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A186511 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) after g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-4+5j^2. Complement of A186512.

%I A186511 #7 Mar 30 2012 18:57:19
%S A186511 2,3,4,6,7,8,10,11,13,14,16,17,18,20,21,23,24,26,27,28,30,31,33,34,36,
%T A186511 37,39,40,42,43,44,46,47,49,50,52,53,55,56,57,59,60,62,63,65,66,68,69,
%U A186511 70,72,73,75,76,78,79,81,82,83,85,86,88,89,91,92,94,95,96,98,99,101,102,104,105,107,108,110,111,112,114,115,117,118,120,121,123,124,125,127,128,130,131,133,134,136,137,138,140,141,143,144
%N A186511 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) after g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-4+5j^2.  Complement of A186512.
%C A186511 See A186219 for a discussion of adjusted joint rank sequences.
%C A186511 The pairs (i,j) for which i^2=-4+5j^2 are (L(2h-1),F(2h-1)), where L=A000032 (Lucas numbers) and F=A000045 (Fibonacci numbers).
%F A186511 a(n)=n+floor(sqrt((n^2)/5+9/10))=A186511(n).
%F A186511 b(n)=n+floor(sqrt(5n^2-9/2))=A186512(n).
%e A186511 First, write
%e A186511 1..4..9..16..25..36..49..... (i^2)
%e A186511 1........16........41........(-4+5j^2)
%e A186511 Then replace each number by its rank, where ties are settled by ranking i^2 after -4+5j^2:
%e A186511 a=(2,3,4,6,7,8,10,11,13,14,16,...)=A186511
%e A186511 b=(1,5,9,12,15,19,22,25,29,32,...)=A186512.
%t A186511 (* adjusted joint rank sequences a and b, using general formula for ranking ui^2+vi+w and xj^2+yj+z *)
%t A186511 d = -1/2; u = 1; v = 0; w = 0; x = 5; y = 0; z =-4;
%t A186511 h[n_] := -y + (4 x (u*n^2 + v*n + w - z - d) + y^2)^(1/2);
%t A186511 a[n_] := n + Floor[h[n]/(2 x)];
%t A186511 k[n_] := -v + (4 u (x*n^2 + y*n + z - w + d) + v^2)^(1/2);
%t A186511 b[n_] := n + Floor[k[n]/(2 u)];
%t A186511 Table[a[n], {n, 1, 100}]  (* A186511 *)
%t A186511 Table[b[n], {n, 1, 100}]  (* A186512 *)
%Y A186511 Cf. A186219, A186499, A186500, A186512.
%K A186511 nonn
%O A186511 1,1
%A A186511 _Clark Kimberling_, Feb 22 2011