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A108306
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Expansion of (3*x+1)/(1-3*x-3*x^2).
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8
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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
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OFFSET
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0,2
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COMMENTS
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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
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)
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LINKS
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FORMULA
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Recurrence: a(0)=1; a(1)=6; a(n) = 3a(n-1) + 3a(n-2) - N-E. Fahssi, Apr 20 2008
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MAPLE
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seriestolist(series((3*x+1)/(1-3*x-3*x^2), x=0, 25));
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MATHEMATICA
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CoefficientList[Series[(3 x + 1) / (1 - 3 x - 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 01 2016 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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