A192804 - OEIS

%I #42 Jul 23 2021 09:02:49

%S 1,1,1,2,3,5,9,16,29,53,97,178,327,601,1105,2032,3737,6873,12641,

%T 23250,42763,78653,144665,266080,489397,900141,1655617,3045154,

%U 5600911,10301681,18947745,34850336,64099761,117897841,216847937,398845538

%N Constant term in the reduction of the polynomial 1+x+x^2+...+x^n by x^3->x^2+x+1. See Comments.

%C For discussions of polynomial reduction, see A192232 and A192744.

%C This sequence provides the most-significant place-values in the construction of a tribonacci code. - _James Dow Allen_, Jul 12 2021

%H James Dow Allen, <a href="https://fabpedigree.com/james/tribocod.htm">Constructing the Tribonacci Code</a>.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,0,-1).

%F a(n) = 2*a(n-1) - a(n-4).

%F a(n) = a(n-1) + a(n-2) + a(n-3) - 1. - _Alzhekeyev Ascar M_, Feb 05 2012

%F G.f.: ( 1-x-x^2 ) / ( (x-1)*(x^3+x^2+x-1) ). - _R. J. Mathar_, May 06 2014

%F a(n) - a(n-1) = A000073(n-1). - _R. J. Mathar_, May 06 2014

%e The first five polynomials p(n,x) and their reductions:

%e   p(1,x)=1 -> 1,

%e   p(2,x)=x+1 -> x+1,

%e   p(3,x)=x^2+x+1 -> x^2+x+1,

%e   p(4,x)=x^3+x^2+x+1 -> 2x^2+2x+2,

%e   p(5,x)=x^4+x^3+x^2+x+1 -> 4x^2+4*x+3, so that

%e A192804=(1,1,1,2,3,...), A000073=(0,1,1,2,4,...), A008937=(0,0,1,2,4,...).

%t q = x^3; s = x^2 + x + 1; z = 40;

%t p[0, x_] := 1; p[n_, x_] := x^n + p[n - 1, x];

%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   (* A192804 *)

%t u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]

%t   (* A000073 *)

%t u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}]

%t   (* A008937 *)

%Y Cf. A192744, A192232, A000073.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Jul 10 2011