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A329975 Beatty sequence for 1 + x + x^2, where x is the real solution of 1/x + 1/(1+x+x^2) = 1.

%I A329975 #13 Feb 16 2025 08:33:58
%S A329975 4,8,12,16,20,24,28,32,36,40,44,48,53,57,61,65,69,73,77,81,85,89,93,
%T A329975 97,101,106,110,114,118,122,126,130,134,138,142,146,150,155,159,163,
%U A329975 167,171,175,179,183,187,191,195,199,203,208,212,216,220,224,228,232
%N A329975 Beatty sequence for 1 + x + x^2, where x is the real solution of 1/x + 1/(1+x+x^2) = 1.
%C A329975 Let x be the real solution of 1/x + 1/(1+x+x^2) = 1. Then (floor(n*x)) and (floor(n*(x^2 + x + 1))) 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 A329975 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BeattySequence.html">Beatty Sequence.</a>
%H A329975 <a href="/index/Be#Beatty">Index entries for sequences related to Beatty sequences</a>
%F A329975 a(n) = floor(n*(1+x+x^2)), where x = 1.324717... is the constant in A060006.
%t A329975 Solve[1/x + 1/(1 + x + x^2) == 1, x]
%t A329975 u = 1/3 (27/2 - (3 Sqrt[69])/2)^(1/3) + (1/2 (9 + Sqrt[69]))^(1/3)/3^(2/3);
%t A329975 u1 = N[u, 150]
%t A329975 RealDigits[u1, 10][[1]]  (* A060006 *)
%t A329975 Table[Floor[n*u], {n, 1, 50}]              (* A329974 *)
%t A329975 Table[Floor[n*(1 + u + u^2)], {n, 1, 50}]  (* A329975 *)
%t A329975 Plot[1/x + 1/(1 + x + x^2) - 1, {x, -2, 2}]
%Y A329975 Cf. A329825, A060006, A329974 (complement).
%K A329975 nonn,easy
%O A329975 1,1
%A A329975 _Clark Kimberling_, Jan 02 2020