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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. 2
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified August 8 14:04 EDT 2020. Contains 336298 sequences. (Running on oeis4.)