This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A100679 #10 Feb 16 2025 08:32:55
%S A100679 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,
%T A100679 16,17,17,18,18,19,19,20,20,21,22,22,23,23,24,24,25,25,26,26,27,28,28,
%U A100679 29,29,30,30,31,31,32,33,33,34
%N A100679 Floor of cube root of tetrahedral numbers.
%C A100679 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.
%D A100679 J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 83.
%D A100679 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.
%H A100679 R. Jovanovic, <a href="http://milan.milanovic.org/math/Math.php?akcija=SviTetra">First 2500 Tetrahedral numbers</a>.
%H A100679 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/TetrahedralNumber.html">Tetrahedral Number</a>
%F A100679 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)).
%e A100679 a(18) = 10 because floor((18*19*20/6)^(1/3)) = floor(1140^(1/3)) = 10.
%t A100679 Table[Floor[Binomial[n + 2, 3]^(1/3)], {n, 0, 61}] (* _Giovanni Resta_, Jun 17 2016 *)
%Y A100679 Cf. A000292, A099179, A099179.
%K A100679 easy,nonn
%O A100679 0,4
%A A100679 _Jonathan Vos Post_, Dec 06 2004
%E A100679 Edited by _Giovanni Resta_, Jun 17 2016