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%I #10 Jan 08 2013 11:15:17
%S 1,3,5,7,8,10,11,12,14,15,16,18,19,20,22,23,24,25,27,28,29,30,31,33,
%T 34,35,36,37,39,40,41,42,43,45,46,47,48,49,50,52,53,54,55,56,57,58,60,
%U 61,62,63,64,65,66,68,69,70,71,72,73,74,76,77,78,79,80,81,82,83,85,86,87,88,89,90,91,92,93,95,96,97,98,99,100,101,102,103,105,106,107,108,109,110,111,112,113,115,116,117,118,119,120,121,122,123,124,126,127,128,129,130,131,132,133,134,135,136,138,139,140,141
%N Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f and g are the odd numbers and the triangular numbers. Complement of A186351.
%C Suppose that f and g are strictly increasing functions for which (f(i)) and (g(j)) are integer sequences. If 0<|d|<1, the sets F={f(i): i>=1} and G={g(j)+d: j>=1} are clearly disjoint. Let f^=(inverse of f) and g^=(inverse of g). When the numbers in F and G are jointly ranked, the rank of f(n) is a(n):=n+floor(g^(f(n))-d), and the rank of g(n)+d is b(n):=n+floor(f^(g(n))+d). Therefore, the sequences a and b are a complementary pair.
%C Although the sequences (f(i)) and (g(j)) may not be disjoint, the sequences (f(i)) and (g(j)+d) are disjoint, and this observation enables two types of adjusted joint rankings:
%C (1) if 0<d<1, we call a and b the "adjusted joint rank sequences of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j)"; (2) if -1<d<0, we call a and b the "adjusted joint rank sequences of (f(i)) and (g(j)) with f(i) after g(j) when f(i)=g(j)".
%C Using f(i)=ui+v, g(j)=xj^2+yj+z, we find a and b given by
%C a(n)=n+floor((-y+sqrt(4x(un+v-d)+y^2))/(2x)),
%C b(n)=n+floor((xn^2+yn-v+d)/(2u))),
%C where a(n) is the rank of un+v and b(n) is the rank
%C xn^2+yn+z+d, and d must be chosen small enough, in
%C absolute value, that the sets F and G are disjoint.
%C Example: f=A000217 (odd numbers) and g=A000290 (triangular numbers) yield adjusted joint rank sequences a=A186350 and b=A186351 for d=1/2 and a=A186352 and b=A186353 for d=-1/2.
%C For other classes of adjusted joint rank sequences, see A186145 and A186219.
%F a(n)=n+floor(-1/2+sqrt(4n-9/4))=A186350(n).
%F b(n)=n+floor((n^2+n+3)/4)=A186351(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 before the triangjular:
%e a=(1,3,5,7,8,10,11,12,14,....)=A186350
%e b=(2,4,6,9,13,17,21,26,32,...)=A186351.
%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}] (* A186350 *)
%t Table[b[n],{n,1,100}] (* A186351 *)
%Y Cf. A186145, A186219, A186351, A186352, A186353,
%Y A005408 (odd numbers), A000217 (triangular numbers).
%K nonn
%O 1,2
%A _Clark Kimberling_, Feb 18 2011
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