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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 November 30 12:38 EST 2023. Contains 367461 sequences. (Running on oeis4.)