A186148 Rank of (1/4)n^3 when {(1/4)i^3: i>=1} and {j^2>: j>=1} are jointly ranked with (1/4)i^3 before j^2 when (1/4)i^3=j^2. Complement of A186149.

1, 3, 5, 7, 10, 13, 16, 19, 22, 25, 29, 32, 36, 40, 44, 47, 52, 56, 60, 64, 69, 73, 78, 82, 87, 92, 97, 102, 107, 112, 117, 122, 127, 133, 138, 143, 149, 155, 160, 166, 172, 178, 183, 189, 195, 201, 208, 214, 220, 226, 233, 239, 245, 252, 258, 265, 272, 278, 285, 292, 299, 306, 313, 319, 327, 334, 341, 348, 355, 362, 370, 377, 384, 392, 399, 407, 414, 422, 430, 437, 445, 453, 461, 468, 476, 484, 492, 500

Offset: 1

Author

Clark Kimberling, Feb 13 2011

Keywords

• nonn
• easy

Comments

See A187645.

Examples

			Write preliminary separate rankings:
1/4...2....27/4....16.....125/4...
....1...4.......9..16..25........36..49
Then replace each number by its rank, where ties are settled by ranking the top number before the bottom.
		

Crossrefs

Cf. A186145, A186149.

Programs

Mathematica

d=1/8; u=1/4; v=1; 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}] (* A186148 *)
Table[b[n],{n,1,100}] (* A186149 *)

Formula

a(n) = n + floor(((1/4)*n^3 - 1/8)^(1/2)).