%I #22 Mar 11 2019 10:01:20
%S 0,0,0,1,2,4,8,13,20,30,42,57,76,98,124,155,190,230,276,327,384,448,
%T 518,595,680,772,872,981,1098,1224,1360,1505,1660,1826,2002,2189,2388,
%U 2598,2820,3055,3302,3562,3836,4123,4424,4740,5070,5415,5776,6152,6544,6953,7378,7820
%N a(n) is the number of solutions to x+y+z = 0 mod 3, where 1 <= x < y < z <= n.
%C (1+x)*(1+x^2)*(1+x^3) / ( (1-x)*(1-x^2)*(1-x^3)*(1-x^4)) is the Poincaré series [or Poincare series] (or Molien series) for H^*(O_4(q); F_2).
%D A. Adem and R. J. Milgram, Cohomology of Finite Groups, Springer-Verlag, 2nd. ed., 2004; p. 233.
%H D. Z. Dokovic, B. H. Smith, <a href="https://doi.org/10.1016/j.laa.2007.08.022">Quaternionic matrices: Unitary similarity, simultaneous triangularization and some trace identities</a>, Lin. Alg. Applic. 428 (4) (2008) 890-910
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,2,-3,3,-1).
%F G.f.: x^3*(1+x)*(1+x^2)*(1+x^3) / ( (1-x)*(1-x^2)*(1-x^3)*(1-x^4)). - _N. J. A. Sloane_, Mar 17 2004
%F a(n) = (binomial(n,3)+2*floor(n/3))/3. - _Claude Morin_, Mar 06 2012
%F G.f.: x^3*(1-x+x^2) / ( (1+x+x^2)*(x-1)^4 ). - _R. J. Mathar_, Dec 18 2014
%F a(n+5) = A014125(n)-A014125(n+1)+A014125(n+2). - _R. J. Mathar_, Mar 11 2019
%t LinearRecurrence[{3,-3,2,-3,3,-1},{0,0,0,1,2,4},60] (* _Harvey P. Dale_, Nov 22 2014 *)
%Y The third diagonal of A061865.
%K nonn,easy
%O 0,5
%A _Antti Karttunen_, May 11 2001