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.
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.