 mersenneforum.org Why can't I find this formula in Wiki or Wolfram? (pet peeve)
 Register FAQ Search Today's Posts Mark Forums Read 2021-02-17, 01:42 #1 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·5·23·41 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
LaurV
Romulan Interpreter

Jun 2011
Thailand

5×1,889 Posts Quote:
 Originally Posted by Batalov $${F_{3n} \over F_n} = L_n^2-(-1)^n$$
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   2021-02-24, 17:19 #3 Dr Sardonicus   Feb 2017 Nowhere 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.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Hugo1177 Miscellaneous Math 0 2021-01-20 23:42 MathDoggy Miscellaneous Math 13 2019-03-03 17:11 mshelikoff Aliquot Sequences 1 2015-05-15 07:40 Mini-Geek Lounge 25 2009-05-23 11:03 dini Puzzles 0 2009-03-22 03:39

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