OFFSET
0,3
COMMENTS
Old definition was "Constant term in the reduction of (x^2+x+1)^n by x^3 -> x^2+x+1." For discussions of polynomial reduction, see A192232 and A192744.
From Bob Selcoe, Jun 10 2014: (Start)
a(n) is the trinomial transform of tribonacci numbers. (i.e., A027907(n) transform of A000073(n+2)).
Let the m-nacci numbers be denoted by M"(n). Examples: Fibonacci numbers are 2"(n); tribonacci numbers are 3"(n); 137-nacci numbers are 137"(n). Then the m-nomial transform of M" is M"(m*n), where M"(0)=1 and M"(n)=0 when n<0. Therefore a(n) = 3"(3n). (End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
László Németh, The trinomial transform triangle, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also arXiv:1807.07109 [math.NT], 2018.
Index entries for linear recurrences with constant coefficients, signature (7,-5,1).
FORMULA
G.f.: (1 - 6*x + 2*x^2)/(1 - 7*x + 5*x^2 - x^3). - R. J. Mathar, May 06 2014
a(n) = A000073(3n+2), n>0. - Bob Selcoe, Jun 10 2014
EXAMPLE
G.f. = 1 + x + 4*x^2 + 24*x^3 + 149*x^4 + 927*x^5 + 5768*x^6 + ...
MATHEMATICA
q = x^3; s = x^2 + x + 1; z = 40;
p[n_, x_] := (x^2 + x + 1)^n;
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}] (* A192806 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192807 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}] (* A099464 *)
LinearRecurrence[{7, -5, 1}, {1, 1, 4}, 50] (* G. C. Greubel, Jan 02 2019 *)
PROG
(PARI) {a(n) = polcoeff( lift( (1 + x + x^2)^n * Mod(1, x^3 - x^2 - x - 1)), 0)}; /* Michael Somos, Jun 17 2014 */
(PARI) my(x='x+O('x^30)); Vec((1-6*x+2*x^2)/(1-7*x+5*x^2-x^3)) \\ G. C. Greubel, Jan 02 2019
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-6*x+2*x^2)/(1-7*x+5*x^2-x^3) )); // G. C. Greubel, Jan 02 2019
(Sage) ((1-6*x+2*x^2)/(1-7*x+5*x^2-x^3)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 02 2019
(GAP) a:=[1, 1, 4];; for n in [4..25] do a[n]:=7*a[n-1]-5*a[n-2]+a[n-3]; od; Print(a); # Muniru A Asiru, Jan 02 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 10 2011
EXTENSIONS
Edited by N. J. A. Sloane, Jun 03 2018
STATUS
approved