Lucas sequence: Difference between revisions

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The terms of Lucas sequences satisfy relations that are generalizations of those between [[Fibonacci number]]s <math>F_n=U_n(1,-1)</math> and [[Lucas number]]s <math>L_n=V_n(1,-1)</math>. For example:

:<math>

\begin{array}{r|l}

\begin{array}{|l}

\text{General case} & (P,Q) = (1,-1), D = P^2 – 4Q = 5

\hline

D U_n = {V_{n+1} – Q V_{n-1}}=2V_{n+1}-P V_n & 5F_n = {L_{n+1} + L_{n-1}}=2L_{n+1} – L_{n}

V_n = U_{n+1} – Q U_{n-1}=2U_{n+1}-PU_n & L_n = F_{n+1} + F_{n-1}=2F_{n+1}-F_n

U_{m+n} = U_n U_{m+1} – Q U_m U_{n-1} = U_mV_n-Q^nU_{m-n} & F_{m+n} = F_n F_{m+1} + F_m F_{n-1} =F_mL_n-(-1)^nF_{m-n}

U_{2n} = U_n (U_{n+1} – QU_{n-1}) = U_n V_n & F_{2n} = F_n (F_{n+1} + F_{n-1}) = F_n L_n

U_{2n+1} = U_{n+1}^2 – Q U_n^2 & F_{2n+1} = F_{n+1}^2 + F_n^2

V_{m+n} = V_m V_n – Q^n V_{m-n} = D U_m U_n + Q^n V_{m-n} & L_{m+n} = L_m L_n – (-1)^n L_{m-n} = 5 F_m F_n + (-1)^n L_{m-n}

V_{2n} = V_n^2 – 2Q^n = D U_n^2 + 2Q^n & L_{2n} = L_n^2 – 2(-1)^n = 5 F_n^2 + 2(-1)^n

U_{m+n} = \frac{U_mV_n+U_nV_m}{2} & F_{m+n} = \frac{F_mL_n+F_nL_m}{2}

V_{m+n}=\frac{V_mV_n+DU_mU_n}{2} & L_{m+n}=\frac{L_mL_n+5F_mF_n}{2}

V_n^2-DU_n^2=4Q^n & L_n^2-5F_n^2=4(-1)^n

U_n^2-U_{n-1}U_{n+1}=Q^{n-1} & F_n^2-F_{n-1}F_{n+1}=(-1)^{n-1}

V_n^2-V_{n-1}V_{n+1}=DQ^{n-1} & L_n^2-L_{n-1}L_{n+1}=5(-1)^{n-1}

2^{n-1}U_n={n \choose 1}P^{n-1}+{n \choose 3}P^{n-3}D+\cdots & 2^{n-1}F_n={n \choose 1}+5{n \choose 3}+\cdots

2^{n-1}V_n=P^n+{n \choose 2}P^{n-2}D+{n \choose 4}P^{n-4}D^2+\cdots & 2^{n-1}L_n=1+5{n \choose 2}+5^2{n \choose 4}+\cdots

\end{array}

</math>

Of these, (6) and (7) allow fast calculation of ”V” independent of ”U” in a way analogous to [[exponentiation by squaring]]. The relation <math>V_{mn} = V_{m}(P = V_n, Q = Q_n)</math> (which belongs to the section above, “relations between sequences with different parameters”) is also useful for this purpose.<ref>{{cite web |last1=Atnashev |first1=Pavel |title=A simpler alternative to Lucas–Lehmer–Riesel primality test |url=https://eprint.iacr.org/2023/195 |website=Cryptology ePrint Archive}}</ref>

=== Divisibility properties ===

Certain constant-recursive integer sequences

In mathematics, the Lucas sequences $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ and $V n ( P , Q ) {\displaystyle V_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40dfdb812d2de843618bac1786e8887494aba43e" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)}" width="3875.3" height="1223.9">$ are certain constant-recursive integer sequences that satisfy the recurrence relation

x n = P ⋅ x n − 1 − Q ⋅ x n − 2 {\displaystyle x_{n}=P\cdot x_{n-1}-Q\cdot x_{n-2}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/2f001cb3047032efac2b4bf6ed7b4d5a5d7ec065" aria-hidden="true" alt="{\displaystyle x_{n}=P\cdot x_{n-1}-Q\cdot x_{n-2}}" width="1200" height="400">

where $P {\displaystyle P} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b4dc73bf40314945ff376bd363916a738548d40a" aria-hidden="true" alt="{\displaystyle P}" width="1200" height="400">$ and $Q {\displaystyle Q} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8752c7023b4b3286800fe3238271bbca681219ed" aria-hidden="true" alt="{\displaystyle Q}" width="1200" height="400">$ are fixed integers. Any sequence satisfying this recurrence relation can be represented as a linear combination of the Lucas sequences $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ and $V n ( P , Q ) . {\displaystyle V_{n}(P,Q).} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d676acac90fb321d4c12f3f1c2a286b8c517b761" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q).}" width="1200" height="400">$

More generally, Lucas sequences $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ and $V n ( P , Q ) {\displaystyle V_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40dfdb812d2de843618bac1786e8887494aba43e" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)}" width="3875.3" height="1223.9">$ represent sequences of polynomials in $P {\displaystyle P} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b4dc73bf40314945ff376bd363916a738548d40a" aria-hidden="true" alt="{\displaystyle P}" width="1200" height="400">$ and $Q {\displaystyle Q} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8752c7023b4b3286800fe3238271bbca681219ed" aria-hidden="true" alt="{\displaystyle Q}" width="1200" height="400">$ with integer coefficients.

Famous examples of Lucas sequences include the Fibonacci numbers, Mersenne numbers, Pell numbers, Lucas numbers, Jacobsthal numbers, and a superset of Fermat numbers (see below). Lucas sequences are named after the French mathematician Édouard Lucas.

Recurrence relations

[edit]

Given two integer parameters $P {\displaystyle P} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b4dc73bf40314945ff376bd363916a738548d40a" aria-hidden="true" alt="{\displaystyle P}" width="1200" height="400">$ and $Q {\displaystyle Q} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8752c7023b4b3286800fe3238271bbca681219ed" aria-hidden="true" alt="{\displaystyle Q}" width="1200" height="400">$ , the Lucas sequences of the first kind $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ and of the second kind $V n ( P , Q ) {\displaystyle V_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40dfdb812d2de843618bac1786e8887494aba43e" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)}" width="3875.3" height="1223.9">$ are defined by the recurrence relations:

U 0 ( P , Q ) = 0 , U 1 ( P , Q ) = 1 , U n ( P , Q ) = P ⋅ U n − 1 ( P , Q ) − Q ⋅ U n − 2 ( P , Q )  for  n > 1 , {\displaystyle {\begin{aligned}U_{0}(P,Q)&=0,\\U_{1}(P,Q)&=1,\\U_{n}(P,Q)&=P\cdot U_{n-1}(P,Q)-Q\cdot U_{n-2}(P,Q){\mbox{ for }}n>1,\end{aligned}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/39b8067dbb65782e6f6908f6f128b1e6c86185ba" aria-hidden="true" alt="{\displaystyle {\begin{aligned}U_{0}(P,Q)&amp;=0,\\U_{1}(P,Q)&amp;=1,\\U_{n}(P,Q)&amp;=P\cdot U_{n-1}(P,Q)-Q\cdot U_{n-2}(P,Q){\mbox{ for }}n&gt;1,\end{aligned}}}" width="1200" height="400">

and

V 0 ( P , Q ) = 2 , V 1 ( P , Q ) = P , V n ( P , Q ) = P ⋅ V n − 1 ( P , Q ) − Q ⋅ V n − 2 ( P , Q )  for  n > 1. {\displaystyle {\begin{aligned}V_{0}(P,Q)&=2,\\V_{1}(P,Q)&=P,\\V_{n}(P,Q)&=P\cdot V_{n-1}(P,Q)-Q\cdot V_{n-2}(P,Q){\mbox{ for }}n>1.\end{aligned}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/4558d7e27cf5c77c030303baf083e6b91542fc64" aria-hidden="true" alt="{\displaystyle {\begin{aligned}V_{0}(P,Q)&amp;=2,\\V_{1}(P,Q)&amp;=P,\\V_{n}(P,Q)&amp;=P\cdot V_{n-1}(P,Q)-Q\cdot V_{n-2}(P,Q){\mbox{ for }}n&gt;1.\end{aligned}}}" width="1200" height="400">

It is not hard to show that for $n > 0 {\displaystyle n>0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/27a6a5d982d54202a14f111cb8a49210501b2c96" aria-hidden="true" alt="{\displaystyle n>0}" width="2435.1" height="936.9">$ ,

U n ( P , Q ) = P ⋅ U n − 1 ( P , Q ) + V n − 1 ( P , Q ) 2 , V n ( P , Q ) = ( P 2 − 4 Q ) ⋅ U n − 1 ( P , Q ) + P ⋅ V n − 1 ( P , Q ) 2 . {\displaystyle {\begin{aligned}U_{n}(P,Q)&={\frac {P\cdot U_{n-1}(P,Q)+V_{n-1}(P,Q)}{2}},\\V_{n}(P,Q)&={\frac {(P^{2}-4Q)\cdot U_{n-1}(P,Q)+P\cdot V_{n-1}(P,Q)}{2}}.\end{aligned}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/38de939c7fd884bcdebc764e6dead95552a4669e" aria-hidden="true" alt="{\displaystyle {\begin{aligned}U_{n}(P,Q)&amp;={\frac {P\cdot U_{n-1}(P,Q)+V_{n-1}(P,Q)}{2}},\\V_{n}(P,Q)&amp;={\frac {(P^{2}-4Q)\cdot U_{n-1}(P,Q)+P\cdot V_{n-1}(P,Q)}{2}}.\end{aligned}}}" width="1200" height="400">

The above relations can be stated in matrix form as follows:

[ U n ( P , Q ) U n + 1 ( P , Q ) ] = [ 0 1 − Q P ] ⋅ [ U n − 1 ( P , Q ) U n ( P , Q ) ] , {\displaystyle {\begin{bmatrix}U_{n}(P,Q)\\U_{n+1}(P,Q)\end{bmatrix}}={\begin{bmatrix}0&1\\-Q&P\end{bmatrix}}\cdot {\begin{bmatrix}U_{n-1}(P,Q)\\U_{n}(P,Q)\end{bmatrix}},} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ad9d8779bd84dbd4fee57e86e215008a07702cb9" aria-hidden="true" alt="{\displaystyle {\begin{bmatrix}U_{n}(P,Q)\\U_{n+1}(P,Q)\end{bmatrix}}={\begin{bmatrix}0&amp;1\\-Q&amp;P\end{bmatrix}}\cdot {\begin{bmatrix}U_{n-1}(P,Q)\\U_{n}(P,Q)\end{bmatrix}},}" width="1200" height="400">

[ V n ( P , Q ) V n + 1 ( P , Q ) ] = [ 0 1 − Q P ] ⋅ [ V n − 1 ( P , Q ) V n ( P , Q ) ] , {\displaystyle {\begin{bmatrix}V_{n}(P,Q)\\V_{n+1}(P,Q)\end{bmatrix}}={\begin{bmatrix}0&1\\-Q&P\end{bmatrix}}\cdot {\begin{bmatrix}V_{n-1}(P,Q)\\V_{n}(P,Q)\end{bmatrix}},} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/4a4a3686290aa145d3736a67a5c703259f349e03" aria-hidden="true" alt="{\displaystyle {\begin{bmatrix}V_{n}(P,Q)\\V_{n+1}(P,Q)\end{bmatrix}}={\begin{bmatrix}0&amp;1\\-Q&amp;P\end{bmatrix}}\cdot {\begin{bmatrix}V_{n-1}(P,Q)\\V_{n}(P,Q)\end{bmatrix}},}" width="1200" height="400">

[ U n ( P , Q ) V n ( P , Q ) ] = [ P / 2 1 / 2 ( P 2 − 4 Q ) / 2 P / 2 ] ⋅ [ U n − 1 ( P , Q ) V n − 1 ( P , Q ) ] . {\displaystyle {\begin{bmatrix}U_{n}(P,Q)\\V_{n}(P,Q)\end{bmatrix}}={\begin{bmatrix}P/2&1/2\\(P^{2}-4Q)/2&P/2\end{bmatrix}}\cdot {\begin{bmatrix}U_{n-1}(P,Q)\\V_{n-1}(P,Q)\end{bmatrix}}.} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/47dcf73b6e25bfe75d0a072ceaced310f986274a" aria-hidden="true" alt="{\displaystyle {\begin{bmatrix}U_{n}(P,Q)\\V_{n}(P,Q)\end{bmatrix}}={\begin{bmatrix}P/2&amp;1/2\\(P^{2}-4Q)/2&amp;P/2\end{bmatrix}}\cdot {\begin{bmatrix}U_{n-1}(P,Q)\\V_{n-1}(P,Q)\end{bmatrix}}.}" width="1200" height="400">

Initial terms of Lucas sequences $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ and $V n ( P , Q ) {\displaystyle V_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40dfdb812d2de843618bac1786e8887494aba43e" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)}" width="3875.3" height="1223.9">$ are given in the table:

n U n ( P , Q ) V n ( P , Q ) 0 0 2 1 1 P 2 P P 2 − 2 Q 3 P 2 − Q P 3 − 3 P Q 4 P 3 − 2 P Q P 4 − 4 P 2 Q + 2 Q 2 5 P 4 − 3 P 2 Q + Q 2 P 5 − 5 P 3 Q + 5 P Q 2 6 P 5 − 4 P 3 Q + 3 P Q 2 P 6 − 6 P 4 Q + 9 P 2 Q 2 − 2 Q 3 {\displaystyle {\begin{array}{r|l|l}n&U_{n}(P,Q)&V_{n}(P,Q)\\\hline 0&0&2\\1&1&P\\2&P&{P}^{2}-2Q\\3&{P}^{2}-Q&{P}^{3}-3PQ\\4&{P}^{3}-2PQ&{P}^{4}-4{P}^{2}Q+2{Q}^{2}\\5&{P}^{4}-3{P}^{2}Q+{Q}^{2}&{P}^{5}-5{P}^{3}Q+5P{Q}^{2}\\6&{P}^{5}-4{P}^{3}Q+3P{Q}^{2}&{P}^{6}-6{P}^{4}Q+9{P}^{2}{Q}^{2}-2{Q}^{3}\end{array}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b2c8407d03f454e8b641aff876dddb6005d62dfe" aria-hidden="true" alt="{\displaystyle {\begin{array}{r|l|l}n&amp;U_{n}(P,Q)&amp;V_{n}(P,Q)\\\hline 0&amp;0&amp;2\\1&amp;1&amp;P\\2&amp;P&{P}^{2}-2Q\\3&{P}^{2}-Q&{P}^{3}-3PQ\\4&{P}^{3}-2PQ&{P}^{4}-4{P}^{2}Q+2{Q}^{2}\\5&{P}^{4}-3{P}^{2}Q+{Q}^{2}&{P}^{5}-5{P}^{3}Q+5P{Q}^{2}\\6&{P}^{5}-4{P}^{3}Q+3P{Q}^{2}&{P}^{6}-6{P}^{4}Q+9{P}^{2}{Q}^{2}-2{Q}^{3}\end{array}}}" width="1200" height="400">

Explicit expressions

[edit]

The characteristic equation of the recurrence relation for Lucas sequences $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ and $V n ( P , Q ) {\displaystyle V_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40dfdb812d2de843618bac1786e8887494aba43e" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)}" width="3875.3" height="1223.9">$ is:

x 2 − P x + Q = 0 {\displaystyle x^{2}-Px+Q=0\,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ee4c9d05c963f53aa52587cfbcc0a45fcc1c9650" aria-hidden="true" alt="{\displaystyle x^{2}-Px+Q=0\,}" width="1200" height="400">

It has the discriminant $D = P 2 − 4 Q {\displaystyle D=P^{2}-4Q} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ab28c050150ad67de72802cc6e9decdac0aec5bf" aria-hidden="true" alt="{\displaystyle D=P^{2}-4Q}" width="5915.7" height="1295.7">$ and the roots:

a = P + D 2 and b = P − D 2 . {\displaystyle a={\frac {P+{\sqrt {D}}}{2}}\quad {\text{and}}\quad b={\frac {P-{\sqrt {D}}}{2}}.\,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/af15de5e67e8eb3c2c381388ea7d92279551a97b" aria-hidden="true" alt="{\displaystyle a={\frac {P+{\sqrt {D}}}{2}}\quad {\text{and}}\quad b={\frac {P-{\sqrt {D}}}{2}}.\,}" width="15678.7" height="2515.6">

Thus:

a + b = P , {\displaystyle a+b=P\,,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5a38b49c0426736742e66f805b0648951607ce99" aria-hidden="true" alt="{\displaystyle a+b=P\,,}" width="4712.7" height="1080.4">

a b = 1 4 ( P 2 − D ) = Q , {\displaystyle ab={\frac {1}{4}}(P^{2}-D)=Q\,,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/2c254ed8edcadda83fce6dfe2d8882bc3b9ea761" aria-hidden="true" alt="{\displaystyle ab={\frac {1}{4}}(P^{2}-D)=Q\,,}" width="9792.9" height="2228.5">

a − b = D . {\displaystyle a-b={\sqrt {D}}\,.} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/1ffe2cbe60ffa734128337f4ac46950a466fe6d7" aria-hidden="true" alt="{\displaystyle a-b={\sqrt {D}}\,.}" width="5623.2" height="1295.7">

Note that the sequence $a n {\displaystyle a^{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/2687184a7698e75db65a25bea7afd207bff3d03b" aria-hidden="true" alt="{\displaystyle a^{n}}" width="1054.1" height="1008.6">$ and the sequence $b n {\displaystyle b^{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b8f8f52cd26bb201e02c8d1b3619a3a682f44dbc" aria-hidden="true" alt="{\displaystyle b^{n}}" width="954.1" height="1008.6">$ also satisfy the recurrence relation. However these might not be integer sequences.

When $D ≠ 0 {\displaystyle D\neq 0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9f87f6bad2ab96ddd45ea3745aa6efa0c54bd2a1" aria-hidden="true" alt="{\displaystyle D\neq 0}" width="2663.1" height="1152.1">$ , a and b are distinct and one quickly verifies that

a n = V n + U n D 2 {\displaystyle a^{n}={\frac {V_{n}+U_{n}{\sqrt {D}}}{2}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e54d235001ff706176add341e193f3d753c1c6f3" aria-hidden="true" alt="{\displaystyle a^{n}={\frac {V_{n}+U_{n}{\sqrt {D}}}{2}}}" width="7949.4" height="2515.6">

b n = V n − U n D 2 . {\displaystyle b^{n}={\frac {V_{n}-U_{n}{\sqrt {D}}}{2}}.} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/786814d0d99fb98f6510a773949879d5c5f2e205" aria-hidden="true" alt="{\displaystyle b^{n}={\frac {V_{n}-U_{n}{\sqrt {D}}}{2}}.}" width="8127.9" height="2515.6">

It follows that the terms of Lucas sequences can be expressed in terms of a and b as follows

U n = a n − b n a − b = a n − b n D {\displaystyle U_{n}={\frac {a^{n}-b^{n}}{a-b}}={\frac {a^{n}-b^{n}}{\sqrt {D}}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/52df8cdb047ec38c4dd573e6598176f2f636b1b7" aria-hidden="true" alt="{\displaystyle U_{n}={\frac {a^{n}-b^{n}}{a-b}}={\frac {a^{n}-b^{n}}{\sqrt {D}}}}" width="11058.6" height="2730.8">

V n = a n + b n {\displaystyle V_{n}=a^{n}+b^{n}\,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/10a06a80b8357f9eb3986899dbfc3dc2bd48aed4" aria-hidden="true" alt="{\displaystyle V_{n}=a^{n}+b^{n}\,}" width="5840" height="1152.1">

The case $D = 0 {\displaystyle D=0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d375dfda80ee8df1d1d7aa8b962114044e464305" aria-hidden="true" alt="{\displaystyle D=0}" width="2663.1" height="936.9">$ occurs exactly when $P = 2 S and Q = S 2 {\displaystyle P=2S{\text{ and }}Q=S^{2}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a310acdc37e1185968175c2853b913cc01cc7d7d" aria-hidden="true" alt="{\displaystyle P=2S{\text{ and }}Q=S^{2}}" width="8579.7" height="1295.7">$ for some integer S so that $a = b = S {\displaystyle a=b=S} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a1ef845ca3b63d3e7ca3dfa0e2284f9f15f3d3b8" aria-hidden="true" alt="{\displaystyle a=b=S}" width="4272.6" height="936.9">$ . In this case one easily finds that

U n ( P , Q ) = U n ( 2 S , S 2 ) = n S n − 1 {\displaystyle U_{n}(P,Q)=U_{n}(2S,S^{2})=nS^{n-1}\,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e6b959f60c8c7c382e9520ad766f257ffcc75a73" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)=U_{n}(2S,S^{2})=nS^{n-1}\,}" width="14182.1" height="1367.4">

V n ( P , Q ) = V n ( 2 S , S 2 ) = 2 S n . {\displaystyle V_{n}(P,Q)=V_{n}(2S,S^{2})=2S^{n}.\,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/dc40c1569fd7cd5d1c6c7fcde6f8035b25c75e03" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)=V_{n}(2S,S^{2})=2S^{n}.\,}" width="13256.2" height="1367.4">

Generating functions

[edit]

The ordinary generating functions are

∑ n ≥ 0 U n ( P , Q ) z n = z 1 − P z + Q z 2 ; {\displaystyle \sum _{n\geq 0}U_{n}(P,Q)z^{n}={\frac {z}{1-Pz+Qz^{2}}};} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c5225d218e300f47926bb01dd2cc16852664b27f" aria-hidden="true" alt="{\displaystyle \sum _{n\geq 0}U_{n}(P,Q)z^{n}={\frac {z}{1-Pz+Qz^{2}}};}" width="14434.3" height="2587.3">

∑ n ≥ 0 V n ( P , Q ) z n = 2 − P z 1 − P z + Q z 2 . {\displaystyle \sum _{n\geq 0}V_{n}(P,Q)z^{n}={\frac {2-Pz}{1-Pz+Qz^{2}}}.} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/68704966d35020994e6e530da6c091039bb2f578" aria-hidden="true" alt="{\displaystyle \sum _{n\geq 0}V_{n}(P,Q)z^{n}={\frac {2-Pz}{1-Pz+Qz^{2}}}.}" width="14334.3" height="2802.6">

When $Q = ± 1 {\displaystyle Q=\pm 1} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/866ac3939d3b70aa2d39a96374c342e036eab9f1" aria-hidden="true" alt="{\displaystyle Q=\pm 1}" width="3404.6" height="1080.4">$ , the Lucas sequences $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ and $V n ( P , Q ) {\displaystyle V_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40dfdb812d2de843618bac1786e8887494aba43e" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)}" width="3875.3" height="1223.9">$ satisfy certain Pell equations:

V n ( P , 1 ) 2 − D ⋅ U n ( P , 1 ) 2 = 4 , {\displaystyle V_{n}(P,1)^{2}-D\cdot U_{n}(P,1)^{2}=4,} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/0c8574371b65e83d93cc3dfa8047a0b1e4146c1e" aria-hidden="true" alt="{\displaystyle V_{n}(P,1)^{2}-D\cdot U_{n}(P,1)^{2}=4,}" width="13063.8" height="1367.4">

V n ( P , − 1 ) 2 − D ⋅ U n ( P , − 1 ) 2 = 4 ( − 1 ) n . {\displaystyle V_{n}(P,-1)^{2}-D\cdot U_{n}(P,-1)^{2}=4(-1)^{n}.} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/32e09a95293d2be6adef57064ea644eebdd3cc80" aria-hidden="true" alt="{\displaystyle V_{n}(P,-1)^{2}-D\cdot U_{n}(P,-1)^{2}=4(-1)^{n}.}" width="17203.4" height="1367.4">

Relations between sequences with different parameters

[edit]

For any number c, the sequences $U n ( P ′ , Q ′ ) {\displaystyle U_{n}(P’,Q’)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/17ad14bfa4cf5d0d7ab8b8bc9293ee11a4906bb2" aria-hidden="true" alt="{\displaystyle U_{n}(P',Q')}" width="4597.7" height="1295.7">$ and $V n ( P ′ , Q ′ ) {\displaystyle V_{n}(P’,Q’)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/fd4e2dc9db74d9dead124a0887ead1268ef4332f" aria-hidden="true" alt="{\displaystyle V_{n}(P',Q')}" width="4497.7" height="1295.7">$ with

P ′ = P + 2 c {\displaystyle P’=P+2c} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e12d0d3999ee128167ef8e26c9cdbff52a54eedc" aria-hidden="true" alt="{\displaystyle P'=P+2c}" width="5321.6" height="1152.1">

Q ′ = c P + Q + c 2 {\displaystyle Q’=cP+Q+c^{2}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9d74d94ef57a806519394b776d56e04b87bf57a3" aria-hidden="true" alt="{\displaystyle Q'=cP+Q+c^{2}}" width="7730.2" height="1295.7">

have the same discriminant as

U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">

and

V n ( P , Q ) {\displaystyle V_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40dfdb812d2de843618bac1786e8887494aba43e" aria-hidden="true" alt="{\displaystyle V_{n}(P,Q)}" width="3875.3" height="1223.9">

P ′ 2 − 4 Q ′ = ( P + 2 c ) 2 − 4 ( c P + Q + c 2 ) = P 2 − 4 Q = D . {\displaystyle P’^{2}-4Q’=(P+2c)^{2}-4(cP+Q+c^{2})=P^{2}-4Q=D.} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e0628456a82fd6efdf772d38fd9f900979cafd3a" aria-hidden="true" alt="{\displaystyle P'^{2}-4Q'=(P+2c)^{2}-4(cP+Q+c^{2})=P^{2}-4Q=D.}" width="25058.6" height="1367.4">

For any number c, we also have

U n ( c P , c 2 Q ) = c n − 1 ⋅ U n ( P , Q ) , {\displaystyle U_{n}(cP,c^{2}Q)=c^{n-1}\cdot U_{n}(P,Q),} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/86d151659ceb328bdf2815915a9c9dc49d59fa89" aria-hidden="true" alt="{\displaystyle U_{n}(cP,c^{2}Q)=c^{n-1}\cdot U_{n}(P,Q),}" width="13469.5" height="1367.4">

V n ( c P , c 2 Q ) = c n ⋅ V n ( P , Q ) . {\displaystyle V_{n}(cP,c^{2}Q)=c^{n}\cdot V_{n}(P,Q).} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/7c80045c50964cb133feb03d3b9bea2137769b56" aria-hidden="true" alt="{\displaystyle V_{n}(cP,c^{2}Q)=c^{n}\cdot V_{n}(P,Q).}" width="12365.1" height="1367.4">

The terms of Lucas sequences satisfy relations that are generalizations of those between Fibonacci numbers $F n = U n ( 1 , − 1 ) {\displaystyle F_{n}=U_{n}(1,-1)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f62dd65d54d501fe5a6237b88785e3da12e9975d" aria-hidden="true" alt="{\displaystyle F_{n}=U_{n}(1,-1)}" width="6714" height="1223.9">$ and Lucas numbers $L n = V n ( 1 , − 1 ) {\displaystyle L_{n}=V_{n}(1,-1)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/66bf0247e739da63c1115fd9cca17a16ffca54e7" aria-hidden="true" alt="{\displaystyle L_{n}=V_{n}(1,-1)}" width="6652" height="1223.9">$ . For example:

General case ( P , Q ) = ( 1 , − 1 ) , D = P 2 − 4 Q = 5 D U n = V n + 1 − Q V n − 1 = 2 V n + 1 − P V n 5 F n = L n + 1 + L n − 1 = 2 L n + 1 − L n ( 1 ) V n = U n + 1 − Q U n − 1 = 2 U n + 1 − P U n L n = F n + 1 + F n − 1 = 2 F n + 1 − F n ( 2 ) U m + n = U n U m + 1 − Q U m U n − 1 = U m V n − Q n U m − n F m + n = F n F m + 1 + F m F n − 1 = F m L n − ( − 1 ) n F m − n ( 3 ) U 2 n = U n ( U n + 1 − Q U n − 1 ) = U n V n F 2 n = F n ( F n + 1 + F n − 1 ) = F n L n ( 4 ) U 2 n + 1 = U n + 1 2 − Q U n 2 F 2 n + 1 = F n + 1 2 + F n 2 ( 5 ) V m + n = V m V n − Q n V m − n = D U m U n + Q n V m − n L m + n = L m L n − ( − 1 ) n L m − n = 5 F m F n + ( − 1 ) n L m − n ( 6 ) V 2 n = V n 2 − 2 Q n = D U n 2 + 2 Q n L 2 n = L n 2 − 2 ( − 1 ) n = 5 F n 2 + 2 ( − 1 ) n ( 7 ) U m + n = U m V n + U n V m 2 F m + n = F m L n + F n L m 2 ( 8 ) V m + n = V m V n + D U m U n 2 L m + n = L m L n + 5 F m F n 2 ( 9 ) V n 2 − D U n 2 = 4 Q n L n 2 − 5 F n 2 = 4 ( − 1 ) n ( 10 ) U n 2 − U n − 1 U n + 1 = Q n − 1 F n 2 − F n − 1 F n + 1 = ( − 1 ) n − 1 ( 11 ) V n 2 − V n − 1 V n + 1 = D Q n − 1 L n 2 − L n − 1 L n + 1 = 5 ( − 1 ) n − 1 ( 12 ) 2 n − 1 U n = ( n 1 ) P n − 1 + ( n 3 ) P n − 3 D + ⋯ 2 n − 1 F n = ( n 1 ) + 5 ( n 3 ) + ⋯ ( 13 ) 2 n − 1 V n = P n + ( n 2 ) P n − 2 D + ( n 4 ) P n − 4 D 2 + ⋯ 2 n − 1 L n = 1 + 5 ( n 2 ) + 5 2 ( n 4 ) + ⋯ ( 14 ) {\displaystyle {\begin{array}{l|l|r}{\text{General case}}&(P,Q)=(1,-1),D=P^{2}-4Q=5\\\hline DU_{n}={V_{n+1}-QV_{n-1}}=2V_{n+1}-PV_{n}&5F_{n}={L_{n+1}+L_{n-1}}=2L_{n+1}-L_{n}&(1)\\V_{n}=U_{n+1}-QU_{n-1}=2U_{n+1}-PU_{n}&L_{n}=F_{n+1}+F_{n-1}=2F_{n+1}-F_{n}&(2)\\U_{m+n}=U_{n}U_{m+1}-QU_{m}U_{n-1}=U_{m}V_{n}-Q^{n}U_{m-n}&F_{m+n}=F_{n}F_{m+1}+F_{m}F_{n-1}=F_{m}L_{n}-(-1)^{n}F_{m-n}&(3)\\U_{2n}=U_{n}(U_{n+1}-QU_{n-1})=U_{n}V_{n}&F_{2n}=F_{n}(F_{n+1}+F_{n-1})=F_{n}L_{n}&(4)\\U_{2n+1}=U_{n+1}^{2}-QU_{n}^{2}&F_{2n+1}=F_{n+1}^{2}+F_{n}^{2}&(5)\\V_{m+n}=V_{m}V_{n}-Q^{n}V_{m-n}=DU_{m}U_{n}+Q^{n}V_{m-n}&L_{m+n}=L_{m}L_{n}-(-1)^{n}L_{m-n}=5F_{m}F_{n}+(-1)^{n}L_{m-n}&(6)\\V_{2n}=V_{n}^{2}-2Q^{n}=DU_{n}^{2}+2Q^{n}&L_{2n}=L_{n}^{2}-2(-1)^{n}=5F_{n}^{2}+2(-1)^{n}&(7)\\U_{m+n}={\frac {U_{m}V_{n}+U_{n}V_{m}}{2}}&F_{m+n}={\frac {F_{m}L_{n}+F_{n}L_{m}}{2}}&(8)\\V_{m+n}={\frac {V_{m}V_{n}+DU_{m}U_{n}}{2}}&L_{m+n}={\frac {L_{m}L_{n}+5F_{m}F_{n}}{2}}&(9)\\V_{n}^{2}-DU_{n}^{2}=4Q^{n}&L_{n}^{2}-5F_{n}^{2}=4(-1)^{n}&(10)\\U_{n}^{2}-U_{n-1}U_{n+1}=Q^{n-1}&F_{n}^{2}-F_{n-1}F_{n+1}=(-1)^{n-1}&(11)\\V_{n}^{2}-V_{n-1}V_{n+1}=DQ^{n-1}&L_{n}^{2}-L_{n-1}L_{n+1}=5(-1)^{n-1}&(12)\\2^{n-1}U_{n}={n \choose 1}P^{n-1}+{n \choose 3}P^{n-3}D+\cdots &2^{n-1}F_{n}={n \choose 1}+5{n \choose 3}+\cdots &(13)\\2^{n-1}V_{n}=P^{n}+{n \choose 2}P^{n-2}D+{n \choose 4}P^{n-4}D^{2}+\cdots &2^{n-1}L_{n}=1+5{n \choose 2}+5^{2}{n \choose 4}+\cdots &(14)\end{array}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/22463c3007f6af9b7beee3c01189a14bd0da7db3" aria-hidden="true" alt="{\displaystyle {\begin{array}{l|l|r}{\text{General case}}&amp;(P,Q)=(1,-1),D=P^{2}-4Q=5\\\hline DU_{n}={V_{n+1}-QV_{n-1}}=2V_{n+1}-PV_{n}&amp;5F_{n}={L_{n+1}+L_{n-1}}=2L_{n+1}-L_{n}&amp;(1)\\V_{n}=U_{n+1}-QU_{n-1}=2U_{n+1}-PU_{n}&amp;L_{n}=F_{n+1}+F_{n-1}=2F_{n+1}-F_{n}&amp;(2)\\U_{m+n}=U_{n}U_{m+1}-QU_{m}U_{n-1}=U_{m}V_{n}-Q^{n}U_{m-n}&amp;F_{m+n}=F_{n}F_{m+1}+F_{m}F_{n-1}=F_{m}L_{n}-(-1)^{n}F_{m-n}&amp;(3)\\U_{2n}=U_{n}(U_{n+1}-QU_{n-1})=U_{n}V_{n}&amp;F_{2n}=F_{n}(F_{n+1}+F_{n-1})=F_{n}L_{n}&amp;(4)\\U_{2n+1}=U_{n+1}^{2}-QU_{n}^{2}&amp;F_{2n+1}=F_{n+1}^{2}+F_{n}^{2}&amp;(5)\\V_{m+n}=V_{m}V_{n}-Q^{n}V_{m-n}=DU_{m}U_{n}+Q^{n}V_{m-n}&amp;L_{m+n}=L_{m}L_{n}-(-1)^{n}L_{m-n}=5F_{m}F_{n}+(-1)^{n}L_{m-n}&amp;(6)\\V_{2n}=V_{n}^{2}-2Q^{n}=DU_{n}^{2}+2Q^{n}&amp;L_{2n}=L_{n}^{2}-2(-1)^{n}=5F_{n}^{2}+2(-1)^{n}&amp;(7)\\U_{m+n}={\frac {U_{m}V_{n}+U_{n}V_{m}}{2}}&amp;F_{m+n}={\frac {F_{m}L_{n}+F_{n}L_{m}}{2}}&amp;(8)\\V_{m+n}={\frac {V_{m}V_{n}+DU_{m}U_{n}}{2}}&amp;L_{m+n}={\frac {L_{m}L_{n}+5F_{m}F_{n}}{2}}&amp;(9)\\V_{n}^{2}-DU_{n}^{2}=4Q^{n}&amp;L_{n}^{2}-5F_{n}^{2}=4(-1)^{n}&amp;(10)\\U_{n}^{2}-U_{n-1}U_{n+1}=Q^{n-1}&amp;F_{n}^{2}-F_{n-1}F_{n+1}=(-1)^{n-1}&amp;(11)\\V_{n}^{2}-V_{n-1}V_{n+1}=DQ^{n-1}&amp;L_{n}^{2}-L_{n-1}L_{n+1}=5(-1)^{n-1}&amp;(12)\\2^{n-1}U_{n}={n \choose 1}P^{n-1}+{n \choose 3}P^{n-3}D+\cdots &amp;2^{n-1}F_{n}={n \choose 1}+5{n \choose 3}+\cdots &amp;(13)\\2^{n-1}V_{n}=P^{n}+{n \choose 2}P^{n-2}D+{n \choose 4}P^{n-4}D^{2}+\cdots &amp;2^{n-1}L_{n}=1+5{n \choose 2}+5^{2}{n \choose 4}+\cdots &amp;(14)\end{array}}}" width="49486.7" height="23469.2">

Of these, (6) and (7) allow fast calculation of V independent of U in a way analogous to exponentiation by squaring. The relation $V m n = V m ( P = V n , Q = Q n ) {\displaystyle V_{mn}=V_{m}(P=V_{n},Q=Q_{n})} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/066dafbabd0763adfaec9d4e5c345202606d6b10" aria-hidden="true" alt="{\displaystyle V_{mn}=V_{m}(P=V_{n},Q=Q_{n})}" width="12227.6" height="1223.9">$ (which belongs to the section above, “relations between sequences with different parameters”) is also useful for this purpose.^[1]

Divisibility properties

[edit]

Among the consequences is that $U k m ( P , Q ) {\displaystyle U_{km}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/0bfdad63270c7fd35027fb4cbde13d284d7303e7" aria-hidden="true" alt="{\displaystyle U_{km}(P,Q)}" width="4540.6" height="1223.9">$ is a multiple of $U m ( P , Q ) {\displaystyle U_{m}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5f5b4bbbcadc7614ceb3264997bbd4b0fe801b45" aria-hidden="true" alt="{\displaystyle U_{m}(P,Q)}" width="4171.9" height="1223.9">$ , i.e., the sequence $( U m ( P , Q ) ) m ≥ 1 {\displaystyle (U_{m}(P,Q))_{m\geq 1}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/14881d24e248b91089df20f3b0e98de6f1191037" aria-hidden="true" alt="{\displaystyle (U_{m}(P,Q))_{m\geq 1}}" width="6576.4" height="1223.9">$
is a divisibility sequence. This implies, in particular, that $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ can be prime only when n is prime.
Another consequence is an analog of exponentiation by squaring that allows fast computation of $U n ( P , Q ) {\displaystyle U_{n}(P,Q)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c9acf94bbc2c8e51474c0b61e79469aa6049fcf0" aria-hidden="true" alt="{\displaystyle U_{n}(P,Q)}" width="3975.3" height="1223.9">$ for large values of n.
Moreover, if $gcd ( P , Q ) = 1 {\displaystyle \gcd(P,Q)=1} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/cdf3c258be81b06321fff3d7ebf9ced206a5f3be" aria-hidden="true" alt="{\displaystyle \gcd(P,Q)=1}" width="6103.2" height="1223.9">$ , then $( U m ( P , Q ) ) m ≥ 1 {\displaystyle (U_{m}(P,Q))_{m\geq 1}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/14881d24e248b91089df20f3b0e98de6f1191037" aria-hidden="true" alt="{\displaystyle (U_{m}(P,Q))_{m\geq 1}}" width="6576.4" height="1223.9">$ is a strong divisibility sequence.

Other divisibility properties are as follows:^[2]

If n is an odd multiple of m, then $V m {\displaystyle V_{m}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40836783b20c374654ab03ddd6b01586ad9b4fb8" aria-hidden="true" alt="{\displaystyle V_{m}}" width="1304.7" height="1080.4">$ divides $V n {\displaystyle V_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3ebc5a637019ce3415183f06995aeeca93547767" aria-hidden="true" alt="{\displaystyle V_{n}}" width="1108.1" height="1080.4">$ .
Let N be an integer relatively prime to 2Q. If the smallest positive integer r for which N divides $U r {\displaystyle U_{r}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3eaec4ad37894dbbbf185dad9af9dcb365ab8916" aria-hidden="true" alt="{\displaystyle U_{r}}" width="1102.8" height="1080.4">$ exists, then the set of n for which N divides $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ is exactly the set of multiples of r.
If P and Q are even, then $U n , V n {\displaystyle U_{n},V_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9e52adeac48c57c0f39c06696b1bb6c4863ef18c" aria-hidden="true" alt="{\displaystyle U_{n},V_{n}}" width="2761.4" height="1080.4">$ are always even except $U 1 {\displaystyle U_{1}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/bc9e7f892894bc50c32ce1b9f9a68a15562146ac" aria-hidden="true" alt="{\displaystyle U_{1}}" width="1137.4" height="1080.4">$ .
If P is odd and Q is even, then $U n , V n {\displaystyle U_{n},V_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9e52adeac48c57c0f39c06696b1bb6c4863ef18c" aria-hidden="true" alt="{\displaystyle U_{n},V_{n}}" width="2761.4" height="1080.4">$ are always odd for every $n > 0 {\displaystyle n>0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/27a6a5d982d54202a14f111cb8a49210501b2c96" aria-hidden="true" alt="{\displaystyle n>0}" width="2435.1" height="936.9">$ .
If P is even and Q is odd, then the parity of $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ is the same as n and $V n {\displaystyle V_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3ebc5a637019ce3415183f06995aeeca93547767" aria-hidden="true" alt="{\displaystyle V_{n}}" width="1108.1" height="1080.4">$ is always even.
If P and Q are odd, then $U n , V n {\displaystyle U_{n},V_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9e52adeac48c57c0f39c06696b1bb6c4863ef18c" aria-hidden="true" alt="{\displaystyle U_{n},V_{n}}" width="2761.4" height="1080.4">$ are even if and only if n is a multiple of 3.
If p is an odd prime, then $U p ≡ ( D p ) , V p ≡ P ( mod p ) {\displaystyle U_{p}\equiv \left({\tfrac {D}{p}}\right),V_{p}\equiv P{\pmod {p}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/0c1f76682e01759973b494461cde2790f3ac2009" aria-hidden="true" alt="{\displaystyle U_{p}\equiv \left({\tfrac {D}{p}}\right),V_{p}\equiv P{\pmod {p}}}" width="13024.3" height="2085">$ (see Legendre symbol).
If p is an odd prime which divides P and Q, then p divides $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ for every $n > 1 {\displaystyle n>1} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ee74e1cc07e7041edf0fcbd4481f5cd32ad17b64" aria-hidden="true" alt="{\displaystyle n>1}" width="2435.1" height="936.9">$ .
If p is an odd prime which divides P but not Q, then p divides $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ if and only if n is even.
If p is an odd prime which divides Q but not P, then p never divides $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ for any $n > 0 {\displaystyle n>0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/27a6a5d982d54202a14f111cb8a49210501b2c96" aria-hidden="true" alt="{\displaystyle n>0}" width="2435.1" height="936.9">$ .
If p is an odd prime which divides D but not PQ, then p divides $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ if and only if p divides n.
If p is an odd prime which does not divide PQD, then p divides $U l {\displaystyle U_{l}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f53e16b0b55bada3e179b785270f2d45f0c20ed9" aria-hidden="true" alt="{\displaystyle U_{l}}" width="994.6" height="1080.4">$ , where $l = p − ( D p ) {\displaystyle l=p-\left({\tfrac {D}{p}}\right)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/63b609dc031d12e6d3d1ff0ce58a5561d4796fd1" aria-hidden="true" alt="{\displaystyle l=p-\left({\tfrac {D}{p}}\right)}" width="5499.8" height="2085">$ .

The last fact generalizes Fermat’s little theorem. These facts are used in the Lucas–Lehmer primality test.
Like Fermat’s little theorem, the converse of the last fact holds often, but not always; there exist composite numbers n relatively prime to D and dividing $U l {\displaystyle U_{l}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f53e16b0b55bada3e179b785270f2d45f0c20ed9" aria-hidden="true" alt="{\displaystyle U_{l}}" width="994.6" height="1080.4">$ , where $l = n − ( D n ) {\displaystyle l=n-\left({\tfrac {D}{n}}\right)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3306aa1018e022445fc9dea7e008b429d0fea70e" aria-hidden="true" alt="{\displaystyle l=n-\left({\tfrac {D}{n}}\right)}" width="5596.8" height="2085">$ . Such composite numbers are called Lucas pseudoprimes.

A prime factor of a term in a Lucas sequence which does not divide any earlier term in the sequence is called primitive.
Carmichael’s theorem states that all but finitely many of the terms in a Lucas sequence have a primitive prime factor.^[3] Indeed, Carmichael (1913) showed that if D is positive and n is not 1, 2 or 6, then $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ has a primitive prime factor. In the case D is negative, a deep result of Bilu, Hanrot, Voutier and Mignotte^[4] shows that if n > 30, then $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ has a primitive prime factor and determines all cases $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ has no primitive prime factor.

The Lucas sequences for some values of P and Q have specific names:

U n (1, -1)

: Fibonacci numbers

V n (1, -1)

: Lucas numbers

U n (2, -1)

: Pell numbers

V n (2, -1)

: Pell–Lucas numbers (companion Pell numbers)

U n (1, -2)

: Jacobsthal numbers

V n (1, -2)

: Jacobsthal–Lucas numbers

U n (3, 2)

: Mersenne numbers 2ⁿ − 1

V n (3, 2)

: Numbers of the form 2ⁿ + 1, which include the Fermat numbers^[3]

U n (6, 1)

: The square roots of the square triangular numbers.

U n (x, -1)

: Fibonacci polynomials

V n (x, -1)

: Lucas polynomials

U n (2 x, 1)

: Chebyshev polynomials of second kind

V n (2 x, 1)

: Chebyshev polynomials of first kind multiplied by 2

U n (x +1, x)

: Repunits in base x

V n (x +1, x)

: xⁿ + 1

Some Lucas sequences have entries in the On-Line Encyclopedia of Integer Sequences:

Lucas sequences are used in probabilistic Lucas pseudoprime tests, which are part of the commonly used Baillie–PSW primality test.
Lucas sequences are used in some primality proof methods, including the Lucas–Lehmer and Lucas–Lehmer–Riesel tests and the hybrid N−1/N+1 methods such as those in Brillhart-Lehmer-Selfridge 1975.^[5]^[6]
LUC is a public-key cryptosystem based on Lucas sequences^[7] that implements the analogs of ElGamal (LUCELG), Diffie–Hellman (LUCDIF), and RSA (LUCRSA). The encryption of the message in LUC is computed as a term of certain Lucas sequence, instead of using modular exponentiation as in RSA or Diffie–Hellman. However, a paper by Bleichenbacher et al.^[8] shows that many of the supposed security advantages of LUC over cryptosystems based on modular exponentiation are either not present, or not as substantial as claimed.

SageMath implements $U n {\displaystyle U_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/67b641ce10cb5e865397c4efe647cbffed98a024" aria-hidden="true" alt="{\displaystyle U_{n}}" width="1208.1" height="1080.4">$ and $V n {\displaystyle V_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3ebc5a637019ce3415183f06995aeeca93547767" aria-hidden="true" alt="{\displaystyle V_{n}}" width="1108.1" height="1080.4">$ as functions lucas_number1() and lucas_number2(), respectively.^[9]

Carmichael, R. D. (1913), “On the numerical factors of the arithmetic forms αⁿ±βⁿ“, Annals of Mathematics, 15 (1/4): 30–70, doi:10.2307/1967797, JSTOR 1967797
Lehmer, D. H. (1930). “An extended theory of Lucas’ functions”. Annals of Mathematics. 31 (3): 419–448. Bibcode:1930AnMat..31..419L. doi:10.2307/1968235. JSTOR 1968235.
Ward, Morgan (1954). “Prime divisors of second order recurring sequences”. Duke Math. J. 21 (4): 607–614. doi:10.1215/S0012-7094-54-02163-8. hdl:10338.dmlcz/137477. MR 0064073.
Somer, Lawrence (1980). “The divisibility properties of primary Lucas Recurrences with respect to primes” (PDF). Fibonacci Quarterly. 18 (4): 316–334. doi:10.1080/00150517.1980.12430140.
Lagarias, J. C. (1985). “The set of primes dividing Lucas Numbers has density 2/3”. Pac. J. Math. 118 (2): 449–461. CiteSeerX 10.1.1.174.660. doi:10.2140/pjm.1985.118.449. MR 0789184.
Hans Riesel (1994). Prime Numbers and Computer Methods for Factorization. Progress in Mathematics. Vol. 126 (2nd ed.). Birkhäuser. pp. 107–121. ISBN 0-8176-3743-5.
Ribenboim, Paulo; McDaniel, Wayne L. (1996). “The square terms in Lucas Sequences”. J. Number Theory. 58 (1): 104–123. doi:10.1006/jnth.1996.0068.
Joye, M.; Quisquater, J.-J. (1996). “Efficient computation of full Lucas sequences” (PDF). Electronics Letters. 32 (6): 537–538. Bibcode:1996ElL….32..537J. doi:10.1049/el:19960359. Archived from the original (PDF) on 2015-02-02.
Ribenboim, Paulo (1996). The New Book of Prime Number Records (eBook ed.). Springer-Verlag, New York. doi:10.1007/978-1-4612-0759-7. ISBN 978-1-4612-0759-7.
Ribenboim, Paulo (2000). My Numbers, My Friends: Popular Lectures on Number Theory. New York: Springer-Verlag. pp. 1–50. ISBN 0-387-98911-0.
Luca, Florian (2000). “Perfect Fibonacci and Lucas numbers”. Rend. Circ Matem. Palermo. 49 (2): 313–318. doi:10.1007/BF02904236. S2CID 121789033.
Yabuta, M. (2001). “A simple proof of Carmichael’s theorem on primitive divisors” (PDF). Fibonacci Quarterly. 39 (5): 439–443. doi:10.1080/00150517.2001.12428701.
Benjamin, Arthur T.; Quinn, Jennifer J. (2003). Proofs that Really Count: The Art of Combinatorial Proof. Dolciani Mathematical Expositions. Vol. 27. Mathematical Association of America. p. 35. ISBN 978-0-88385-333-7.
Lucas sequence at Encyclopedia of Mathematics.
Weisstein, Eric W. “Lucas Sequence”. MathWorld.
Wei Dai. “Lucas Sequences in Cryptography”.

Recurrence relations

Explicit expressions

Generating functions

Relations between sequences with different parameters

Divisibility properties

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