login
A192761
Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.
4
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
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 09 2011
STATUS
approved