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%I #38 Dec 31 2023 10:30:50
%S 1,6,21,81,306,1161,4401,16686,63261,239841,909306,3447441,13070241,
%T 49553046,187869861,712268721,2700415746,10238053401,38815407441,
%U 147160382526,557927369901,2115263257281,8019571881546,30404505416481,115272231894081
%N Expansion of (3*x+1)/(1-3*x-3*x^2).
%C 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.
%C 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
%C From _Klaus Purath_, Mar 09 2023: (Start)
%C 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.
%C 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)
%H Vincenzo Librandi, <a href="/A108306/b108306.txt">Table of n, a(n) for n = 0..1000</a>
%H Martin Burtscher, Igor Szczyrba and Rafał Szczyrba, <a href="http://www.emis.de/journals/JIS/VOL18/Szczyrba/sz3.html">Analytic Representations of the n-anacci Constants and Generalizations Thereof</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,3).
%F Recurrence: a(0)=1; a(1)=6; a(n) = 3a(n-1) + 3a(n-2) - _N-E. Fahssi_, Apr 20 2008
%p seriestolist(series((3*x+1)/(1-3*x-3*x^2), x=0,25));
%t CoefficientList[Series[(3 x + 1) / (1 - 3 x - 3 x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Aug 01 2016 *)
%o (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
%Y Cf. A055271.
%Y Cf. A084057.
%K easy,nonn
%O 0,2
%A _Creighton Dement_, Jul 24 2005
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