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%I #15 Sep 08 2022 08:45:31
%S 2,5,7,14,18,29,35,50,58,77,87,110,122,149,163,194,210,245,263,302,
%T 322,365,387,434,458,509,535,590,618,677,707,770,802,869,903,974,1010,
%U 1085,1123,1202,1242,1325,1367,1454,1498,1589,1635,1730,1778,1877,1927,2030
%N Antidiagonal sums of the triangle A133128.
%H Vincenzo Librandi, <a href="/A133146/b133146.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 (1, 2, -2, -1, 1).
%F First differences: a(n+1) - a(n) = A059029(n+1).
%F Bisections: a(2n+1) = A005918(n+1). a(2n) = A141631(n+1).
%F G.f.: (1+2*x)(2 - x + x^3)/((1-x)^3*(1+x)^2). - _R. J. Mathar_, Oct 15 2008
%F a(n) = 19/8 + 5*n/4 + 3*n^2/4 - (-1)^n*(n/4 + 3/8). - _R. J. Mathar_, Oct 15 2008
%F From _Harvey P. Dale_, Aug 26 2013: (Start)
%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5); a(0)=2, a(1)=5, a(2)=7, a(3)=14, a(4)=18. (End)
%e a(2) = A133128(2,0) + A133128(1,1) = 10 - 3 = 7.
%e a(3) = A133128(3,0) + A133128(2,1) = 17 - 3 = 14.
%t CoefficientList[Series[(1+2x)(2-x+x^3)/((1-x)^3(1+x)^2),{x,0,60}],x] (* or *) LinearRecurrence[{1,2,-2,-1,1},{2,5,7,14,18},60] (* _Harvey P. Dale_, Aug 26 2013 *)
%o (Magma) [19/8 +5*n/4 +3*n^2/4 -(-1)^n*(n/4+3/8): n in [0..60]]; // _Vincenzo Librandi_, Aug 10 2011
%K nonn,less
%O 0,1
%A _Paul Curtz_, Aug 27 2008
%E Edited and extended by _R. J. Mathar_, Oct 15 2008
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