|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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}]
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|