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A094683
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Juggler sequence: if n mod 2 = 0 then floor(sqrt(n)) else floor(n^(3/2)).
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21
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0, 1, 1, 5, 2, 11, 2, 18, 2, 27, 3, 36, 3, 46, 3, 58, 4, 70, 4, 82, 4, 96, 4, 110, 4, 125, 5, 140, 5, 156, 5, 172, 5, 189, 5, 207, 6, 225, 6, 243, 6, 262, 6, 281, 6, 301, 6, 322, 6, 343, 7, 364, 7, 385, 7, 407, 7, 430, 7, 453, 7, 476, 7, 500, 8, 524, 8, 548, 8, 573, 8, 598, 8, 623, 8, 649
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OFFSET
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0,4
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COMMENTS
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Interspersion of A000093 and A000196. - Michel Marcus, Nov 11 2013
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REFERENCES
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C. Pickover, Computers and the Imagination, St. Martin's Press, NY, 1991, p. 233.
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LINKS
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Karl V. Boddy, Table of n, a(n) for n = 0..10000
H. J. Smith, Juggler Sequence
Eric Weisstein's World of Mathematics, Juggler Sequence
Wikipedia, Juggler sequence
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MAPLE
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A094683 :=proc(n) if n mod 2 = 0 then RETURN(floor(sqrt(n))) else RETURN(floor(n^(3/2))); end if; end proc;
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MATHEMATICA
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Table[If[EvenQ[n], Floor[Sqrt[n]], Floor[n^(3/2)]], {n, 0, 100}] (* Indranil Ghosh, Apr 07 2017 *)
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PROG
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(PARI) for(n=0, 100, print1(if(n%2, sqrtint(n^3), sqrtint(n)), ", ")) \\ Indranil Ghosh, Apr 08 2017
(Python)
import math
from sympy import sqrt
def a(n): return int(math.floor(sqrt(n))) if n%2 == 0 else int(math.floor(n**(3/2)))
print [a(n) for n in xrange(0, 101)] # Indranil Ghosh, Apr 08 2017
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CROSSREFS
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Cf. A007320, A007321, A094684, A094685, A094716.
Cf. A000093, A000196.
Sequence in context: A111187 A257323 A293559 * A094685 A095396 A051308
Adjacent sequences: A094680 A094681 A094682 * A094684 A094685 A094686
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane, Jun 09 2004
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STATUS
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approved
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