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A283234 2*A001950.

%I A283234 #13 Aug 10 2022 02:58:32
%S A283234 4,10,14,20,26,30,36,40,46,52,56,62,68,72,78,82,88,94,98,104,108,114,
%T A283234 120,124,130,136,140,146,150,156,162,166,172,178,182,188,192,198,204,
%U A283234 208,214,218,224,230,234,240,246,250,256,260,266,272,276,282,286,292
%N A283234 2*A001950.
%C A283234 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 A283234 a(n)=n+[ns/r]+[nt/r],
%C A283234 b(n)=n+[nr/s]+[nt/s],
%C A283234 c(n)=n+[nr/t]+[ns/t], where [ ]=floor.
%C A283234 Taking r=1, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2 gives
%C A283234 a=A283233, b=A283234, c=A005843.
%H A283234 Clark Kimberling, <a href="/A283234/b283234.txt">Table of n, a(n) for n = 1..10000</a>
%F A283234 a(n) = 2*floor(n*s), where r = (-1+sqrt(5))/2.
%t A283234 r = 1; s = (-1 + 5^(1/2))/2; t = (1 + 5^(1/2))/2;
%t A283234 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A283234 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A283234 c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t A283234 Table[a[n], {n, 1, 120}]  (* A283233 *)
%t A283234 Table[b[n], {n, 1, 120}]  (* A283234 *)
%t A283234 Table[c[n], {n, 1, 120}]  (* A005408 *)
%o A283234 (Python)
%o A283234 from math import isqrt
%o A283234 def A283234(n): return ((n+isqrt(5*n**2))&-2)+(n<<1) # _Chai Wah Wu_, Aug 10 2022
%Y A283234 Cf. A000201, A283233, A005843.
%K A283234 nonn,easy
%O A283234 1,1
%A A283234 _Clark Kimberling_, Mar 03 2017