%I #4 Mar 30 2012 18:57:18
%S 2,4,5,7,8,10,11,13,14,15,17,18,19,20,22,23,24,25,27,28,29,30,32,33,
%T 34,35,36,38,39,40,41,42,43,45,46,47,48,49,50,52,53,54,55,56,57,59,60,
%U 61,62,63,64,65,67,68,69,70,71,72,73,74,76,77,78,79,80,81,82,83,85,86,87,88,89,90,91,92,94,95,96,97,98,99,100,101,102,104,105,106,107,108,109,110,111,112,113,115,116,117,118,119,120,121,122,123,124,126,127,128,129,130,131,132,133,134,135,137,138,139,140,141
%N Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) after g(j) when f(i)=g(j), where f and g are the odd numbers and the triangular numbers. Complement of A186353.
%F a(n)=n+floor(-1/2+sqrt(4n-3/4))=A186352(n).
%F b(n)=n+floor((n^2+n+1)/4)=A186353(n).
%e First, write
%e 1..3..5..7..9..11..13..15..17..21..23.. (odds)
%e 1..3....6.....10.......15......21.... (triangular)
%e Then replace each number by its rank, where ties are settled by ranking the odd number after the triangular:
%e a=(2,4,5,7,8,10,11,13,14,15,....)=A186352
%e b=(1,3,6,9,12,16,21,26,31,37,...)=A186353.
%t (* adjusted joint rank sequences a and b, using general formula for ranking 1st degree u*n+v and 2nd degree x*n^2+y*n+z *)
%t d=-1/2; u=2; v=-1; x=1/2; y=1/2; (* odds and triangular *)
%t h[n_]:=(-y+(4x(u*n+v-d)+y^2)^(1/2))/(2x);
%t a[n_]:=n+Floor[h[n]]; (* rank of u*n+v *)
%t k[n_]:=(x*n^2+y*n-v+d)/u;
%t b[n_]:=n+Floor[k[n]]; (* rank of x*n^2+y*n+d *)
%t Table[a[n], {n, 1, 120}] (* A186352 *)
%t Table[b[n], {n, 1, 100}] (* A186353 *)
%Y Cf. A186350, A186351, A186353.
%K nonn
%O 1,1
%A _Clark Kimberling_, Feb 18 2011