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

1,2

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<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.

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.

Links

Table of n, a(n) for 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).

a(n) = n + floor(n^(2/3))-1 if n is a cube and a(n) = n + floor(n^(2/3)) otherwise. - Chai Wah Wu, Oct 15 2025

Example

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,...)

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}]

Prog

(Python)
from sympy import integer_nthroot
def A186145(n): return n+(lambda x:x[0]-x[1])(integer_nthroot(n**2, 3)) # Chai Wah Wu, Oct 15 2025

Crossrefs

Cf. A186146.

Sequence in context: A336497 A116883 A256543 * A364058 A335740 A381439

Adjacent sequences: A186142 A186143 A186144 * A186146 A186147 A186148

Keyword

nonn

Author

Clark Kimberling, Feb 13 2011

Status

approved