A100679 Floor of cube root of tetrahedral numbers.

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

Table of n, a(n) for n=0..61.

R. Jovanovic, First 2500 Tetrahedral numbers.

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

Cf. A000292, A099179, A099179.

Sequence in context: A222422 A194204 A348110 * A226190 A195182 A225875

Adjacent sequences: A100676 A100677 A100678 * A100680 A100681 A100682

Keyword

easy,nonn

Author

Jonathan Vos Post, Dec 06 2004

Extensions

Edited by Giovanni Resta, Jun 17 2016

Status

approved