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

1,2

Comments

See A187645.

Links

Table of n, a(n) for n=1..88.

Formula

a(n)=n+floor(((1/4)n^3-1/8))^(1/2)), A186148 .

b(n)=n+floor((4n^3+1/2)^(1/3)), A186149.

Example

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.

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 *)

Crossrefs

Cf. A186145, A186149.

Sequence in context: A213510 A130257 A276498 * A079511 A092757 A062430

Adjacent sequences: A186145 A186146 A186147 * A186149 A186150 A186151

Keyword

nonn

Author

Clark Kimberling, Feb 13 2011

Status

approved