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%I #10 May 19 2020 16:19:05
%S 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,106,111,
%T 116,121,126,131,136,141,146,151,156,161,166,171,176,181,186,191,196,
%U 201,207,212,217,222,227,232,237,242,247,252,257,262,267,272,277,282
%N Beatty sequence for (x+1)^2, where 1/x + 1/(x+1)^2 = 1.
%C Let x be the solution of 1/x + 1/(x+1)^2 = 1. Then (floor(n x) and (floor(n (x+1)^2))) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BeattySequence.html">Beatty Sequence.</a>
%H <a href="/index/Be#Beatty">Index entries for sequences related to Beatty sequences</a>
%F a(n) = floor(n x), where x = 1.24697960371... is the constant in A255249.
%t r = x /. FindRoot[1/x + 1/(x+1)^2 == 1, {x, 2, 10}, WorkingPrecision -> 120]
%t RealDigits[r][[1]] (* A255249 *)
%t Table[Floor[n*r]], {n, 1, 250}] (* A330002 *)
%t Table[Floor[n*(1+r)^2], {n, 1, 250}] (* A330003 *)
%Y Cf. A329825, A255249, A330002 (complement).
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jan 04 2020
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