How to count all the solutions for $\sum\limits_{i=1}^{n} \frac{1}{2^{k_i}}= 1$ for $k_i\in \Bbb{N}$ and $n$ a fixed positive integer?

Question

After reading this question, I would like to just count all solutions for:

$$\frac{1}{2^{k_1}} + \frac{1}{2^{k_2}} + \frac{1}{2^{k_3}} + \dots + \frac{1}{2^{k_n}}=1$$

for $k_i\in \Bbb{N}$ (we can include $0$) with $n$ a fixed positive integer.

I noticed that if we denote with $f(k)$ the number of times the value $k$ appears in the sequence $k_i$ then:

$$2^n=\sum_{k=0}^{n}{2^{n-k}f(k)}$$

and also

$$n=\sum_{k=0}^{n}{f(k)}$$

So the problem is equivalent to count all $f(0), ... ,f(n)$ solutions to the system of the last two equations.

I tried to apply the star and bars method and inclusion-exclusion principle, but with no success so far.

EDIT 2019-08-14

After reading this answer to another question, I found that the number of solutions is the coefficient of $x^ny^{2^n}$ of the generating function:

$$\frac{1}{(1-xy)(1-xy^2)(1-xy^4)\ldots(1-xy^{2^{n-1}})(1-xy^{2^n})}$$

but is it possible to get a formula?

Answer 1

Using the solution scheme by Crostul in the mentioned foregoing question I let Mathematica compute the number $a(n)$ of solutions $k_1\leq k_2\leq\ldots\leq k_n$ for $1\leq n\leq11$ and obtained the sequence $$1, 1, 1, 2, 3, 5, 9, 16, 28, 50, 89, \dots\quad.$$ This is sequence A002572 at OEIS, where reference is made to this problem, and additional information is given.