Modulus Summation

Haskell, 22 bytes

f x=sum$mod x<$>[1..x]

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Explanation

Yes.

Ruby, 28 27 23 bytes

-4 bytes thanks to @daniero.

->n{(1..n).sum{|i|n%i}}

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Ruby, 28 bytes

f=->n{n>($.+=1)?n%$.+f[n]:0}

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05AB1E, 3 bytes

L%O

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