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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.

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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.

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

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