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Expansion of 4/sqrt(phi) in base phi, where phi=(1+sqrt(5))/2.
2

%I #40 Mar 23 2022 11:10:44

%S 1,0,0,0,1,0,0,1,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,

%T 0,1,0,0,0,0,1,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,

%U 0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,0,0,0,0,0,1,0,1,0

%N Expansion of 4/sqrt(phi) in base phi, where phi=(1+sqrt(5))/2.

%C 4/sqrt(phi) = 3.14460551102... is an approximation to Pi obtained by dividing a square into 16 parts.

%e sqrt(16/((sqrt(5)+1)/2)) = 10.00100101010100101010000000000010010000101010000000010000100000\

%e 10000100101001000100101001000100010010000001010... in base phi.

%t RealDigits[Sqrt[16/GoldenRatio], GoldenRatio, 111][[1]]

%Y Cf. A202142.

%K base,cons,nonn

%O 2,1

%A Ville Takio, Dec 11 2011

%E Entry revised by _N. J. A. Sloane_, Dec 12 2011 and Apr 03 2016.

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