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Is there a manipulation that can be performed on the natural number sequence (1,2,3,...) in order to give the Fibonacci sequence?

I know the recurrence relation starting from 1,1 and successively adding the two previous numbers of the sequence. However, while this gives the Fibonacci sequence, it is not a manipulation of the natural number sequence as a whole. My question is whether there is a certain operation that can be repeatedly applied to the natural numbers as a whole (ie the sequence of natural numbers) to give the Fibonacci numbers.

Raghib
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  • It's unclear what you mean when you say a "manipulation." Why is the traditional recurrence relation not a "manipulation?" – Franklin Pezzuti Dyer Oct 27 '18 at 21:36
  • @Frpzzd: It isn't applied to the natural numbers as a whole sequence. It is only applied to the two previous terms at a time, starting from 1,1. It doesn't start from 1,2,3,... (the natural numbers as a whole). – Raghib Oct 27 '18 at 21:37
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    Perhaps it would help if you provided an example of a sequence which was obtained via a manipulation of the natural numbers as a whole. – lulu Oct 27 '18 at 21:39
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    As you can see from the comments, it isn't clear what you mean. Please provide an example to illustrate. – lulu Oct 27 '18 at 21:40
  • There's the Binet Formula which gives the Fibonacci numbers, for instance. Is that an example of what you want? – lulu Oct 27 '18 at 21:41
  • @lulu: Sorry, I pressed enter accidentally. Here is my response: If we select the operation "average of adjacent terms" and apply it to the natural numbers, we get the sequence (1+2)/2, (2+3)/2, ... ie 1.5,2.5,3.5,... This is a very simple example of an operation applied to the natural numbers as a whole to give a new sequence. – Raghib Oct 27 '18 at 21:41
  • Ok...so then I'd have thought the Binet formula was good. It provides a simple, closed formula for $F_n$, given $n$. – lulu Oct 27 '18 at 21:42
  • I think perhaps that Raghib just wants an explicit function $f:\Bbb N\rightarrow\Bbb N$ that maps $n$ to the $n$th Fibonacci number. For instance, if it were the square numbers instead of the Fibonacci numbers, the function $f(n)=n^2$ would do. Raghib, can you confirm or deny this? – TonyK Oct 27 '18 at 21:43
  • @TonyK: Yes, but it has to be simple as well. It can't have "n" as an exponent as in the Binet formula, as that doesn't really count as a manipulation on the natural numbers. It is hard to explain as I don't actually know a lot of maths, but I hope that what I have said about n being in the exponent should clarify what I mean a little. – Raghib Oct 27 '18 at 21:46
  • @TonyK: By the way, I am also happy with a series of manipulations, even an infinite number of manipulations. Eg apply "simple" manipulation 1 to the natural numbers to get sequence 1. Then apply manipulation 2 to get sequence 2, etc. And finally, sequence N for some large number N starts to approximate the Fibonacci sequence. But each manipulation must be simple and not have n as an exponent. – Raghib Oct 27 '18 at 21:48
  • I am afraid your requirements are just too vague, Raghib. And if the Binet formula doesn't meet them, I think you are doomed to disappointment. – TonyK Oct 27 '18 at 22:16

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