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A186156 Rank of n^3 when {i^3: i>=1} and {2j^2: j>=1} are jointly ranked with i^3 before 2j^2 when i^3=2j^2. Complement of A186157.

%I A186156 #4 Mar 30 2012 18:57:18
%S A186156 1,3,6,9,12,16,20,23,28,32,36,41,46,51,56,61,66,71,77,83,89,94,100,
%T A186156 107,113,119,126,132,139,146,153,159,167,174,181,188,196,203,211,218,
%U A186156 226,234,242,250,258,266,274,283,291,299,308,317,325,334,343,352,361,370,379,388,397,407,416,426,435,445,454,464,474,484,494,503,514,524,534,544,554,565,575,585,596,607,617,628,639,649,660,671,682
%N A186156 Rank of n^3 when {i^3: i>=1} and {2j^2: j>=1} are jointly ranked with i^3 before 2j^2 when i^3=2j^2.  Complement of A186157.
%C A186156 See A186145 for a discussion of adjusted joint rank sequences.
%F A186156 a(n)=n+floor(((n^3-1/2)/2)^(1/2)), A186156.
%F A186156 b(n)=n+floor((2n^2+1/2)^(1/3)), A186157.
%e A186156 Write separate rankings as
%e A186156 1....8.....27........64........125...
%e A186156 ..2..8..18....32..50....72..98.....128...
%e A186156 Then replace each number by its rank, where ties are settled by ranking i^3 before 2j^2.
%t A186156 d=1/2; u=1; v=2; p=3; q=2;
%t A186156 h[n_]:=((u*n^p-d)/v)^(1/q);
%t A186156 a[n_]:=n+Floor[h[n]]; (* rank of u*n^p *)
%t A186156 k[n_]:=((v*n^q+d)/u)^(1/p);
%t A186156 b[n_]:=n+Floor[k[n]]; (* rank of v*n^q *)
%t A186156 Table[a[n],{n,1,100}] (* A186156 *)
%t A186156 Table[b[n],{n,1,100}] (* A186157 *)
%Y A186156 Cf. A186145, A186157, A186158, A186159.
%K A186156 nonn
%O A186156 1,2
%A A186156 _Clark Kimberling_, Feb 13 2011