# 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?

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.

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.

• So there is no hope for a formula, right? – mbjoe Mar 7 at 14:35
• At A002572 there is no formula; but I would not use the words "no hope". – Christian Blatter Mar 7 at 14:40