A192804 Constant term in the reduction of the polynomial 1+x+x^2+...+x^n by x^3->x^2+x+1. See Comments.
1, 1, 1, 2, 3, 5, 9, 16, 29, 53, 97, 178, 327, 601, 1105, 2032, 3737, 6873, 12641, 23250, 42763, 78653, 144665, 266080, 489397, 900141, 1655617, 3045154, 5600911, 10301681, 18947745, 34850336, 64099761, 117897841, 216847937, 398845538
Offset: 0
Examples
The first five polynomials p(n,x) and their reductions: p(1,x)=1 -> 1, p(2,x)=x+1 -> x+1, p(3,x)=x^2+x+1 -> x^2+x+1, p(4,x)=x^3+x^2+x+1 -> 2x^2+2x+2, p(5,x)=x^4+x^3+x^2+x+1 -> 4x^2+4*x+3, so that A192804=(1,1,1,2,3,...), A000073=(0,1,1,2,4,...), A008937=(0,0,1,2,4,...).
Links
- James Dow Allen, Constructing the Tribonacci Code.
- Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).
Programs
-
Mathematica
q = x^3; s = x^2 + x + 1; z = 40; p[0, x_] := 1; p[n_, x_] := x^n + p[n - 1, x]; 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}] (* A192804 *) u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A000073 *) u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}] (* A008937 *)
Formula
a(n) = 2*a(n-1) - a(n-4).
a(n) = a(n-1) + a(n-2) + a(n-3) - 1. - Alzhekeyev Ascar M, Feb 05 2012
G.f.: ( 1-x-x^2 ) / ( (x-1)*(x^3+x^2+x-1) ). - R. J. Mathar, May 06 2014
a(n) - a(n-1) = A000073(n-1). - R. J. Mathar, May 06 2014
Comments