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Question Number 211414 by AlagaIbile last updated on 08/Sep/24

Answered by Frix last updated on 08/Sep/24

$$\mathrm{Let}{x}_j⩽x_{j+1}$$$$\frac{1}{x_1}=1\Rightarrow \max \left(x_1\right)=1$$$$\frac{1}{x_1}+\frac{1}{x_2}=1\Rightarrow \max \left(x_2\right)=1\times (1+1)=2$$$$\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=1\Rightarrow \max \left(x_3\right)=2\times (2+1)=6$$$$...$$$$\max \left(x_{k+1}\right)=\max \left(x_k\right)\times (\max \left(x_k\right)+1)$$$$x_6=3263442$$

Commented by AlagaIbile last updated on 08/Sep/24

$$Yes.{Sylvester}^{\prime }sSequenceisthehint$$

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