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A100679
Floor of cube root of tetrahedral numbers.
0
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 33, 33, 34
OFFSET
0,4
COMMENTS
Tetrahedral numbers Tet(n) = A000292(n) = C(n+2, 3) = n(n+1)(n+2)/6 are obviously of order n^3, varying approximately with the cube of n. Taking the cube root and rounding down, we get the new sequence.
REFERENCES
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 83.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 4.
LINKS
Eric Weisstein's World of Mathematics, Tetrahedral Number
FORMULA
a(n) = floor((A000292(n))^(1/3)) = floor(Tet(n)^(1/3)) = floor(C(n+2, 3)^(1/3)) = floor((n(n+1)(n+2)/6)^(1/3)).
EXAMPLE
a(18) = 10 because floor((18*19*20/6)^(1/3)) = floor(1140^(1/3)) = 10.
MATHEMATICA
Table[Floor[Binomial[n + 2, 3]^(1/3)], {n, 0, 61}] (* Giovanni Resta, Jun 17 2016 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Dec 06 2004
EXTENSIONS
Edited by Giovanni Resta, Jun 17 2016
STATUS
approved