A094683 Juggler sequence: if n mod 2 = 0 then floor(sqrt(n)) else floor(n^(3/2)).

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

Offset

0,4

Comments

Interspersion of A000093 and A000196. - Michel Marcus, Nov 11 2013

References

C. Pickover, Computers and the Imagination, St. Martin's Press, NY, 1991, p. 233.

Links

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

Maple

A094683 :=proc(n) if n mod 2 = 0 then RETURN(floor(sqrt(n))) else RETURN(floor(n^(3/2))); end if; end proc;

Mathematica

Table[If[EvenQ[n], Floor[Sqrt[n]], Floor[n^(3/2)]], {n, 0, 100}] (* Indranil Ghosh, Apr 07 2017 *)

Prog

(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 range(51)]) # Indranil Ghosh, Apr 08 2017
(Python)
from math import isqrt
def A094683(n): return isqrt(n**3 if n % 2 else n) # Chai Wah Wu, Feb 18 2022

Crossrefs

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

Keyword

nonn

Author

N. J. A. Sloane, Jun 09 2004

Status

approved