\displaystyle L_{n}=\varphi ^{n}+(1-\varphi )^{n}=\varphi ^{n}+{\bigl (}{-\varphi ^{-1}}{\bigr )}^{n}=\left({1+{\sqrt {5}} \over 2}\right)^{n}+\left({1-{\sqrt {5}} \over 2}\right)^{n},}

where ${\displaystyle \varphi }${\displaystyle \varphi } is the golden ratio. Alternatively, as for ${\displaystyle n>1}${\displaystyle n>1} the magnitude of the term ${\displaystyle \textstyle {\bigl (}{-\varphi ^{-1}}{\bigr )}^{n}}${\displaystyle \textstyle {\bigl (}{-\varphi ^{-1}}{\bigr )}^{n}} is less than 1/2, ${\displaystyle L_{n}}${\displaystyle L_{n}} is the closest integer to ${\displaystyle \varphi ^{n}}${\displaystyle \varphi ^{n}} or, equivalently, the integer part of ${\displaystyle \varphi ^{n}+1/2}${\displaystyle \varphi ^{n}+1/2}, also written as ${\displaystyle \lfloor \varphi ^{n}+1/2\rfloor }${\displaystyle \lfloor \varphi ^{n}+1/2\rfloor }.

Combining the above with Binet's formula,

${\displaystyle F_{n}={\frac {\varphi ^{n}-(1-\varphi )^{n}}{\sqrt {5}}},,円}${\displaystyle F_{n}={\frac {\varphi ^{n}-(1-\varphi )^{n}}{\sqrt {5}}},,円}

a formula for ${\displaystyle \varphi ^{n}}${\displaystyle \varphi ^{n}} is obtained:

${\displaystyle \varphi ^{n}={{L_{n}+F_{n}{\sqrt {5}}} \over 2},円.}${\displaystyle \varphi ^{n}={{L_{n}+F_{n}{\sqrt {5}}} \over 2},円.}

For integers n ≥ 2, we also get:

${\displaystyle \varphi ^{n}=L_{n}-{\bigl (}{-\varphi ^{-1}}{\bigr )}^{n}=L_{n}-(-1)^{n}L_{n}^{-1}-L_{n}^{-3}+R}${\displaystyle \varphi ^{n}=L_{n}-{\bigl (}{-\varphi ^{-1}}{\bigr )}^{n}=L_{n}-(-1)^{n}L_{n}^{-1}-L_{n}^{-3}+R}

with remainder R satisfying

${\displaystyle \vert R\vert <3L_{n}^{-5}}${\displaystyle \vert R\vert <3L_{n}^{-5}}.

Many of the Fibonacci identities have parallels in Lucas numbers. For example, the Cassini identity becomes

${\displaystyle L_{n}^{2}-L_{n-1}L_{n+1}=(-1)^{n}5}${\displaystyle L_{n}^{2}-L_{n-1}L_{n+1}=(-1)^{n}5}

Also

${\displaystyle \sum _{k=0}^{n}L_{k}=L_{n+2}-1}${\displaystyle \sum _{k=0}^{n}L_{k}=L_{n+2}-1}

${\displaystyle \sum _{k=0}^{n}L_{k}^{2}=L_{n}L_{n+1}+2}${\displaystyle \sum _{k=0}^{n}L_{k}^{2}=L_{n}L_{n+1}+2}

${\displaystyle 2L_{n-1}^{2}+L_{n}^{2}=L_{2n+1}+5F_{n-2}^{2}}${\displaystyle 2L_{n-1}^{2}+L_{n}^{2}=L_{2n+1}+5F_{n-2}^{2}}

where ${\displaystyle \textstyle F_{n}={\frac {L_{n-1}+L_{n+1}}{5}}}${\displaystyle \textstyle F_{n}={\frac {L_{n-1}+L_{n+1}}{5}}}.

${\displaystyle L_{n}^{k}=\sum _{j=0}^{\lfloor {\frac {k}{2}}\rfloor }(-1)^{nj}{\binom {k}{j}}L'_{(k-2j)n}}${\displaystyle L_{n}^{k}=\sum _{j=0}^{\lfloor {\frac {k}{2}}\rfloor }(-1)^{nj}{\binom {k}{j}}L'_{(k-2j)n}}

where ${\displaystyle L'_{n}=L_{n}}${\displaystyle L'_{n}=L_{n}} except for ${\displaystyle L'_{0}=1}${\displaystyle L'_{0}=1}.

For example if n is odd, ${\displaystyle L_{n}^{3}=L'_{3n}-3L'_{n}}${\displaystyle L_{n}^{3}=L'_{3n}-3L'_{n}} and ${\displaystyle L_{n}^{4}=L'_{4n}-4L'_{2n}+6L'_{0}}${\displaystyle L_{n}^{4}=L'_{4n}-4L'_{2n}+6L'_{0}}

Checking, ${\displaystyle L_{3}=4,4^{3}=64=76-3(4)}${\displaystyle L_{3}=4,4^{3}=64=76-3(4)}, and ${\displaystyle 256=322-4(18)+6}${\displaystyle 256=322-4(18)+6}

The ordinary generating function of the sequence of Lucas numbers is the power series

$${\displaystyle \Phi (x)=\sum _{k=0}^{\infty }L_{k}x^{k}=2+x+3x^{2}+4x^{3}+7x^{4}+11x^{5}+\cdots .}$${\displaystyle \Phi (x)=\sum _{k=0}^{\infty }L_{k}x^{k}=2+x+3x^{2}+4x^{3}+7x^{4}+11x^{5}+\cdots .}

This series is convergent for any complex number ${\displaystyle x}${\displaystyle x} satisfying ${\displaystyle |x|<1/\varphi \approx 0.618,}${\displaystyle |x|<1/\varphi \approx 0.618,} and its sum has a simple closed form: $${\displaystyle \Phi (x)={\frac {2-x}{1-x-x^{2}}}.}$${\displaystyle \Phi (x)={\frac {2-x}{1-x-x^{2}}}.}

This can be proved by multiplying by ${\textstyle (1-x-x^{2})}${\textstyle (1-x-x^{2})}: $${\displaystyle {\begin{aligned}(1-x-x^{2})\Phi (x)&=\sum _{k=0}^{\infty }L_{k}x^{k}-\sum _{k=0}^{\infty }L_{k}x^{k+1}-\sum _{k=0}^{\infty }L_{k}x^{k+2}\\&=\sum _{k=0}^{\infty }L_{k}x^{k}-\sum _{k=1}^{\infty }L_{k-1}x^{k}-\sum _{k=2}^{\infty }L_{k-2}x^{k}\\&=2x^{0}+1x^{1}-2x^{1}+\sum _{k=2}^{\infty }(L_{k}-L_{k-1}-L_{k-2})x^{k}\\&=2-x,\end{aligned}}}$${\displaystyle {\begin{aligned}(1-x-x^{2})\Phi (x)&=\sum _{k=0}^{\infty }L_{k}x^{k}-\sum _{k=0}^{\infty }L_{k}x^{k+1}-\sum _{k=0}^{\infty }L_{k}x^{k+2}\\&=\sum _{k=0}^{\infty }L_{k}x^{k}-\sum _{k=1}^{\infty }L_{k-1}x^{k}-\sum _{k=2}^{\infty }L_{k-2}x^{k}\\&=2x^{0}+1x^{1}-2x^{1}+\sum _{k=2}^{\infty }(L_{k}-L_{k-1}-L_{k-2})x^{k}\\&=2-x,\end{aligned}}} where all terms involving ${\displaystyle x^{k}}${\displaystyle x^{k}} for ${\displaystyle k\geq 2}${\displaystyle k\geq 2} cancel out because of the defining Lucas numbers recurrence relation.

${\displaystyle \Phi \!\left(-{\frac {1}{x}}\right)}${\displaystyle \Phi \!\left(-{\frac {1}{x}}\right)} gives the generating function for the negative indexed Lucas numbers, ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}L_{n}x^{-n}=\sum _{n=0}^{\infty }L_{-n}x^{-n}}${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}L_{n}x^{-n}=\sum _{n=0}^{\infty }L_{-n}x^{-n}}, and

${\displaystyle \Phi \!\left(-{\frac {1}{x}}\right)={\frac {x+2x^{2}}{1-x-x^{2}}}}${\displaystyle \Phi \!\left(-{\frac {1}{x}}\right)={\frac {x+2x^{2}}{1-x-x^{2}}}}

${\displaystyle \Phi (x)}${\displaystyle \Phi (x)} satisfies the functional equation

${\displaystyle \Phi (x)-\Phi \!\left(-{\frac {1}{x}}\right)=2}${\displaystyle \Phi (x)-\Phi \!\left(-{\frac {1}{x}}\right)=2}

As the generating function for the Fibonacci numbers is given by

${\displaystyle s(x)={\frac {x}{1-x-x^{2}}}}${\displaystyle s(x)={\frac {x}{1-x-x^{2}}}}

we have

${\displaystyle s(x)+\Phi (x)={\frac {2}{1-x-x^{2}}}}${\displaystyle s(x)+\Phi (x)={\frac {2}{1-x-x^{2}}}}

which proves that

${\displaystyle F_{n}+L_{n}=2F_{n+1},}${\displaystyle F_{n}+L_{n}=2F_{n+1},}

and

${\displaystyle 5s(x)+\Phi (x)={\frac {2}{x}}\Phi (-{\frac {1}{x}})=2{\frac {1}{1-x-x^{2}}}+4{\frac {x}{1-x-x^{2}}}}${\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

${\displaystyle 5F_{n}+L_{n}=2L_{n+1}}${\displaystyle 5F_{n}+L_{n}=2L_{n+1}}

The partial fraction decomposition is given by

${\displaystyle \Phi (x)={\frac {1}{1-\phi x}}+{\frac {1}{1-\psi x}}}${\displaystyle \Phi (x)={\frac {1}{1-\phi x}}+{\frac {1}{1-\psi x}}}

where ${\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}}}${\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}}} is the golden ratio and ${\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}}${\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}} is its conjugate.

This can be used to prove the generating function, as

${\displaystyle \sum _{n=0}^{\infty }L_{n}x^{n}=\sum _{n=0}^{\infty }(\phi ^{n}+\psi ^{n})x^{n}=\sum _{n=0}^{\infty }\phi ^{n}x^{n}+\sum _{n=0}^{\infty }\psi ^{n}x^{n}={\frac {1}{1-\phi x}}+{\frac {1}{1-\psi x}}=\Phi (x)}${\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)}

Using ${\displaystyle x}${\displaystyle x} equal to any of 0.01, 0.001, 0.0001, etc. lays out the first Lucas numbers in the decimal expansion of ${\displaystyle \Phi (x)}${\displaystyle \Phi (x)}. For example, ${\displaystyle \Phi (0.001)={\frac {1.999}{0.998999}}={\frac {1999000}{998999}}=2.001003004007011018029047\ldots .}${\displaystyle \Phi (0.001)={\frac {1.999}{0.998999}}={\frac {1999000}{998999}}=2.001003004007011018029047\ldots .}

If ${\displaystyle F_{n}\geq 5}${\displaystyle F_{n}\geq 5} is a Fibonacci number then no Lucas number is divisible by ${\displaystyle F_{n}}${\displaystyle F_{n}}.

The Lucas numbers satisfy Gauss congruence. This implies that ${\displaystyle L_{n}}${\displaystyle L_{n}} is congruent to 1 modulo ${\displaystyle n}${\displaystyle n} if ${\displaystyle n}${\displaystyle n} is prime. The composite values of ${\displaystyle n}${\displaystyle n} which satisfy this property are known as Fibonacci pseudoprimes.

${\displaystyle L_{n}-L_{n-4}}${\displaystyle L_{n}-L_{n-4}} is congruent to 0 modulo 5.

A Lucas prime is a Lucas number that is prime. The first few Lucas primes are

2, 3, 7, 11, 29, 47, 199, 521, 2207, 3571, 9349, 3010349, 54018521, 370248451, 6643838879, ... (sequence A005479 in the OEIS).

The indices of these primes are (for example, L₄ = 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, ... (sequence A001606 in the OEIS).

As of September 2015^[update] , the largest confirmed Lucas prime is L₁₄₈₀₉₁, which has 30950 decimal digits.^[4] As of August 2022^[update] , the largest known Lucas probable prime is L_5466311, with 1,142,392 decimal digits.^[5]

If L_n is prime then n is 0, prime, or a power of 2.^[6] L_2^m is prime for m = 1, 2, 3, and 4 and no other known values of m.

In the same way as Fibonacci polynomials are derived from the Fibonacci numbers, the Lucas polynomials ${\displaystyle L_{n}(x)}${\displaystyle L_{n}(x)} are a polynomial sequence derived from the Lucas numbers.

For all but the smallest values of n, the integer L_n very closely approximates the n-th power of the golden ratio, ⁠${\displaystyle \varphi ^{n}}${\displaystyle \varphi ^{n}}⁠. Furthermore, close rational approximations for powers of the golden ratio can be obtained from their continued fractions.

For positive integers n, the continued fractions are:

${\displaystyle \varphi ^{2n-1}=[L_{2n-1};L_{2n-1},L_{2n-1},L_{2n-1},\ldots ]}${\displaystyle \varphi ^{2n-1}=[L_{2n-1};L_{2n-1},L_{2n-1},L_{2n-1},\ldots ]}

${\displaystyle \varphi ^{2n}=[L_{2n}-1;1,L_{2n}-2,1,L_{2n}-2,1,L_{2n}-2,1,\ldots ]}${\displaystyle \varphi ^{2n}=[L_{2n}-1;1,L_{2n}-2,1,L_{2n}-2,1,L_{2n}-2,1,\ldots ]}.

For example:

${\displaystyle \varphi ^{5}=[11;11,11,11,\ldots ]}${\displaystyle \varphi ^{5}=[11;11,11,11,\ldots ]}

is the limit of

${\displaystyle {\frac {11}{1}},{\frac {122}{11}},{\frac {1353}{122}},{\frac {15005}{1353}},\ldots }${\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

${\displaystyle \varphi ^{6}=[18-1;1,18-2,1,18-2,1,18-2,1,\ldots ]=[17;1,16,1,16,1,16,1,\ldots ]}${\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

${\displaystyle {\frac {17}{1}},{\frac {18}{1}},{\frac {305}{17}},{\frac {323}{18}},{\frac {5473}{305}},{\frac {5796}{323}},{\frac {98209}{5473}},{\frac {104005}{5796}},\ldots }${\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 the second previous term.

Lucas numbers are the second most common pattern in sunflowers after Fibonacci numbers, when clockwise and counter-clockwise spirals are counted, according to an analysis of 657 sunflowers in 2016.^[7]

• Generalizations of Fibonacci numbers

• "Lucas polynomials", Encyclopedia of Mathematics , EMS Press, 2001 [1994]
• Weisstein, Eric W. "Lucas Number". MathWorld .
• Weisstein, Eric W. "Lucas Polynomial". MathWorld .
• "The Lucas Numbers", Dr Ron Knott
• Lucas numbers and the Golden Section
• A Lucas Number Calculator can be found here.
• OEIS sequence A000032 (Lucas numbers beginning at 2)

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