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 A330520 #35 Dec 29 2021 11:38:47 %S A330520 0,0,2,8,24,54,108,192,320,500,750,1080,1512,2058,2744,3584,4608,5832, %T A330520 7290,9000,11000,13310,15972,19008,22464,26364,30758,35672,41160, %U A330520 47250,54000,61440,69632,78608,88434,99144,110808,123462,137180,152000,168000,185220,203742,223608,244904,267674 %N A330520 Sum of even integers <= n times the sum of odd integers <= n. %C A330520 Number of crossings in a grid formed by drawing n parallel infinite-length lines perpendicular to the previous number of lines. %C A330520 The sum of odd integers <= n is m^2 where m = round(n/2) is their number. The sum of even integers <= n is k(k+1) where k = floor(n/2) is their number. So a(n) = m^2*k(k+1), where the factor m appears three times. - _M. F. Hasler_, Dec 19 2019 %H A330520 Harvey P. Dale, <a href="/A330520/b330520.txt">Table of n, a(n) for n = 0..1000</a> %H A330520 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1,-5,5,1,-3,1). %F A330520 G.f.: 2*(x^2+x+1)*x^2/((x+1)^2*(1-x)^5). %F A330520 a(n) = 2 * A007009(n-1) for n>1. %F A330520 a(2k+i) = (k+i)^3 (k+1-i), with i = 0 or 1. - _M. F. Hasler_, Dec 19 2019 %F A330520 a(n) = A002378(floor(n/2)) * A000290(ceiling(n/2)). - _Bernard Schott_, Dec 22 2019 %t A330520 CoefficientList[Series[2 (x^2 + x + 1) x^2/((x + 1)^2*(1 - x)^5), {x, 0, 45}], x] (* _Michael De Vlieger_, Dec 22 2019 *) %t A330520 LinearRecurrence[{3,-1,-5,5,1,-3,1},{0,0,2,8,24,54,108},50] (* _Harvey P. Dale_, Dec 29 2021 *) %o A330520 (PARI) apply( A330520(n)=n\2*(n\2+1)*(n\/2)^2, [0..99]) \\ _M. F. Hasler_, Dec 19 2019 %Y A330520 Cf. A179824, A085537, A007009, A002620. %Y A330520 Cf. A000290 (sum of odd integers), A002378 (sum of even integers). %K A330520 nonn,easy %O A330520 0,3 %A A330520 _J. Stauduhar_, Dec 17 2019