> For fun (no guarantee of accuracy, but works for 4_000_000)...
>
> include Math
> R = Math.sqrt(5) #radical
> P = (1 + R) / 2 #phi
> my_number = 4_000_000
> #inverting my_number to see approximately where n sits for F(n)
> max_inverted = (log((my_number - 0.5) * R) / log(P)).to_i
> #adding up F(n) for 0 to n
> even_sum = (0..max_inverted).inject do |s, i|

When you do (max_inverted + 1) in the previous line, this works for 4_000_000.

>  #next line is for direct F(n)
>  #the inversion above is the opposite of this
>  fib = (((P**i) - (1-P)**i) / R).floor
>  #see if it's even
>  fib % 2 == 0 ? (s + fib) : s
> end
> p even_sum
>
>
> ...It would be interesting to see what ceilings (like 4_000_000) this fails for.

Revisiting old emails.  Sorry for noise, but important to be concise.

Todd