|
| |
|
|
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
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,4
|
|
|
COMMENTS
| Tetrahedral numbers Tet(n) = C(n+3,3) = (n+1)(n+2)(n+3)/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
| 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+3, 3) 1/3)) = Floor(((n+1)(n+2)(n+3)/6)^(1/3))
|
|
|
EXAMPLE
| a(18) = 10 because Floor((19*20*21/6)^(1/3)) = Floor(1330^(1/3)) = Floor(10.9972445) = 10.
|
|
|
CROSSREFS
| Cf. A000292, A099179, A099179.
Sequence in context: A026233 A147954 A194204 * A195182 A189688 A092982
Adjacent sequences: A100676 A100677 A100678 * A100680 A100681 A100682
|
|
|
KEYWORD
| easy,nonn
|
|
|
AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), Dec 06 2004
|
| |
|
|
