On Fri, Jan 11, 2013 at 9:47 AM, John Benediktsson wrote:
> This should be a direct translation of that method, I think
>
> :: fibonacci? ( n )
> 1 [ dup fibonacci n < ] [ 1 + ] while fibonacci n = ;
>
Awesome.
--
Ma
This should be a direct translation of that method, I think
:: fibonacci? ( n )
1 [ dup fibonacci n < ] [ 1 + ] while fibonacci n = ;
On Thu, Jan 10, 2013 at 10:03 PM, Leonard P wrote:
> On Thu, Jan 10, 2013 at 11:44 AM, Alex Vondrak wrote:
>
>> Haven't looked at the code, but to answer yo
