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  1. f=@(x)x(~mod(sum(dec2base(x,10)'-48).^.5,1))
  2.  
  3. f([1,4,9,16,25,1111])
  4. f([1431,2,0,22,999999999])
  5. f([22228,4,113125,22345])
  6. f([421337,99,123456789,1133557799])
Success #stdin #stdout 0.51s 121792KB
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f =

@(x) x (!mod (sum (dec2base (x, 10)' - 48) .^ .5, 1))

ans =

      1      4      9   1111

ans =

        1431           0          22   999999999

ans =

   22228       4   22345

ans = [](1x0)
https://ideone.com/Dde5kh
language:
Octave (octave 4.4.1)
created:
7 years ago
visibility:
public

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