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A283233 2*A000201.

%I A283233 #16 Aug 10 2022 02:59:59
%S A283233 2,6,8,12,16,18,22,24,28,32,34,38,42,44,48,50,54,58,60,64,66,70,74,76,
%T A283233 80,84,86,90,92,96,100,102,106,110,112,116,118,122,126,128,132,134,
%U A283233 138,142,144,148,152,154,158,160,164,168,170,174,176,180,184,186
%N A283233 2*A000201.
%C A283233 This is one of three sequences that partition the positive integers.  In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint.  Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked.  Define b(n) and c(n) as the ranks of n/s and n/t.  It is easy to prove that
%C A283233 a(n)=n+[ns/r]+[nt/r],
%C A283233 b(n)=n+[nr/s]+[nt/s],
%C A283233 c(n)=n+[nr/t]+[ns/t], where [ ]=floor.
%C A283233 Taking r=1, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2 gives a=A283233, b=A283234, c=A005843.
%H A283233 Clark Kimberling, <a href="/A283233/b283233.txt">Table of n, a(n) for n = 1..10000</a>
%F A283233 a(n) = 2*floor(n*r), where r = (1+sqrt(5))/2.
%t A283233 r = 1; s = (-1 + 5^(1/2))/2; t = (1 + 5^(1/2))/2;
%t A283233 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A283233 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A283233 c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t A283233 Table[a[n], {n, 1, 120}]  (* A283233 *)
%t A283233 Table[b[n], {n, 1, 120}]  (* A283234 *)
%t A283233 Table[c[n], {n, 1, 120}]  (* A005408 *)
%o A283233 (Python)
%o A283233 from math import isqrt
%o A283233 def A283233(n): return (n+isqrt(5*n**2))&-2 # _Chai Wah Wu_, Aug 10 2022
%Y A283233 Cf. A000201, A283234, A005843 (sequential union of A283233 and A283234), A005408.
%K A283233 nonn,easy
%O A283233 1,1
%A A283233 _Clark Kimberling_, Mar 03 2017