The Truly Remarkable Lucas Sequence

[@myjonpol]

Although I visited this video to learn about lucas sequence for my programming lessons, it seems I'm being pulled towards loving mathematics. Why do you make things so easy and enjoyable! Love it!

[@chrisg3030]

You can "lucasate" other sequences too and create equally remarkable ratio constant power sequences. For example take Narayana's Cows going 1 1 1 2 3 4 6 13 19 28 41 60 88 129 . . . (OEIS A000930) where a(n)=a(n-3) + a(n-1) and the ratio constant is approx 1.4656, which I call Moo instead of Phi. Moo raised to the power of a Narayana index number gives you a number in the corresponding lucasated sequence 3 1 1 4 5 6 10 15 21 31 46 67 98 144 211 309. So for example Moo^13 = approx 143.9464. How do you lucasate a sequence? One way seems to be take the number of 1's that a member of this family of sequences (such as Fibonacci) begins with, reduce it by 1, and start with that number. So 1 1 becomes 2 1, and 111 becomes 3 1 1. There are other ways.

[@karstenmeyer1729]

The sequence of Fibonacci numbers is stronger bond with phi, as any sequence with the same rules but differen starting numbers, The proof you can find, if you use continued fraction.You can represent phi as an infinite continued fraction. But if you broke this infinate continued fraction at any point, you get the ratio of two consecutive fibonacci numbers!

[@peawormsworth]

lucas(n) = phi^n+(-1/phi)^n It gets closer and closer to an integer, because the remainder is plus then minus 1/phi to the power of the lucas number. ie: the 1/phi difference between phi powers and lucas numbers gets smaller and smaller by phi at each step.

[@chrisg3030]

Let's perform the Lucas magic on other sequences. Let's alter them so they retain the same ratio constant (like Phi or 1.6180339987... for Fibonacci and also Lucas) and the same recurrence relation (like a(n) = a(n-1) + a (n-2) for Fibonacci and also Lucas), but start with different positive integers so that each successive term gets much closer to the ratio constant raised to the power of the position of that term, as Eddie demonstrated for the Lucas sequence. One example is the Pell sequence (OEIS A000129) with a recurrence of a(n) = 2a(n-1) + a(n-2) and a ratio constant of (2 + sqrt8)/2 or 2.414213562... otherwise known as the "Silver Ratio". It conventionally goes 0 1 2 5 12 29 70 169 408 985 2378 5741 13860 and so on. But although 2378 is the 10th term, 2.414213562^10 = 6725.9999..., which is pretty far off it or any other of those numbers. How can we magic the sequence up then? Easy. Start 2 2 instead of 1 2 and continue 6 14 34 82 198 478 1154 2786 6726 ... Exercise: do the same for the next "metallic" sequence (OEIS A006190), which has the so-called bronze ratio, (3+sqrt13)/2, and recurrence a(n) = 3a(n-1) + a(n-2) .

[@mainrig4981]

What are the Lukas equivalent numbers for this fibonacci numbers: 23.6%, 38.2%, 61.8%. 78.6%, 50%

[@jamesbacon6825]

another fun way to get phi on excel is start with any positive number and then (x^-1)+1

[@underagedlolkid2kelo]

hi what school does mr woo teach?

[@priyanshukushwaha318]

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