Permalink Submitted by Anonymous on January 17, 2015

Now let's pretend 5 is the first Fibonacci number instead of the usual 1, but still use the same addition algorithm: 5 5 10 15 25. That 25 is of course 5 squared.

Try this with 4 however: 4 4 8 12 20, and we don't land on 4 squared. 16 isn't in the sequence generated. The same goes for 6: 6 6 12 18 30 48 doesn't include 36. But then 4 and 6 aren't in the Fibonacci sequence either.

In fact the only start numbers we can hit the square with seem to be the Fibonaccis and no others:

1

2 2 4

3 3 6 9

5 5 10 15 25

8 8 16 24 40 64

13 13 26 39 65 104 169

etc etc.

(My treatment of 1 seems a bit anomalous here since although it 's a perfect square, I haven't presented it as the result of any addition. This can be remedied perhaps with: 0 1 1)

Note also the number of numbers in each sequence, which is equal to the position of the start number in the standard Fibonacci sequence. For example 13 takes 7 numbers to get to its square, and 13 occupies position 7.

Lastly although I came across these results concerning Fibonacci powers on my own (see also my previous comment about 5), I daresay they aren't new discoveries. So please tell me of any similar work.

## Hitting that square

Now let's pretend 5 is the first Fibonacci number instead of the usual 1, but still use the same addition algorithm: 5 5 10 15 25. That 25 is of course 5 squared.

Try this with 4 however: 4 4 8 12 20, and we don't land on 4 squared. 16 isn't in the sequence generated. The same goes for 6: 6 6 12 18 30 48 doesn't include 36. But then 4 and 6 aren't in the Fibonacci sequence either.

In fact the only start numbers we can hit the square with seem to be the Fibonaccis and no others:

1

2 2 4

3 3 6 9

5 5 10 15 25

8 8 16 24 40 64

13 13 26 39 65 104 169

etc etc.

(My treatment of 1 seems a bit anomalous here since although it 's a perfect square, I haven't presented it as the result of any addition. This can be remedied perhaps with: 0 1 1)

Note also the number of numbers in each sequence, which is equal to the position of the start number in the standard Fibonacci sequence. For example 13 takes 7 numbers to get to its square, and 13 occupies position 7.

Lastly although I came across these results concerning Fibonacci powers on my own (see also my previous comment about 5), I daresay they aren't new discoveries. So please tell me of any similar work.