A186156 Rank of n^3 when {i^3: i>=1} and {2j^2: j>=1} are jointly ranked with i^3 before 2j^2 when i^3=2j^2. Complement of A186157.

1, 3, 6, 9, 12, 16, 20, 23, 28, 32, 36, 41, 46, 51, 56, 61, 66, 71, 77, 83, 89, 94, 100, 107, 113, 119, 126, 132, 139, 146, 153, 159, 167, 174, 181, 188, 196, 203, 211, 218, 226, 234, 242, 250, 258, 266, 274, 283, 291, 299, 308, 317, 325, 334, 343, 352, 361, 370, 379, 388, 397, 407, 416, 426, 435, 445, 454, 464, 474, 484, 494, 503, 514, 524, 534, 544, 554, 565, 575, 585, 596, 607, 617, 628, 639, 649, 660, 671, 682

Offset

1,2

Comments

See A186145 for a discussion of adjusted joint rank sequences.

Links

Table of n, a(n) for n=1..89.

Formula

a(n)=n+floor(((n^3-1/2)/2)^(1/2)), A186156.

b(n)=n+floor((2n^2+1/2)^(1/3)), A186157.

Example

Write separate rankings as
1....8.....27........64........125...
..2..8..18....32..50....72..98.....128...
Then replace each number by its rank, where ties are settled by ranking i^3 before 2j^2.

Mathematica

d=1/2; u=1; v=2; p=3; q=2;
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}] (* A186156 *)
Table[b[n], {n, 1, 100}] (* A186157 *)

Crossrefs

Cf. A186145, A186157, A186158, A186159.

Sequence in context: A194146 A276854 A184906 * A211267 A084515 A084525

Adjacent sequences: A186153 A186154 A186155 * A186157 A186158 A186159

Keyword

nonn

Author

Clark Kimberling, Feb 13 2011

Status

approved