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

Author

Clark Kimberling, Feb 13 2011

Keywords

• nonn

Comments

See A186145 for a discussion of adjusted joint rank sequences.

Examples

			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.
		

Crossrefs

Cf. A186145, A186157, A186158, A186159.

Programs

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 *)

Formula

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

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