Lucas Number

Author

Eric W. Weisstein
April 24, 2009

This notebook downloaded from http://mathworld.wolfram.com/notebooks/IntegerSequences/LucasNumber.nb.

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

©2009 Wolfram Research, Inc. except for portions noted otherwise

Notation

LucasL[n]//TraditionalForm

Enumeration

LucasL[Range[10]]//Expand

Powers of 10

Direct

Table[Digits[LucasL[10^n]//Expand],{n,0,4}]//Timing

lucasL[1]:=1
lucasL[n_]:=Module[{l1=1,l2=3,tmp},Do[tmp=l1;l1=l2;l2+=tmp,{n-2}];l2]

Table[Digits[lucasL[10^n]],{n,0,7}]//Timing

Table[Digits[lucasL[10^n]],{n,0,8}]//Timing

Limit

Limit[(1+((Log[((((1/2) (1-(Sqrt[5]))))^(10^n))+((((1/2) (1+(Sqrt[5]))))^(10^n))])/(Log[10])))/10^n,n->Infinity]

FullSimplify[%]

Congruences

(fm=Table[Mod[Table[LucasL[n],{n,15}],m],{m,2,8}])//ColumnForm

p=Period/@Table[Mod[Table[LucasL[n],{n,200}],k],{k,2,8}]

f=LucasL/@Range[10^5];//Timing

Table[Period[Mod[f,10^n]],{n,4}]//Timing

Table[Period[Mod[f,10^n]],{n,5}]//Timing

Primes

Mod[LucasL[#],#]&/@Prime[Range[10^4]]//Union//Timing

Complement[Select[Range[10^4],Mod[LucasL[#],#]==1&],Prime[Range[PrimePi[10^4]]]]

Solution

V5

RSolve[{L[n]==L[n-1]+L[n-2],L[1]==1,L[2]==3},L[n],n]

Lucas[n_]:=((((1/2) (1-(Sqrt[5]))))^n)+((((1/2) (1+(Sqrt[5]))))^n)

Table[Lucas[n],{n,10}]//Expand

V7

RSolve[{L[n]==L[n-1]+L[n-2],L[1]==1,L[2]==3},L[n],n]

Other Forms

LucasL[Range[10]]

((((1/2) (1-(Sqrt[5]))))^n)+((((1/2) ((Sqrt[5])+1)))^n)

FullSimplify[((((1/2) (1-(Sqrt[5]))))^n)+((((1/2) (1+(Sqrt[5]))))^n),n∈Integers&&n≥0]

FullSimplify[((((1/2) (1-(Sqrt[5]))))^n)+((((1/2) (1+(Sqrt[5]))))^n)-LucasL[n],n∈Integers&&n≥0]

Table[%,{n,0,10}]//FullSimplify

Round((ϕ^n))

For n ≥ 2

Table[Round[GoldenRatio^n],{n,2,10}]

FullSimplify[Round[GoldenRatio^n],n∈Integers&&n≥2]

⌊(1/2) (((Sqrt[5])+1) (Subscript[L, n-1])+1)⌋

For n ≥ 4

Table[LucasL[n+1]-Floor[(LucasL[n](1+Sqrt[5])+1)/2],{n,15}]//Expand

Table[Floor[(LucasL[n-1](1+Sqrt[5])+1)/2],{n,4,10}]

FullSimplify[Floor[(LucasL[n-1](1+Sqrt[5])+1)/2],n∈Integers&&n≥4]

Identities

(((-1))^n) (LucasL[n])==(LucasL[-n])

FullSimplify[LucasL[-n]-(((-1))^n)LucasL[n],n∈Integers]

FullSimplify[(((-1))^n)LucasL[n],n∈Integers]

(LucasL[n-1]) (LucasL[n+1])+5 (((-1))^n)==0

V6

FullSimplify[Lucas[n]^2-Lucas[n-1]Lucas[n+1]-5(-1)^n,n∈Integers]

V7

FullSimplify[LucasL[n-1]LucasL[n+1]+5(-1)^n,n∈Integers]

(1/2) (5 (Fibonacci[m]) (Fibonacci[n])+(LucasL[m]) (LucasL[n]))==0

FullSimplify[FunctionExpand[LucasL[m+n]-(5 Fibonacci[m]Fibonacci[n]+LucasL[m]LucasL[n])/2],(m|n)∈Integers]

V7

FullSimplify[(5 Fibonacci[m]Fibonacci[n]+LucasL[m]LucasL[n])/2,(m|n)∈Integers]

(LucasL[m-n])-(1/2) (((-1))^n) ((LucasL[m]) (LucasL[n])-5 (Fibonacci[m]) (Fibonacci[n]))==0

LucasL[m-n]-(-1)^n(LucasL[m]LucasL[n]-5Fibonacci[m]Fibonacci[n])/2

V6

FullSimplify[FunctionExpand[LucasL[m-n]-(-1)^n(LucasL[m]LucasL[n]-5Fibonacci[m]Fibonacci[n])/2],(m|n)∈Integers]

V7

FullSimplify[(-1)^n(LucasL[m]LucasL[n]-5Fibonacci[m]Fibonacci[n])/2,(m|n)∈Integers]

-5 ((((Fibonacci[n])))^2)+((((LucasL[n])))^2)-4 (((-1))^n)==0

FullSimplify[FunctionExpand[LucasL[n]^2-5Fibonacci[n]^2-4(-1)^n],n∈Integers]

V7

FullSimplify[5Fibonacci[n]^2+4(-1)^n,n∈Integers]

((((LucasL[n])))^2)-((LucasL[2 n])+2 (((-1))^n))==0

FullSimplify[FunctionExpand[LucasL[n]^2-(LucasL[2n]+2(-1)^n)],n∈Integers]

V7

FullSimplify[(LucasL[2n]+2(-1)^n),n∈Integers]

Sums

(∑_(k=1)^n)((Subscript[L, k])^2)==(Subscript[L, n]) (Subscript[L, n+1])-2

V6

FullSimplify[Sum[LucasL[k]^2,{k,n}],n∈Integers&&n>0]

Simplify[Subtract@@{Sum[LucasL[k]^2,{k,n}],LucasL[n]LucasL[n+1]-2}]

V7

Sum[LucasL[k]^2,{k,n}]

FullSimplify[%,n∈Integers&&n>0]

Table[Sum[LucasL[k]^2,{k,n}]-(LucasL[n]LucasL[n+1]-2),{n,20}]//Union

(Subscript[L, n k])=(∑_(i=0)^k)(({{k}, {i}})) (Subscript[L, i]) (Subscript[F, n]^i) (Subscript[F, n-1]^(k-i))

Table[Sum[Binomial[k,i]LucasL[i]Fibonacci[n]^i Fibonacci[n-1]^(k-i),{i,0,k}],{n,2,5},{k,0,5}]

Table[LucasL[n k],{n,2,5},{k,0,5}]

V6

id=Simplify[Sum[Binomial[k,i]LucasL[i]Fibonacci[n]^i Fibonacci[n-1]^(k-i),{i,0,k}],k>0]

V7

Sum[Binomial[k,i]LucasL[i]Fibonacci[n]^i Fibonacci[n-1]^(k-i),{i,0,k}]

FullSimplify[%,n>1&&n∈Integers]

(1/2) (∑_(i=0)^k)(((-1))^(i n)) (Binomial[k, i]) (LucasL[(k-2 i) n])==((((LucasL[n])))^k)

(1/2)Sum[Binomial[k,i](((-1))^(i n))LucasL[(k-2i)n],{i,0,k}]

FullSimplify[%,(k|n)∈Integers&&k≥0&&n≥0]

Subtract@@@Flatten[Table[{((LucasL[n])^k),(2^(-1+k n)) (((-1-(Sqrt[5])))^(-k n)) ((((-(1/4)))^(k n)) (((1+(Sqrt[5])))^(2 k n)) (((1+(((-4))^n) (((1+(Sqrt[5])))^(-2 n))))^k)+(((1+(4^(-n)) (((1+(Sqrt[5])))^(2 n)) (E^(3 I n π))))^k))},
{k,0,5},{n,0,5}],1]//FullSimplify//Union

(∑_(k=0)^n)(LucasL[k]) (LucasL[n-k])==(Fibonacci[n])+(n+2) (LucasL[n])

Sum[LucasL[k]LucasL[n-k],{k,0,n}]

FullSimplify[Subtract@@{%,(n+2)LucasL[n]+Fibonacci[n]},n∈Integers&&n≥0]

Table[(((((-2 (1+(Sqrt[5]))))^(-n)) ((4^n) (9+(Sqrt[5]))+(((-2 (3+(Sqrt[5]))))^n) (11+3 (Sqrt[5]))+(5+(Sqrt[5])) ((4^n)+(((-2 (3+(Sqrt[5]))))^n)) n))/(5+(Sqrt[5]))),{n,10}]//FullSimplify

Table[(n+2)LucasL[n]+Fibonacci[n],{n,10}]

(∑_(k=0)^n)(LucasL[k]) (Binomial[n, k])==(LucasL[2 n])

Sum[Binomial[n,k]LucasL[k],{k,0,n}]

FullSimplify[%,n∈Integers&&n≥0]

FullSimplify[%-LucasL[2n],n∈Integers&&n≥0]

FullSimplify[FunctionExpand[%],n∈Integers&&n≥0]

Table[((((1/2) (3-(Sqrt[5]))))^n)-((((1/2) (1+(Sqrt[5]))))^(2 n))-(4^n) (((1+(Sqrt[5])))^(-2 n))+((((1/2) (3+(Sqrt[5]))))^n),{n,0,10}]//FullSimplify

(∑_(k=0)^n)(2^k) (LucasL[k]) (Binomial[n, k])==(LucasL[3 n])

Sum[Binomial[n,k](2^k)LucasL[k],{k,0,n}]

FullSimplify[%,n∈Integers&&n>0]

FullSimplify[%-LucasL[3n],n∈Integers&&n>0]

Table[%,{n,0,10}]//FullSimplify