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# Another form of the Lucas sequence

As before, let’s write the nth tem of the Lucas seqence as

g(n) = an + bn

Just like in the previous post, let's write

z = (n / 2) log (a / b)

so that

2 cos iz = 2 cos [ (i n / 2) log (a / b) ]

e (n/2) log (a/b) + e – (n/2) log (a/b)

= (a / b) n/2 + (a / b) n/2

= (an + bn) / (ab) n/2

Solving for

g(n) = an + bn

we find that

g(n) = 2 (ab) n/2 cos [ (in / 2) log (a / b) ]

which is very similar to the form of the Fibonacci sequence that we found earlier that involved the sine function:

f(n) = - [ 2 i (ab) n/2 / (ab) ] sin [ (in / 2) log (a / b) ]

If we have the recursion

g(n + 2) = g(n + 1) – g(n)

for example, with

g(0) = 2

and

g(1) = 1

we find that we get

g(2) = -1

g(3) = -2

g(4) = -1

g(5) = 1

g(6) = 2

where the sequence 2, 1, -1, -2, -1, 1 starts over again.

Because we can write this sequence as

g(n) = an + bn

where

a = (1 + i √3) / 2

and

b = (1 – i √3) / 2

we can use the fact that

g(n) = 2 (ab) n/2 cos [ (in / 2) log (a / b) ]

to find that

g(n) = 2 cos( n π / 3)

much like we did in one of the previous posts.

Because we have

f(n) = – [ 2 i (ab) n/2 / (a – b) ] sin [ (in / 2) log (a / b) ]

and

g(n) = 2 (ab) n/2 cos [ (in / 2) log (a / b) ]

it seems reasonable to look for a relationship that's much like the complex exponential, where

eiz = cos z + i sin z

perhaps something that contains lots of HTML entities for Greek letters like

ει n = g(n) + ι f(n)

It doesn't quite look like anything nice and clean is possible, although if we write

ι = ab

then it looks like we can write

g(n) + ι f(n) = 2 an

and

g(n) – ι f(n) = 2 bn

That may be as close as we can get.