Revision bd72a146-ee21-4bcd-b484-b7ccfd2a922c - Code Golf Stack Exchange

#J, 34 bytes

`f=:3 :'(+/((i.y)e.y%2+i.y)#i.y)>y'`

Returns 1 if it's abundant and 0 if it's not.

I wonder if I would've gotten less bytes if it was tacit instead.

**Output:**

 f 3
 0
 f 12
 1
 f 11
 0
 f 20
 1