1
$\begingroup$

I was wondering if there is a closed expression for how many ways you can express a power of 2 as sum of $k$ powers of 2. That is, given $n,k \in \mathbb{N}$ is there an explicit formula for $$ \#\{(l_1, \dots, l_k) \in \mathbb{N}^k:\sum_{i=1}^k =2^n \text{ and }\forall i=1, \dots, k, \exists j_i \in \mathbb{N}, l_i = 2^{j_i} \} ? $$

Where $\#$ denotes cardinality of a set.

  • 1
    In your equation statement, did you mean to add that $$\sum_{i=1}^k 2^{l_i} = 2^k$$ – 2017-02-08
  • 0
    Would that just be the partitions of the power of $2$? – 2017-02-08
  • 0
    Yes, i will fix it now. – 2017-02-08
  • 0
    @Mark Fischler Actually, I meant this new statment, not necessarily with the number of parcels on the sum being the power of 2. – 2017-02-08
  • 0
    @hypergeometric no because for example 4 admits the partition 1+3, and 3 is not a power of 2. – 2017-02-08
  • 0
    OK - it would be the partition if the question had been about *products* of power of $2$. – 2017-02-08
  • 0
    Just to check if I understood correctly: you want $k$ powers of two that sum up to another power of two. Are permutations, like, say, $16=8+4+2+1+1=1+2+1+8+4$ to be distinguished or counted as one? – 2017-02-08
  • 0
    I would like to count permutations as different. – 2017-02-08

0 Answers 0