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How do I get the number of possibilities by the length of these 'blocks'? The smallest block has to have a minimum of 2 numbers.

enter image description here

What I already observed is, that it's always + (n-4)...

Example:

4 : 2
+ (5-4)
5 : 3
+ (6-4)
6 : 5

At the moment I don't even know how to start...

EDIT enter image description here

Are there anymore options here?

Janick Fischer
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  • You haven't given the rule for how to construct the blocks, so there cannot be a certain answer. $2,3,5$ makes me think of Fibonacci numbers. Are they compositions by parts greater than $1$? – Ross Millikan Jun 01 '17 at 14:57
  • @RossMillikan The blocks can be created as you wish, except that the smallest one has to be a minimum of 2 numbers... – Janick Fischer Jun 01 '17 at 15:00
  • You should have said that in the question. Please edit it in so people don't have to read the comments to find out. – Ross Millikan Jun 01 '17 at 15:05
  • @RossMillikan I just updated my questions with another picture... Shouldn't it be 13 not 12 options? – Janick Fischer Jun 01 '17 at 15:24
  • @RossMillikan $3,2,3$ is the 3rd one... I also think that when you set 8 in the Fibonacci function, you get 21. (wrote this on my smartphone...) – Janick Fischer Jun 01 '17 at 15:58
  • It is 4,4 that is missing. It would be much easier if you represent these as numbers instead of colored blocks. Yes, next is 21. – Ross Millikan Jun 01 '17 at 16:03

1 Answers1

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These are the Fibonacci numbers. One of the comments in the OEIS entry is that $F(n)$ is the number of compositions of $n+1$ with no part equal to $1.$ To justify it, note that an acceptable composition of length $n$ can either be an acceptable composition of length $n+1$ with the last element lengthened by $1$, or an acceptable composition of length $n+2$ with a $2$ added at the end. This gives the Fibonacci recurrence and doing a couple base cases by hand shows the offset.

Ross Millikan
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