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A186499 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-4+5j^2. Complement of A186500.

%I A186499 #10 Mar 30 2012 18:57:19
%S A186499 1,3,4,5,7,8,10,11,13,14,15,17,18,20,21,23,24,26,27,28,30,31,33,34,36,
%T A186499 37,39,40,41,43,44,46,47,49,50,52,53,55,56,57,59,60,62,63,65,66,68,69,
%U A186499 70,72,73,75,76,78,79,81,82,83,85,86,88,89,91,92,94,95,96,98,99,101,102,104,105,107,108,109,111,112,114,115,117,118,120,121,123,124,125,127,128,130,131,133,134,136,137,138,140,141,143,144
%N A186499 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-4+5j^2.  Complement of A186500.
%C A186499 See A186219 for a discussion of adjusted joint rank sequences.
%C A186499 The pairs (i,j) for which i^2=-4+5j^2 are (L(2h-2),F(2h-1)), where L=A000032 (Lucas numbers) and F=A000045 (Fibonacci numbers); compare this with the comment at A186511.
%F A186499 a(n)=n+floor((1/10)(sqrt(2n^2+7)))=A186499(n).
%F A186499 b(n)=n+floor(sqrt(5n^2-7/2))=A186500(n).
%e A186499 First, write
%e A186499 1..4..9..16..25..36..49..... (i^2)
%e A186499 1........16........41........(-4+5j^2)
%e A186499 Then replace each number by its rank, where ties are settled by ranking i^2 before -4+5j^2:
%e A186499 a=(1,3,4,5,7,8,10,11,13,14,15,17,18...)=A186499
%e A186499 b=(2,6,9,12,16,19,22,25,29,32,35,38,..)=A186500.
%t A186499 (* adjusted joint rank sequences a and b, using general formula for ranking ui^2+vi+w and xj^2+yj+z *)
%t A186499 d=1/2; u=1; v=0; w=0; x=5; y=0; z=4;
%t A186499 h[n_]:=-y+(4x(u*n^2+v*n+w-z-d)+y^2)^(1/2);
%t A186499 a[n_]:=n+Floor[h[n]/(2 x)];
%t A186499 k[n_]:=-v+(4u(x*n^2+y*n+z-w+d)+v^2)^(1/2);
%t A186499 b[n_]:=n + Floor[k[n]/(2 u)];
%t A186499 Table[a[n], {n, 1, 100}]  (* A186499 *)
%t A186499 Table[b[n], {n, 1, 100}]  (* A186500 *)
%Y A186499 Cf. A186219, A186500, A186511, A186512.
%K A186499 nonn
%O A186499 1,2
%A A186499 _Clark Kimberling_, Feb 22 2011