%I #19 Jun 13 2015 00:53:53
%S 0,1,5,11,22,40,70,119,199,329,540,882,1436,2333,3785,6135,9938,16092,
%T 26050,42163,68235,110421,178680,289126,467832,756985,1224845,1981859,
%U 3206734,5188624,8395390,13584047,21979471,35563553,57543060,93106650
%N Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.
%C The titular polynomial is defined recursively by p(n,x) = x*(n-1,x) + n + 3 for n > 0, where p(0,x) = 1. For discussions of polynomial reduction, see A192232 and A192744.
%C Construct a triangle with T(n,0) = n*(n+1)+1 and T(n,n) = (n+1)*(n+2)/2 starting at n=0. Define the interior terms by T(r,c) = T(r-2,c-1) + T(r-1,c). The sequence of its row sums is 1, 6, 17, 39, 79, 149, 268, 467,... and the first differences of these (the sum of the terms in row(n) less those in row(n-1)) equals a(n+1). - _J. M. Bergot_, Mar 10 2013
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-1,1).
%F a(n) = 3*a(n-1)-2*a(n-2)-a(n-3)+a(n-4). G.f.: x*(2*x^2-2*x-1) / ((x-1)^2*(x^2+x-1)). [_Colin Barker_, Dec 08 2012]
%t q = x^2; s = x + 1; z = 40;
%t p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n + 3;
%t Table[Expand[p[n, x]], {n, 0, 7}]
%t reduce[{p1_, q_, s_, x_}] :=
%t FixedPoint[(s PolynomialQuotient @@ #1 +
%t PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
%t t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
%t u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
%t (* A022318 *)
%t u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
%t (* A192761 *)
%Y Cf. A192744, A192232, partial sums of A022318.
%K nonn,easy
%O 0,3
%A _Clark Kimberling_, Jul 09 2011
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