Integers whose Divisor Sum is Cube/Mistake

Source Work

 * The Dictionary
 * $110$
 * $110$

Mistake

 * $\map \sigma {110} = 216 = 6^3$, the first $n$, after $n = 1, 7$, for which $\map \sigma n$ is a cube.

Correction
It is in fact the $2$nd $n$ after $n = 1, 7$, for which $\map \sigma n$ is a cube.

That is, it is the $4$th such $n$ altogether.

From Integers whose Sigma Value is Cube, the sequence is in fact:
 * $1, 7, 102, 110, 142, 159, 187, 381, 690, 714, 770, 994, 1034, \ldots$

So $102$ in particular has been missed.