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A186514 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 A186513. 4

%I #7 Mar 30 2012 18:57:19

%S 4,6,10,13,16,19,22,26,29,32,35,38,42,45,48,51,55,58,61,64,68,71,74,

%T 77,80,84,87,90,93,97,100,103,106,110,113,116,119,122,126,129,132,135,

%U 139,142,145,148,152,155,158,161,165,168,171,174,178,181,184,187,190,194,197,200,203,207,210,213,216,220,223,226,229,233,236,239,242,245,249,252,255,258,262,265

%N 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 A186513.

%C See A186219 for a discussion of adjusted joint rank sequences.

%C The pairs (i,j) for which i^2=4+5j^2 are (L(2h),F(2h)), where L=A000032 (Lucas numbers) and F=A000045 (Fibonacci numbers).

%F a(n)=n+floor((1/10)(-4+sqrt(20n^2+6)))=A186513(n).

%F b(n)=n+floor(sqrt(5n^2+4n+1/2))=A186514(n).

%e First, write

%e 1..4..9..16..25..36..49..... (i^2)

%e ......9.....24.......49.. (4+5j^2)

%e Then replace each number by its rank, where ties are settled by ranking i^2 before 4+5j^2:

%e a=(1,2,3,5,7,8,9,11,12,14,15,17,..)=A186513

%e b=(4,6,10,13,16,19,22,26,29,32,...)=A186514.

%t (See A186513.)

%Y Cf. A186219, A186513, A186515, A186516.

%K nonn

%O 1,1

%A _Clark Kimberling_, Feb 22 2011

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