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For a positive integer n, let  

                                                            \(R_n = \underbrace{111 \dots 1}_{n \text{ 1s}}.\)
For example, \(R_4=1111.\)

Find the polynomial p(x) such that \(p(R_n) = R_{2n + 1}\) for all positive integers n.

 Jun 3, 2020
 #1
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Rn  =  (2n - 1)  written in base 2.

 

p( Rn )  =  (22n + 1 - 1)  written in base 2.

 Jun 3, 2020
 #2
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So that makes p(x) = x^2 + 2x + 1.

 Jun 3, 2020

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