cp's OEIS Frontend

A186512 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 A186511.

%I A186512 #5 Mar 30 2012 18:57:19
%S A186512 1,5,9,12,15,19,22,25,29,32,35,38,41,45,48,51,54,58,61,64,67,71,74,77,
%T A186512 80,84,87,90,93,97,100,103,106,109,113,116,119,122,126,129,132,135,
%U A186512 139,142,145,148,152,155,158,161,165,168,171,174,177,181,184,187,190,194,197,200,203,207,210,213,216,220,223,226,229,232,236,239,242,245,249,252,255,258,262,265
%N A186512 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 A186511.
%C A186512 See A186219 for a discussion of adjusted joint rank sequences.
%C A186512 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 A186512 a(n)=n+floor(sqrt((n^2)/5+9/10))=A186511(n).
%F A186512 b(n)=n+floor(sqrt(5n^2-9/2))=A186512(n).
%e A186512 First, write
%e A186512 1..4..9..16..25..36..49..... (i^2)
%e A186512 1........16........41........(-4+5j^2)
%e A186512 Then replace each number by its rank, where ties are settled by ranking i^2 after -4+5j^2:
%e A186512 a=(2,3,4,6,7,8,10,11,13,14,16,...)=A186511
%e A186512 b=(1,5,9,12,15,19,22,25,29,32,...)=A186512.
%t A186512 (See A186511.)
%Y A186512 Cf. A186219, A186499, A186500, A186511.
%K A186512 nonn
%O A186512 1,2
%A A186512 _Clark Kimberling_, Feb 22 2011