A192761 Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

0, 1, 5, 11, 22, 40, 70, 119, 199, 329, 540, 882, 1436, 2333, 3785, 6135, 9938, 16092, 26050, 42163, 68235, 110421, 178680, 289126, 467832, 756985, 1224845, 1981859, 3206734, 5188624, 8395390, 13584047, 21979471, 35563553, 57543060, 93106650

Offset

0,3

Comments

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.

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

Links

Table of n, a(n) for n=0..35.

Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Formula

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]

Mathematica

q = x^2; s = x + 1; z = 40;
p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n + 3;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] :=
FixedPoint[(s PolynomialQuotient @@ #1 +
       PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
  (* A022318 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192761 *)

Crossrefs

Cf. A192744, A192232, partial sums of A022318.

Sequence in context: A222548 A024921 A189978 * A152533 A228485 A161896

Adjacent sequences: A192758 A192759 A192760 * A192762 A192763 A192764

Keyword

nonn,easy

Author

Clark Kimberling, Jul 09 2011

Status

approved