Lucas Sequence

Author

Eric W. Weisstein
February 17, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/NumberTheory/LucasSequence.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/LucasSequence.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Code

𝒰[n_,{P_,Q_}]:=Module[{k,D=(P^2)-4Q,α,β},
{α,β}=(P+{1,-1}(Sqrt[D]))/2;
((α^n)-(β^n))/(α-β)
]

𝒱[n_,{P_,Q_}]:=Module[{k,D=(P^2)-4Q,α,β},
{α,β}=(P+{1,-1}(Sqrt[D]))/2;
(α^n)+(β^n)
]

Enumeration

Table[(Subscript[U, n])==𝒰[n,{P,Q}],{n,0,10}]//Factor//TableForm//TraditionalForm

Table[(Subscript[U, n])==2^(1-n)*Sum[Binomial[n,2*k+1]*P^(n-2*k-1)*(P^2-4*Q)^k,{k,0,Floor[(n-1)/2]}],{n,0,10}]//Factor//TableForm//TraditionalForm

Table[(Subscript[V, n])==𝒱[n,{P,Q}],{n,0,10}]//Factor//TableForm//TraditionalForm

Table[(Subscript[V, n])==2^(1-n)*Sum[Binomial[n,2*k]*P^(n-2*k)*(P^2-4*Q)^k,{k,0,Floor[n/2]}],{n,0,10}]//Factor//TableForm//TraditionalForm

Closed Forms

2^(1-n)*Sum[Binomial[n,2*k+1]*P^(n-2*k-1)*(P^2-4*Q)^k,{k,0,Floor[(n-1)/2]}]

2^(1-n)*Sum[Binomial[n,2*k]*P^(n-2*k)*(P^2-4*Q)^k,{k,0,Floor[n/2]}]