%I #91 Jul 27 2023 12:14:34
%S 0,-1,2,9,20,35,54,77,104,135,170,209,252,299,350,405,464,527,594,665,
%T 740,819,902,989,1080,1175,1274,1377,1484,1595,1710,1829,1952,2079,
%U 2210,2345,2484,2627,2774,2925,3080,3239,3402,3569,3740,3915,4094,4277
%N a(n) = n*(2*n-3).
%C Positive terms give a bisection of A000096. - _Omar E. Pol_, Dec 16 2016
%H G. C. Greubel, <a href="/A014107/b014107.txt">Table of n, a(n) for n = 0..5000</a>
%H Emily Barnard and Nathan Reading, <a href="https://arxiv.org/abs/1605.03524">Coxeter-biCatalan combinatorics</a>, arXiv:1605.03524 [math.CO], 2016. See p. 51.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = A100345(n, n - 3) for n > 2.
%F a(n) = A033537(n) - 8*n^2; A100035(a(n)) = 2 for n > 1. - _Reinhard Zumkeller_, Oct 31 2004
%F a(n) = A014106(-n) for all n in Z. - _Michael Somos_, Nov 06 2005
%F From _Michael Somos_, Nov 06 2005: (Start)
%F G.f.: x*(-1 + 5*x)/(1 - x)^3.
%F E.g.f: x*(-1 + 2*x)*exp(x). (End)
%F a(n) = A097070(n)/A000108(n - 2), n >= 2. - _Philippe Deléham_, Apr 12 2007
%F a(n) = 2*a(n-1) - a(n-2) + 4, n > 1; a(0) = 0, a(1) = -1, a(2) = 2. - _Zerinvary Lajos_, Feb 18 2008
%F a(n) = a(n-1) + 4*n - 5 with a(0) = 0. - _Vincenzo Librandi_, Nov 20 2010
%F a(n) = (2*n-1)*(n-1) - 1. Also, with an initial offset of -1, a(n) = (2*n-1)*(n+1) = 2*n^2 + n - 1. - _Alonso del Arte_, Dec 15 2012
%F (a(n) + 1)^2 + (a(n) + 2)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2n - 1)^2 starting with a(1) = -1. - _Jeffreylee R. Snow_, Sep 17 2013
%F a(n) = A014105(n-1) - 1 for all n in Z. - _Michael Somos_, Nov 23 2021
%F From _Amiram Eldar_, Feb 20 2022: (Start)
%F Sum_{n>=1} 1/a(n) = -2*(1 - log(2))/3.
%F Sum_{n>=1} (-1)^n/a(n) = Pi/6 + log(2)/3 + 2/3. (End)
%F For n > 0, A002378(a(n)) = A000384(n-1)*A000384(n). - _Charlie Marion_, May 21 2023
%p A014107:=n->n*(2*n-3); seq(A014107(n), n=0..100); # _Wesley Ivan Hurt_, Nov 19 2013
%t Table[2n^2 - 3n, {n, 0, 49}] (* _Alonso del Arte_, Dec 15 2012 *)
%t LinearRecurrence[{3,-3,1},{0,-1,2},50] (* _Harvey P. Dale_, Sep 18 2018 *)
%o (PARI) a(n)=n*(2*n-3)
%Y Cf. A000096, A000108, A014105, A033537, A097070, A100035, A100036, A100037, A100038, A100039, A100345, A002378, A000384.
%K sign,easy
%O 0,3
%A _N. J. A. Sloane_