A108306 Expansion of (3*x+1)/(1-3*x-3*x^2).

1, 6, 21, 81, 306, 1161, 4401, 16686, 63261, 239841, 909306, 3447441, 13070241, 49553046, 187869861, 712268721, 2700415746, 10238053401, 38815407441, 147160382526, 557927369901, 2115263257281, 8019571881546, 30404505416481, 115272231894081

Offset: 0

Author

Creighton Dement, Jul 24 2005

Keywords

• easy
• nonn

Comments

Binomial transform is A055271. May be seen as a ibasefor-transform of the zero-sequence A000004 with respect to the floretion given in the program code.

The sequence is the INVERT transform of (1, 5, 10, 20, 40, 80, 160, ...) and can be obtained by extracting the upper left terms of matrix powers of [(1,5); (1,2)]. These results are a case (a=5, b=2) of the conjecture: The INVERT transform of a sequence starting (1, a, a*b, a*b^2, a*b^3, ...) is equivalent to extracting the upper left terms of powers of the 2x2 matrix [(1,a); (1,b)]. - Gary W. Adamson, Jul 31 2016

From Klaus Purath, Mar 09 2023: (Start)

For any terms (a(n), a(n+1)) = (x, y), -3*x^2 - 3*x*y + y^2 = 15*(-3)^n = A082111(2)*(-3)^n. This is valid in general for all recursive sequences (t) with constant coefficients (3,3) and t(0) = 1: -3*x^2 - 3*x*y + y^2 = A082111(t(1)-4)*(-3)^n.

By analogy to this, for three consecutive terms (x, y, z) of any sequence (t) of the form (3,3) with t(0) = 1: y^2 - x*z = A082111(t(1)-4)*(-3)^n. (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Martin Burtscher, Igor Szczyrba and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (3,3).

Crossrefs

Cf. A055271.

Cf. A084057.

Programs

Magma

I:=[1,6]; [n le 2 select I[n] else 3*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 01 2016

Maple

seriestolist(series((3*x+1)/(1-3*x-3*x^2), x=0,25));

Mathematica

CoefficientList[Series[(3 x + 1) / (1 - 3 x - 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 01 2016 *)

Formula

Recurrence: a(0)=1; a(1)=6; a(n) = 3a(n-1) + 3a(n-2) - N-E. Fahssi, Apr 20 2008