A186145 - OEIS

%I #9 Sep 24 2021 10:42:02

%S 1,3,5,6,7,9,10,11,13,14,15,17,18,19,21,22,23,24,26,27,28,29,31,32,33,

%T 34,35,37,38,39,40,42,43,44,45,46,48,49,50,51,52,54,55,56,57,58,60,61,

%U 62,63,64,65,67,68,69,70,71,72,74,75,76,77,78,79,81,82,83,84,85,86,88,89,90,91,92,93,95,96,97,98,99,100,102,103,104,105,106,107,108,110,111,112,113,114,115,116,118,119,120,121

%N Rank of n^2 when {i^2: i>=1} and {j^3: j>=1} are jointly ranked with i^2 before j^3 when i^2=j^3.  Complement of A186146.

%C Suppose u,v,p,q are positive integers and 0<|d|<1. Let

%C a(n)=n+floor(((u*n^p-d)/v)^(1/q)),

%C b(n)=n+floor(((v*n^q+d)/u)^(1/p)).

%C When the disjoint sets {u*i^p} and {v*j^q+d} are jointly ranked, the rank of u*n^p is a(n) and the rank of v*n^q+d is b(n).  Therefore a and b are a pair of complementary sequences.  Choosing d carefully serves as a basis for two types of adjusted joint rankings of non-disjoint sets {u*i^p} and {v*j^q}.

%C First, if we place u*i^p before v*j^q whenever u*i^p=v*j^q, then with 0<d<1, a(n) and b(n) are the ranks of u*n^p and v*j^q, respectively.  For the second type, if we place u*i^p after v*j^q whenever u*i^p=v*j^q, then with -1<d<0, a(n) and b(n) are ranks of u*n^p and v*j^q, respectively.

%C More generally, if u=h/k and v=s/t are positive rational numbers in lowest terms, then a(n) and b(n) are the respective ranks of u*n^p and v*n^q, adjusted as described above, according as d=1/(2kq) or d=-1/(2kq).  Examples: A186148-A186159.

%F a(n)=n+floor((n^2-1/2)^(1/3)) (A186145).

%F b(n)=n+floor((n^3+1/2)^(1/2)) (A186146).

%e Write the squares and cubes thus:

%e 1..4....9..16..25....36..49..64..81

%e 1.....8...........27.........64.....

%e Replace each by its rank, where ties are settled by ranking the square before the cube:

%e a=(1,3,5,6,7,9,10,11,13,...)

%e b=(2,4,8,12,...)

%t d=1/2;

%t a[n_]:=n+Floor[(n^2-d)^(1/3)]; (* rank of n^2 *)

%t b[n_]:=n+Floor[(n^3+d)^(1/2)]; (* rank of n^3+1/2 *)

%t Table[a[n],{n,1,100}]

%t Table[b[n],{n,1,100}]

%t (* end *)

%t (* A more general program follows. *)

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

%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}]

%t Table[b[n],{n,1,100}]

%Y Cf. A186146.

%K nonn

%O 1,2

%A _Clark Kimberling_, Feb 13 2011