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Old 2021-02-17, 01:42   #1
Batalov's Avatar
Mar 2008

2·5·23·41 Posts
Question Why can't I find this formula in Wiki or Wolfram? (pet peeve)

For Lucas numbers:
\({L(3n) \over L(n)} = L(2n)-(-1)^n\)
...or if n is even, simpler still: \({L(6m) \over L(2m)} = L(4m)-1 \)

For example: this simplifies L2250 (or primV(2250)) cofactor shorthand, or primV(122754)

Well, for sure this is a partial case of (14) in Wolfram (multiple-angle recurrence), but it is sort of elegant to get a separate trivial case line, for convenience.

And similarly
\({F_{3n} \over F_n} = L_n^2-(-1)^n\)
(this one is in Wolfram in disguise, (66))
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Old 2021-02-17, 02:10   #2
Romulan Interpreter
LaurV's Avatar
Jun 2011

5×1,889 Posts

Originally Posted by Batalov View Post
\({F_{3n} \over F_n} = L_n^2-(-1)^n\)
Uh, man, you should say that's Fibonacci!
For a beat, my heart went boom, I thought you just found a way to factor Fermat\(_{24}\)

Last fiddled with by LaurV on 2021-02-17 at 02:11
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Old 2021-02-24, 17:19   #3
Dr Sardonicus
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Feb 2017

26·71 Posts

The Lucas identity follows immediately from (x^3 + y^3)/(x + y) = x^2 - x*y + y^2.

The Fibonacci identity follows immediately from (x^3 - y^3)/(x - y) = x^2 + x*y + y^2.

Clearly the same identities hold for the generalizations of Fibonacci and Lucas numbers for any quadratic u^2 - k*u - 1, k <> 0 an integer.
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