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Posted 13 Dec 2012

Fibonacci Without Loops or Recursion

, 17 Dec 2012
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A method for calculating a Fibonacci number without using loops or recursion.


While reading one of our Insider News posts which linked to Evan Miller's site,  he mentioned a mathematical means of producing a Fibonacci number without using loops or recursion.   I decided to post the C# version of it here, but in no way do I claim credit to creating this.   I thought it was interesting enough to share for those who might not read the Insider News articles.  

You can read more about this closed-form solution on wiki.

The Code

public static long Fibonacci(long n)
    return (long)Math.Round(0.44721359549995682d * Math.Pow(1.6180339887498949d, n));

NOTE: Due to limits of precision, the preceding formula is only accurate up to n = 77. 


Based on YvesDaoust's recommendation, I've updated the formula to use a simpler version of the closed form solution (also found on Wiki), as it proves to be faster and more compact.

Furthermore, I've adjusted the constants slightly to improve the function's accuracy.


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


About the Author

Andrew Rissing
Software Developer (Senior)
United States United States
Since I've begun my profession as a software developer, I've learned one important fact - change is inevitable. Requirements change, code changes, and life changes.

So..If you're not moving forward, you're moving backwards.

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Comments and Discussions

GeneralThoughts Pin
PIEBALDconsult8-Sep-13 13:55
professionalPIEBALDconsult8-Sep-13 13:55 
GeneralRe: Thoughts Pin
Andrew Rissing10-Sep-13 5:08
memberAndrew Rissing10-Sep-13 5:08 
Questionformula's missing part Pin
Member 33856988-Sep-13 11:30
memberMember 33856988-Sep-13 11:30 
AnswerRe: formula's missing part Pin
Andrew Rissing10-Sep-13 5:06
memberAndrew Rissing10-Sep-13 5:06 
GeneralRe: formula's missing part Pin
Yuksel YILDIRIM14-Oct-13 15:56
memberYuksel YILDIRIM14-Oct-13 15:56 
GeneralMy vote of 2 Pin
YvesDaoust17-Dec-12 1:57
memberYvesDaoust17-Dec-12 1:57 
QuestionCan do better Pin
YvesDaoust17-Dec-12 1:51
memberYvesDaoust17-Dec-12 1:51 
AnswerRe: Can do better Pin
Andrew Rissing17-Dec-12 4:48
memberAndrew Rissing17-Dec-12 4:48 
AnswerRe: Can do better Pin
Andrew Rissing17-Dec-12 5:04
memberAndrew Rissing17-Dec-12 5:04 
GeneralRe: Can do better Pin
YvesDaoust17-Dec-12 5:32
memberYvesDaoust17-Dec-12 5:32 
This opens the way to deeper thinking on the problem.

Considering all formulas of the form Round(A.B^N), many values of parameters A and B will return correct integers for a number of values of N. There should be a way to maximize this range.

For instance, let us keep B = Phi exactly. We need to have Fk <= A.Phi^k + 0.5 < Fk + 1 for all k's in range 0..N. In other words, (Fk - 0.5) / Phi^k <= A < (Fk + 0.5) / Phi^k.

Unless I am wrong, the intervals are nested so that the most constraining value is k = N and any A in range (FN - 0.5) / Phi^N <= A < (FN + 0.5) / Phi^N can do.

More freedom is obtained by letting B vary as well. And we can even introduce an extra degree of freedom by using the more general form Floor(A.B^N + C) without complicating the formula so much.

modified 17-Dec-12 11:50am.

GeneralRe: Can do better Pin
Andrew Rissing17-Dec-12 12:04
memberAndrew Rissing17-Dec-12 12:04 
GeneralRe: Can do better Pin
YvesDaoust17-Dec-12 20:34
memberYvesDaoust17-Dec-12 20:34 
GeneralRe: Can do better Pin
Andrew Rissing18-Dec-12 4:24
memberAndrew Rissing18-Dec-12 4:24 
GeneralMy vote of 5 Pin
Manish Choudhary .NET expert14-Dec-12 18:50
memberManish Choudhary .NET expert14-Dec-12 18:50 
BugThis is correct only for n up to 70 Pin
Matt T Heffron13-Dec-12 12:12
memberMatt T Heffron13-Dec-12 12:12 
GeneralRe: This is correct only for n up to 70 Pin
Andrew Rissing13-Dec-12 12:45
memberAndrew Rissing13-Dec-12 12:45 
GeneralRe: This is correct only for n up to 70 Pin
Andrew Rissing17-Dec-12 12:05
memberAndrew Rissing17-Dec-12 12:05 

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