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a(n) = floor(1/((n*r) mod 1)), where r = phi^(-2) = (3 - sqrt(5))/2.
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%I #8 Nov 20 2017 23:31:31

%S 2,1,6,1,1,3,1,17,2,1,4,1,1,2,1,8,2,1,3,1,46,2,1,5,1,1,3,1,12,2,1,4,1,

%T 1,2,1,7,1,1,3,1,23,2,1,5,1,1,2,1,10,2,1,4,1,122,2,1,6,1,1,3,1,15,2,1,

%U 4,1,1,2,1,8,1,1,3,1,33,2,1,5,1,1,3,1,11,2,1,4,1,1,2,1,7,1,1,3,1,19,2,1,5,1

%N a(n) = floor(1/((n*r) mod 1)), where r = phi^(-2) = (3 - sqrt(5))/2.

%C 1. a(n) = 1 iff A024569 is not 1, (A024569 = 1, 4, 1, 2, 11, 1, 3, 1, 1, ...)

%C 2. a(n) = 1 iff A078588 = 0.

%C 3. a(n) = 1 iff A089809 = 1.

%H G. C. Greubel, <a href="/A089808/b089808.txt">Table of n, a(n) for n = 1..5000</a>

%e a(6) = 3. Take 6*r = 2.29179...( mod 1) = 0.29179...; invert = 3.42705... and delete the fractional part, getting 3.

%t r := (3 - Sqrt[5])/2; Table[Floor[1/(Mod[(n*r), 1])], {n, 1, 50}] (* _G. C. Greubel_, Nov 20 2017 *)

%Y Cf. A024569, A078588, A089809.

%K nonn,easy

%O 1,1

%A _Gary W. Adamson_, Nov 11 2003

%E More terms from _Sam Alexander_, Nov 16 2003