Assume [itex]n\in\mathbb{N}[/itex] and [itex]r\in\mathbb{R}[/itex] are some fixed constants and [itex]r>0[/itex]. I want to find some nice lower bound for the amount of elements [itex](k_1,k_2,\ldots,k_n)\in\mathbb{N}^n[/itex] such that [itex]2^{k_1}+\cdots + 2^{k_n}\leq r[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

In other words, if we define a function

[tex]

f:\mathbb{N}^n\to\mathbb{R},\quad f(k_1,\ldots, k_n)=2^{k_1}+\cdots + 2^{k_n}

[/tex]

I want to find some nice formula for a some reasonably large quantity [itex]\varphi(r,n)[/itex] such that

[tex]

\varphi(r,n)\leq \# f^{-1}([1,r])

[/tex]

where # means the number of elements in the set.

I tried to approximate some sums as integrals, but didn't get anywhere.

The special case [itex]n=1[/itex] easy. The number of elements [itex]2^0,2^1,2^2,\ldots[/itex] which are less than [itex]r[/itex] is

[tex]\Big[\frac{\log(r)}{\log(2)}\Big]+1[/tex]

where [itex]x\mapsto [x][/itex] is the floor function defined by [itex][x]=j\in\mathbb{Z}[/itex] and [itex]j\leq x < j+1[/itex].

So if we want [itex]\varphi(r,1)[/itex] "nice", it could be made differentiable by setting

[tex]

\varphi(r,1) = \frac{\log(r)}{\log(2)}.

[/tex]

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# Sum of 2^k less than r

Can you offer guidance or do you also need help?

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