Counting Fibonacci Squares

Probability Level 1

Consider the above sequence. In step $n$ , we add $F_n$ ( Fibonacci number ) unit squares in a column to the right. We want to count the number of squares (of all sizes) in each figure.

If $n = 1$ , then we count 1 square.

If $n = 2$ , then 1 small square is added, so we count 2 1-by-1 squares.

If $n = 3$ , then 2 small squares are added, so we count 4 1-by-1 squares.

If $n = 4$ , then 3 small squares are added, so we count 7 1-by-1 squares and a 2-by-2 square, which altogether makes 8 squares in all.

For $n = 5$ , is the number of all possible different-sized squares 16?

No. Yes.

1 solution

Marta Reece
Aug 10, 2017

Next Fibonacci number is $2+3=5$ .

If we add it, we get $1+1+2+3+5=12$ size 1 squares.

And $3$ size 2 squares.

Total $12+3=15$ squares $\not=16$ squares.

@Michael Huang Your problem needs to be clarified. Marta's solution is misinterpreting your question.

I think it would be helpful to add the n=5 case for people to look at.

Calvin Lin Staff - 3 years, 10 months ago

Her solution is correct, which answers the question.

I couldn't think what to $1-1-2-3-5$ configuration, where there are $12$ small squares and $3$ larger squares.

Let me think about it.

Michael Huang - 3 years, 10 months ago

(Calvin was right, my solution did contain a misunderstanding of the question. I have edited it since then, though, so I think it's okay now.)

Marta Reece - 3 years, 10 months ago

Ah, the solution has been edited. It previously was just $1 + 1 +2+3+5 = 12$ .

Calvin Lin Staff - 3 years, 10 months ago

@Calvin Lin – I edited the question, which should be fine as it is...

Now, I see that the solution is correct. Thanks!

Michael Huang - 3 years, 10 months ago

Counting Fibonacci Squares

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