0 doesn't count, right?

Let's call N(x) the number of ways to write x

* N(1) = 1

* trivial case

* N(2) = 1 + N(1)

* the 2 itself

* sum a 1, then add 1

* = N(1) + N(1) = 2·N(1)

* N(3) = 1 + N(1) + N(2)

* the 3 itself

* sum a 1, then add 2

* sum a 2, then add 1

* = N(2)+N(2) = 2·N(2)

* N(4) = 1 + N(1) + N(2) + N(3)

* the 4 itself

* sum a 1, then add 3

* sum a 2, then add 2

* sum a 3, then add 1

* = N(3)+N(3) = 2·N(3)

* N(5) = 1 + N(1) + N(2) + N(3) + N(4)

* the 5 itself

* sum a 1, then add 4

* sum a 2, then add 3

* sum a 3, then add 2

* sum a 4, then add 1

* = N(4)+N(4) = 2·N(4)

* etc.

So N(x) = 2·N(x-1), or N(x) = 2^(x-1)

It's not a proof but maybe it'll help