[seqfan] Re: Two tenuously related problems concerning squares

[Charles Greathouse]

The sum of the first n Fibonacci numbers is F(n+2) - 1, so once F(n+2) - 1
> n^2 this certainly won't be possible.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Sun, Jun 16, 2013 at 1:15 PM, William Keith <william.keith at gmail.com>wrote:

> On Sun, Jun 16, 2013 at 11:04 AM, Alonso Del Arte
> <alonso.delarte at gmail.com>wrote:
>
> > 1. Is it always possible to express n^2 as a sum of n distinct Fibonacci
> > numbers, or at least almost so but allowing two 1s? e.g.,
> >
> >  1 = 1
> >  4 = 1 + 3
> >  9 = 1 + 3 + 5
> > 16 = 1 + 2 + 5 + 8
> > 25 = 1 + 1 + 2 + 8 + 13
> >
>
> No. It can't be done with 8 Fibonacci numbers for n^2 = 64.  Or apparently
> any higher square; at least certainly not up to 10,000, according to a
> quick Mathematica search.
>
> In[1] := FibSeries = Expand[Product[1 + z q^Fibonacci[n], {n, 1, 21}]];
>
> In[2] := Sum[(z^n) (q^(n^2))*
>   Coefficient[Coefficient[FibSeries, q^(n^2)], z^n], {n, 1, 100}]
>
> Out[2] := 2 q z + 2 q^4 z^2 + 2 q^9 z^3 + 2 q^16 z^4 + q^25 z^5 + q^36 z^6
> +
>  q^49 z^7
>
> In[3] := Coefficient[FibSeries, q^(64)]
>
> Out[3] := 2 z^3 + 4 z^4 + 3 z^5 + 3 z^6 + z^7
>
> There may be a proof with Binet's formula.
>
> William Keith
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>