We are given a space of shifted Fibonacci sequences, Fk, Fk+1, Fk+2, Fk+3, Fk+6, Fk+8, Fk+9, Fk+10, Fk+10, Fk+11….. Given a number,n, what is the probability of this number within this space? And generally what is the probability function of any number to be found in this sufficient space? The size of the Space can be defined and it is flexible in that it should be sufficient to cover the number.
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It would be a good idea to include a link to this related earlier question. – Brian M. Scott Feb 16 '15 at 10:07
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Thank you Brian, being new in here I am trying to find my way. You are right. This question is an extension to an earlier question I posted and kindly a couple of Mathematicians replied. So here it is.[link]http://math.stackexchange.com/questions/1148059/how-many-times-a-positive-number-can-be-found-in-shifted-fibonacci-sequences – Anthony Feb 16 '15 at 10:24
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You have to define a probability space. "A random number" does not mean anything. By the way, a number $n$ belongs to the Fibonacci sequence if and only if $5n^2+4$ or $5n^2-4$ is a square, so your probability, once suitably defined, will be probably zero. – Jack D'Aurizio Feb 16 '15 at 10:25
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Thank you for your time Jack, but does this formula hold for a shifted one too? I dont think so. I think I defined the problem sufficently. For example if my chosen number is 5 there is no need to have the shifted space starting with 6. So my space is defined precicely. – Anthony Feb 16 '15 at 10:33
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Given a Number n, the number of times a positive number can be found in shifted Fibonacci Sequences have been worked out in this blog, [link]How many times a positive number can be found in shifted Fibonacci Sequences?. In order to work out the probability of a number to occur we need to define the space.The space should be sufficient/minimum to cover the existence of all the numbers n within the two dimensional space defined by: One dimension is the minimum fibonacci order required k, such that $F_k\le n$, and the other dimension is the number (shift l+ 1 = n+1). Plus one because we start from zero. So $$P(n)=\frac{1+⌊ln_φ(n√5+\frac{1}{2})⌋}{(⌊ln_φ(n√5+\frac{1}{2})⌋)(n+1)}$$ If $ F_k=n$ then $$P(n)=\frac{1}{(n+1)}$$
