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Can any natural number be expressed in an increasing series (whole numbers, greater than 2 elements) in which the next element is the sum of the previous two elements? If so, what is the proof and how does one find some of these elements (such as Binets formula for the original Fibonacci sequence)?

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    https://en.wikipedia.org/wiki/Lucas_number – 2017-01-06
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    @SimpleArt May you elaborate on the use of Lucas numbers to answer this question – 2017-01-06
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    Their definition is your Fibonacci-like sequence. – 2017-01-06
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    @SimpleArt Yes, but every natural number is not present in them – 2017-01-06
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    Depends how you want the question. Take the below answer for example. – 2017-01-06
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    @DanyilBee I feel you find the following subjects interesting that maybe are related to your question. 1)[Complete sequence](https://en.wikipedia.org/wiki/Complete_sequence). 2) [Fibonacci coding](https://en.wikipedia.org/wiki/Fibonacci_coding). 3) [Zeckendorf's theorem](https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem). – 2017-01-06

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The answer is trivially yes: Say you have the number $N$. Then take the sequence $$ A_0 = 1 \\ A_1 = N-1 \\ \forall k>1: A_k = A_{k-1}+A_{k-2} $$

Then $A_2 = N$.

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    Are you basically saying that any natural number $n$ greater than $1$ has predecessor $n - 1$ and that $(n - 1) + 1 = n$? – 2017-01-06
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    Why do you say $\forall k$? – 2017-01-06
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    Edited question. – 2017-01-06