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%I #57 Mar 15 2023 11:48:41
%S 0,1,2,5,8,11,16,21,26,33,40,47,56,65,74,85,96,107,120,133,146,161,
%T 176,191,208,225,242,261,280,299,320,341,362,385,408,431,456,481,506,
%U 533,560,587,616,645,674,705,736,767,800,833,866,901,936,971,1008
%N a(n) = floor(n*(n+1)*(n+2) / (n + n+1 + n+2)), which equals floor(n*(n+2)/3).
%C Satisfies a(n+1) - 2*a(n) + a(n-1) = (2/3)*(1 + w^(n+1) + w^(2n+2)), a(0)=0 & a(1)=1 where w is the imaginary cubic root of unity. - _Robert G. Wilson v_, Jun 24 2002
%C First differences have this pattern: (+1) +1 +1 +3 +3 +3 +5 +5 +5 +7 +7 +7 +9 +9 +9. - _Alexandre Wajnberg_, Dec 19 2005
%C In Duistermaat (2010) section 11.3 The Planar Four-Bar Link on page 516: "It follows from (4.3.2) that the number of k-periodic fibers of the QRT automorphism, counted with multiplicities, is equal to \nu(\tau^S)^k) = 3n^2 - 1 when k = 3n, 3n^2 + 2n when k = 3n + 1, 3n^2 + 4n + 1 when k = 3n + 2, for every integer n." - _Michael Somos_, Mar 14 2023
%D J. J. Duistermaat, Discrete Integrable Systems, 2010, Springer Science+Business Media.
%H J. Bader, <a href="http://www.tik.ee.ethz.ch/sop/people/baderj/?page=other.php">Kobon Triangles</a>
%H Gilles Clement and Johannes Bader, <a href="http://www.tik.ee.ethz.ch/sop/publicationListFiles/cb2007a.pdf">Tighter Upper Bound for the Number of Kobon Triangles</a>, Unpublished, 2007
%H Gilles Clement and Johannes Bader, <a href="/A006066/a006066.pdf">Tighter Upper Bound for the Number of Kobon Triangles</a>, Unpublished, 2007 [Cached copy, with permission]
%H Taylor Short, <a href="https://arxiv.org/abs/1807.11355">The saturation number of carbon nanocones and nanotubes</a>, arXiv:1807.11355 [math.CO], 2018.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/KobonTriangle.html">Kobon Triangle</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,1).
%F From _Ralf Stephan_, May 05 2004: (Start)
%F a(n) = n^2 - ceiling(n*(n-1)/3).
%F G.f.: x*(1+2x^2-x^3)/((1+x+x^2)(1-x)^3). (End)
%F a(n) = floor(n*(n+2)/3). - Saburo Tamura, sent by _Alexandre Wajnberg_, Dec 19 2005
%F a(3*n - 1) = 3*n^2 - 1, a(3*n) = 3*n^2 + 2*n, a(3*n + 1) = 3*n^2 + 4*n + 1. - _Michael Somos_, Mar 14 2023
%e G.f. = x + 2*x^2 + 5*x^3 + 8*x^4 + 11*x^5 + 16*x^6 + 21*x^7 + 26*x^8 + ... - _Michael Somos_, Mar 14 2023
%p A032765:=n->floor(n*(n+2)/3); seq(A032765(n), n=0..100); # _Wesley Ivan Hurt_, Dec 20 2013
%t Table[ Floor[ n(n + 1)(n + 2)/(n + (n + 1) + (n + 2))], {n, 0, 55}]
%t Table[Floor[n (n + 2)/3], {n, 0, 100}] (* _Wesley Ivan Hurt_, Dec 20 2013 *)
%t LinearRecurrence[{2,-1,1,-2,1},{0,1,2,5,8},60] (* _Harvey P. Dale_, Jun 06 2016 *)
%o (PARI) a(n)=n*(n+2)\3 \\ _Charles R Greathouse IV_, Jun 11 2015
%Y Cf. A001082, A032766.
%K nonn,easy
%O 0,3
%A _Patrick De Geest_, May 15 1998
%E Name change suggested by _Wesley Ivan Hurt_, Dec 20 2013
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