On Tue, Aug 9, 2011 at 4:49 PM, I wrote:
>   (([ = ] +/@#~ 0 = |~) i.&.<:)

As Ric Sherlock points out, you might not need that &.<:

All it's doing is eliminating a 0 from the sum.

Thus:
   (([ = ] +/@#~ 0 = |~) i.)

That said, with this change, 0 is treated as a perfect number where
before it was not.

-- 
Raul
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