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%I A242477 #25 Feb 16 2023 05:07:57
%S A242477 0,0,3,6,12,18,27,36,48,60,75,90,108,126,147,168,192,216,243,270,300,
%T A242477 330,363,396,432,468,507,546,588,630,675,720,768,816,867,918,972,1026,
%U A242477 1083,1140,1200,1260,1323,1386,1452,1518,1587,1656,1728,1800,1875
%N A242477 a(n) = floor(3*n^2/4).
%C A242477 The even-numbered terms are the same as the three - quarter squares; the odd-numbered terms are one less.
%H A242477 Vincenzo Librandi, <a href="/A242477/b242477.txt">Table of n, a(n) for n = 0..1000</a>
%H A242477 Craig Knecht, <a href="/A242477/a242477.png">Maximum number of octiamonds in a hexagon</a>.
%H A242477 Robert Munafo, <a href="http://mrob.com/pub/math/nu-sequences.html#MCS429697">Sequence MCS429697</a>.
%H A242477 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F A242477 a(n) = a(n-2) + 3*(n-1) for n>1, a(0) = a(1) = 0.
%F A242477 From _Bruno Berselli_, May 22 2014: (Start)
%F A242477 G.f.: 3*x^2/((1-x)^2*(1-x^2)).
%F A242477 a(n) = 3*A002620(n). (End)
%F A242477 Sum_{n>=2} 1/a(n) = Pi^2/18 + 1/3. - _Amiram Eldar_, Feb 16 2023
%t A242477 Table[Floor[3 n^2/4], {n, 0, 60}]
%t A242477 LinearRecurrence[{2,0,-2,1},{0,0,3,6},60] (* _Harvey P. Dale_, Sep 07 2019 *)
%o A242477 (Magma) [Floor(3*n^2/4): n in [0..60]];
%o A242477 (Sage) [3*floor(n^2/4) for n in (0..60)] # _Bruno Berselli_, May 22 2014
%Y A242477 Cf. A002620.
%K A242477 nonn,easy
%O A242477 0,3
%A A242477 _Vincenzo Librandi_, May 22 2014

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