# Proofing that the Lucas numbers come closer to the Phi rounded numbers then the Fibonacci numbers.

Morning everyone,

Bit of background, I'm a mid level programmer with very limited mathematics skills. As part of an assessment for a new role I've been asked to complete a technical task which mirrors the title of this question.

The programming side is fine yet it is the mathematics side of things that have confused me. We have been given a sample output to replicate(see below).

Fibonacci: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610

Lucas: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843

Phi Rounded: 2, 2, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843

Now where I come unstuck, I understand how Fibonacci and Lucas work and the sequence begins but it is the rounded Phi sequence that confuses me. From 3 onwards it seems to generate the next number like the previous 2. It is the beginning of 2,2 that confuses me. Obviously Phi rounded is 2 so how is the second 2 generated in this sequence? Is there a rule that I am missing?

Any help would be greatly appreciated.

We have been told to assume Phi is 1.618.

• To proof something in English means to make something (like a boat) water-proof. To prove something is what you want. – Mariano Suárez-Álvarez Apr 29 '16 at 8:53
• Once one has a proof, it is not a bad idea to proof it just to be sure, of course! – Mariano Suárez-Álvarez Apr 29 '16 at 8:53
• How is the sequence named Phi Rounded generated? – gammatester Apr 29 '16 at 8:56
• "Closer" usually implies some kind of metric ("distance function", to put it simply). For infinite sequences, to my knowledge, there is no canonical ("natural", "classical") metric and there is a metric $d$ so that $d(F,\phi) < d(L,\phi)$. – Abstraction Apr 29 '16 at 8:57
• Seems that the Lucas Number are not only close to the phi-rounded numbers, but identical (with a single exception). Probably, it can be shown by induction that this is actually the case. – Peter Apr 29 '16 at 9:04