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%I #20 Sep 08 2022 08:45:49
%S 1,1,3,7,17,33,59,95,145,209,291,391,513,657,827,1023,1249,1505,1795,
%T 2119,2481,2881,3323,3807,4337,4913,5539,6215,6945,7729,8571,9471,
%U 10433,11457,12547,13703,14929,16225,17595,19039,20561
%N a(n) = (4*n^3 - 6*n^2 + 8*n + 9 + 3*(-1)^n)/12.
%C Starting with a(2), the sum of the first and last term in row n-1 of the Janet table A172002.
%H Vincenzo Librandi, <a href="/A168582/b168582.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-2,3,-1).
%F a(n+2) = A168388(n) + A168380(n), n >= 0.
%F a(2n) = A168547(n);
%F a(2n+1) = A168574(n).
%F G.f.: (1 - 2*x + x^4 + 2*x^2 + 2*x^3)/((1+x)*(x-1)^4). - _R. J. Mathar_, Jun 27 2011
%F E.g.f.: (1/12)*((4*x^3 + 6*x^2 + 6*x + 9)*exp(x) + 3*exp(-x)). - _G. C. Greubel_, Jul 26 2016
%t Table[(4*n^3 - 6*n^2 + 8*n + 9 + 3*(-1)^n)/12, {n,0,50}] (* _G. C. Greubel_, Jul 26 2016 *)
%o (Magma) [2*n/3 +3/4 -n^2/2 +n^3/3 +(-1)^n/4: n in [0..40]]; // _Vincenzo Librandi_, Aug 06 2011
%o (PARI) a(n)=(4*n^3-6*n^2+8*n+9+3*(-1)^n)/12 \\ _Charles R Greathouse IV_, Jul 26 2016
%Y Cf. A137928 (first differences).
%K nonn,easy
%O 0,3
%A _Paul Curtz_, Nov 30 2009
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