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%I #6 Mar 30 2012 18:57:18
%S 1,3,5,7,10,13,16,19,22,25,29,32,36,40,44,47,52,56,60,64,69,73,78,82,
%T 87,92,97,102,107,112,117,122,127,133,138,143,149,155,160,166,172,178,
%U 183,189,195,201,208,214,220,226,233,239,245,252,258,265,272,278,285,292,299,306,313,319,327,334,341,348,355,362,370,377,384,392,399,407,414,422,430,437,445,453,461,468,476,484,492,500
%N Rank of (1/4)n^3 when {(1/4)i^3: i>=1} and {j^2>: j>=1} are jointly ranked with (1/4)i^3 before j^2 when (1/4)i^3=j^2. Complement of A186149.
%C See A187645.
%F a(n)=n+floor(((1/4)n^3-1/8))^(1/2)), A186148 .
%F b(n)=n+floor((4n^3+1/2)^(1/3)), A186149.
%e Write preliminary separate rankings:
%e 1/4...2....27/4....16.....125/4...
%e ....1...4.......9..16..25........36..49
%e Then replace each number by its rank, where ties are settled by ranking the top number before the bottom.
%t d=1/8; u=1/4; 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}] (* A186148 *)
%t Table[b[n],{n,1,100}] (* A186149 *)
%Y Cf. A186145, A186149.
%K nonn
%O 1,2
%A _Clark Kimberling_, Feb 13 2011
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