riddle - Powers of two my children be

Powers of two my children be,
Arriving in birth years separately.

Their sum is now the reverse of me
And the years between me and a power of three.

My digit difference is the number you see
And my digits are powers of binary.

And now the question I ask of thee:
Their number and age, and the age that I be.

A little riddle I came up with for my sons birthday.

Answer

Powers of two my children be,

Arriving in birth years separately.

Unspecified number of children but no twins and their age is 2^n.

Their sum is now the reverse of me

Their sum can be any number since any number can be expressed as a sum of powers of two. Your age can be written 10d + u and the sum of their age 10u + d.

And the years between me and a power of three.

10d + u - 3^x = 10u + d or 3^x - (10d + u) = 10u + d
Therefore 3^x = 9(u - d) or 3^x = 11(u + d)
The latter has no solution but the former has many solutions: (u - d) just needs to be a power of three

My digit difference is the number you see

The difference is the number of children: you have either three or nine children.

And my digits are powers of binary.

There are four possible digits: 1, 2, 4 and 8
If their difference is a power of three (= 3), they can only be 1 and 4.

And now the question I ask of thee:
Their number and age, and the age that I be.

You are 41 and the sum of their age is 14 (1110 in binary). Therefore you have three children and they are 8, 4 and 2 years old.