A186145 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.

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

Offset: 1

Author

Clark Kimberling, Feb 13 2011

Keywords

• nonn

Comments

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

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

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

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}.

First, if we place u*i^p before v*j^q whenever u*i^p=v*j^q, then with 0

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.

Examples

			Write the squares and cubes thus:
1..4....9..16..25....36..49..64..81
1.....8...........27.........64.....
Replace each by its rank, where ties are settled by ranking the square before the cube:
a=(1,3,5,6,7,9,10,11,13,...)
b=(2,4,8,12,...)
		

Crossrefs

Cf. A186146.

Programs

Mathematica

d=1/2;
a[n_]:=n+Floor[(n^2-d)^(1/3)]; (* rank of n^2 *)
b[n_]:=n+Floor[(n^3+d)^(1/2)]; (* rank of n^3+1/2 *)
Table[a[n],{n,1,100}]
Table[b[n],{n,1,100}]
(* end *)
(* A more general program follows. *)
d=1/2; u=1; v=1; p=2; q=3;
h[n_]:=((u*n^p-d)/v)^(1/q);
a[n_]:=n+Floor[h[n]]; (* rank of u*n^p *)
k[n_]:=((v*n^q+d)/u)^(1/p);
b[n_]:=n+Floor[k[n]]; (* rank of v*n^q *)
Table[a[n],{n,1,100}]
Table[b[n],{n,1,100}]

Formula

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

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

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