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2022-10-07

Solving $s\le n\mathrm{log}n$ for smallest n

I am given an arbitrary positive integer $s$. I want to find the smallest integer $n$ such that

$$s\le n{\mathrm{log}}_{2}n$$

where ${\mathrm{log}}_{2}$ is log base $2$

Is there an efficient way to compute n?

I am given an arbitrary positive integer $s$. I want to find the smallest integer $n$ such that

$$s\le n{\mathrm{log}}_{2}n$$

where ${\mathrm{log}}_{2}$ is log base $2$

Is there an efficient way to compute n?

Clare Acosta

Beginner2022-10-08Added 7 answers

$$s\le n{\mathrm{log}}_{2}(n)$$

$$s\le \frac{n\mathrm{log}n}{\mathrm{log}2}$$

$$s\mathrm{log}2\le n\mathrm{log}n$$

$$s\mathrm{log}2\le {e}^{\mathrm{log}n}\mathrm{log}n$$

Using the Lambert W function and the fact that you want the smallest n, we have

$$\mathrm{log}n=W(s\mathrm{log}2)$$

$$n={e}^{W(s\mathrm{log}2)}=\frac{s\mathrm{log}2}{W(s\mathrm{log}2)}$$

As others have stated, you'll need a computer to evaluate this.

$$s\le \frac{n\mathrm{log}n}{\mathrm{log}2}$$

$$s\mathrm{log}2\le n\mathrm{log}n$$

$$s\mathrm{log}2\le {e}^{\mathrm{log}n}\mathrm{log}n$$

Using the Lambert W function and the fact that you want the smallest n, we have

$$\mathrm{log}n=W(s\mathrm{log}2)$$

$$n={e}^{W(s\mathrm{log}2)}=\frac{s\mathrm{log}2}{W(s\mathrm{log}2)}$$

As others have stated, you'll need a computer to evaluate this.

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