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A032765
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Floor(n(n+1)(n+2) / (n+ n+1 + n+2)), which equals floor(n(n + 2)/3).
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16
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0, 1, 2, 5, 8, 11, 16, 21, 26, 33, 40, 47, 56, 65, 74, 85, 96, 107, 120, 133, 146, 161, 176, 191, 208, 225, 242, 261, 280, 299, 320, 341, 362, 385, 408, 431, 456, 481, 506, 533, 560, 587, 616, 645, 674, 705, 736, 767, 800, 833, 866, 901, 936, 971, 1008
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OFFSET
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0,3
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COMMENTS
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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
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
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LINKS
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Table of n, a(n) for n=0..54.
J. Bader, Kobon Triangles
Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007
Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007 [Cached copy, with permission]
Eric Weisstein's World of Mathematics, Kobon Triangle
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).
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FORMULA
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a(n) = n^2 - ceiling(n(n-1)/3). G.f.: [x(1+2x^2-x^3)]/[(1+x+x^2)(1-x)^3]. - Ralf Stephan, May 05 2004
a(n) = Floor [n(n+2)/3]. - Saburo Tamura, sent by Alexandre Wajnberg, Dec 19 2005
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MAPLE
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A032765:=n->floor(n*(n+2)/3); seq(A032765(n), n=0..100); # Wesley Ivan Hurt, Dec 20 2013
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MATHEMATICA
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Table[ Floor[ n(n + 1)(n + 2)/(n + (n + 1) + (n + 2))], {n, 0, 55}]
Table[Floor[n (n + 2)/3], {n, 0, 100}] (* Wesley Ivan Hurt, Dec 20 2013 *)
LinearRecurrence[{2, -1, 1, -2, 1}, {0, 1, 2, 5, 8}, 60] (* Harvey P. Dale, Jun 06 2016 *)
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PROG
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(PARI) a(n)=n*(n+2)\3 \\ Charles R Greathouse IV, Jun 11 2015
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CROSSREFS
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Cf. A001082, A032766.
Sequence in context: A184747 A130258 A186496 * A261416 A300272 A192147
Adjacent sequences: A032762 A032763 A032764 * A032766 A032767 A032768
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KEYWORD
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nonn,easy
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AUTHOR
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Patrick De Geest, May 15 1998
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EXTENSIONS
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Name change suggested by Wesley Ivan Hurt, Dec 20 2013
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STATUS
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approved
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