2021-02-17, 01:42 | #1 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×3×1,571 Posts |
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)) |
2021-02-17, 02:10 | #2 |
Romulan Interpreter
Jun 2011
Thailand
10010011100101_{2} Posts |
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 |
2021-02-24, 17:19 | #3 |
Feb 2017
Nowhere
4536_{10} 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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