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POAlgorithm for k-Fibonacci
 

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  1. POAlgorithm for k-Fibonacci
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    <p>We all know fibonacci series, when k = 2.</p> <p>I.e.: <code>1,1,2,3,5,8,13</code></p> <p>But this is the 2-fibonacci. Like this, I can count the third-fibonacci:</p> <pre><code>1,1,2,4,7,13,24 </code></pre> <p>And the 4-fibonacci:</p> <pre><code>1,1,2,4,8,15,29 </code></pre> <p>...and so goes on</p> <p>What I'm asking is an algorithm to calculate an 'n' element inside a k-fibonacci series.</p> <p>Like this: if I ask for <code>fibonacci(n=5,k=4)</code>, the result should be: <code>8</code>, i.e. the fifth element inside the 4-fibonacci series.</p> <p>I didn't found it anywhere web. A resouce to help could be <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html" rel="noreferrer">mathworld</a></p> <p>Anyone? And if you know python, I prefer. But if not, any language or algorithm can help.</p> <p>Tip I think that can help: Let's analyze the k-fibonacci series, where k will go from 1 to 5</p> <pre><code>k fibonacci series 1 1, 1, 1, 1, 1, 1, 1, 1,1, 1, 1, ... 2 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ... 3 1, 1, 2, 4, 7, 13, 24, 44, 81, ... 4 1, 1, 2, 4, 8, 15, 29, 56, 108, ... 5 1, 1, 2, 4, 8, 16, 31, 61, 120, ... </code></pre> <p>Analyzing this, we can see that the array [0:k] on the k-fibonacci series is equal to the previous fibonacci series, and it goes on till the k=1</p> <p>i.e. (I'll try to show, but I'm not finding the right way to say it):</p> <pre><code>k fibonacci series 1 1, 2 1, 1, 3 1, 1, 2, 4 1, 1, 2, 4, 5 1, 1, 2, 4, 8, </code></pre> <p>Hope I've helped somehow to solve this.</p> <p><strong>[SOLUTION in python (if anyone needs)]</strong></p> <pre><code>class Fibonacci: def __init__(self, k): self.cache = [] self.k = k #Bootstrap the cache self.cache.append(1) for i in range(1,k+1): self.cache.append(1 &lt;&lt; (i-1)) def fib(self, n): #Extend cache until it includes value for n. #(If we've already computed a value for n, we won't loop at all.) for i in range(len(self.cache), n+1): self.cache.append(2 * self.cache[i-1] - self.cache[i-self.k-1]) return self.cache[n] #example for k = 5 if __name__ == '__main__': k = 5 f = Fibonacci(k) for i in range(10): print f.fib(i), </code></pre>
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    1. USGabriel L. Oliveira
    1. USGabriel L. Oliveira
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    1. CO@Amber, @Itay: thanks for tips. Any algorithm to solve this? I'm really lost on this problem.
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      1. USGabriel L. Oliveira
    2. CO@ Gabriel - Not really sure what you mean by algorithm? The computation of fibonacci numbers is not really complex...
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    3. COI found some paper about it ***THE GENERALIZED BINET FORMULA***. Posted the link in my answer.
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      1. USItay Karo
 

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