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A379852 a(n) = floor(8*n^3/27).

%I A379852 #16 Jan 30 2025 03:54:55
%S A379852 0,0,2,8,18,37,64,101,151,216,296,394,512,650,813,1000,1213,1455,1728,
%T A379852 2032,2370,2744,3154,3605,4096,4629,5207,5832,6504,7226,8000,8826,
%U A379852 9709,10648,11645,12703,13824,15008,16258,17576,18962,20421,21952,23557,25239
%N A379852 a(n) = floor(8*n^3/27).
%C A379852 a(n) is the integer part of the area of the largest triangle that can be inscribed in the region bounded by the parabola y = x^2, the x-axis, and the line x = n.
%C A379852 To estimate the integral int_{x = 0..n} x^2 dx by means of a triangle, we find that the triangle with the largest area that can be inscribed in the region bounded by the parabola y = x^2, the x-axis and the line x = n is the right triangle with vertices (n/3, 0), (n, 0) and (n, (8/9)*n^2), whose area is (2n/3)^3 and a(n) has been defined as floor((2n/3)^3).
%H A379852 Paolo Xausa, <a href="/A379852/b379852.txt">Table of n, a(n) for n = 0..10000</a>
%H A379852 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,0,0,0,0,1,-3,3,-1).
%F A379852 a(n) = floor(A016743(n)/27).
%F A379852 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-9) - 3*a(n-10) + 3*a(n-11) - a(n-12) for n >= 12. - _Pontus von Brömssen_, Jan 14 2025
%e A379852 If n = 2, the largest triangle that can be inscribed in the region bounded by the parabola y = x^2, the x-axis, and the line x = n is the right triangle with vertices (2/3,0),(2,0) and (2,32/9), whose area is 64/27. Since floor(64/27) = 2, it follows that a(2) = 2.
%t A379852 Floor[8/27*Range[0, 50]^3] (* _Paolo Xausa_, Jan 30 2025 *)
%Y A379852 Cf. A016743, A375473.
%K A379852 nonn,easy
%O A379852 0,3
%A A379852 _Gonzalo Martínez_ and Javier Astudillo, Jan 04 2025