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Rank of (1/8)n^3 when {(1/8)i^3: i>=1} and {j^2>: j>=1} are jointly ranked with (1/8)i^3 before j^2 when (1/8)i^3=j^2. Complement of A186153.
4

%I #6 Mar 30 2012 18:57:18

%S 1,2,4,6,8,11,13,15,18,21,23,26,29,32,35,38,41,44,48,51,55,58,61,65,

%T 69,72,76,80,84,88,92,95,100,104,108,112,116,120,125,129,133,138,142,

%U 147,151,156,160,165,170,174,179,184,189,194,199,204,209,214,219,224,229,234,239,245,250,255,260,266,271,277,282,287,293,299,304,310,315,321,327,332,338,344,350,356,362,367,373,379,385,391,397,403,410,416,422,428,434,440,447,453

%N Rank of (1/8)n^3 when {(1/8)i^3: i>=1} and {j^2>: j>=1} are jointly ranked with (1/8)i^3 before j^2 when (1/8)i^3=j^2. Complement of A186153.

%C See A186145 for a discussion of adjusted joint rank sequences.

%F a(n)=n+floor(((1/8)n^3-1/16)^(1/2)), A186152.

%F b(n)=n+floor(((n^2)/16+1/2)^(1/3)), A186153.

%t d=1/16;u=1/8;v=1;p=3;q=2;

%t h[n_]:=((u*n^p-d)/v)^(1/q);

%t a[n_]:=n+Floor[h[n]]; (* rank of u*n^p *)

%t k[n_]:=((v*n^q+d)/u)^(1/p);

%t b[n_]:=n+Floor[k[n]]; (* rank of v*n^q *)

%t Table[a[n],{n,1,100}] (* A186152 *)

%t Table[b[n],{n,1,100}] (* A186153 *)

%Y Cf. A186145, A186153.

%K nonn

%O 1,2

%A _Clark Kimberling_, Feb 13 2011