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Search: id:A100679
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| A100679 |
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Floor of cube root of tetrahedral numbers. |
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+0
1
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| 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
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OFFSET
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0,4
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COMMENT
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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.
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REFERENCES
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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.
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LINKS
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R. Jovanovic, First 2500 Tetrahedral numbers.
Eric Weisstein's World of Mathematics, Tetrahedral Number
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FORMULA
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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))
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EXAMPLE
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a(18) = 10 because Floor((19*20*21/6)^(1/3)) = Floor(1330^(1/3)) = Floor(10.9972445) = 10.
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CROSSREFS
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Cf. A000292, A099179, A099179.
Sequence in context: A026261 A026233 A147954 this_sequence A092982 A030566 A007963
Adjacent sequences: A100676 A100677 A100678 this_sequence A100680 A100681 A100682
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KEYWORD
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easy,nonn
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AUTHOR
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Jonathan Vos Post (jvospost3(AT)gmail.com), Dec 06 2004
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