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%I #72 Sep 08 2022 08:45:13
%S 1,2,3,5,9,17,33,65,129,257,513,1025,2049,4097,8193,16385,32769,65537,
%T 131073,262145,524289,1048577,2097153,4194305,8388609,16777217,
%U 33554433,67108865,134217729,268435457,536870913,1073741825,2147483649,4294967297,8589934593
%N Expansion of (1-x-x^2)/((1-x)*(1-2*x)).
%C Partial sum of 1,1,1,2,4,8,...
%C Binomial transform of abs(A073097).
%C Binomial transform is A094374.
%C Partial sums are in A006127. - _Paul Barry_, Aug 05 2004
%C An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 2, 8, 32 and 128, lead to this sequence. For the central square these vectors lead to the companion sequence A011782. - _Johannes W. Meijer_, Aug 15 2010
%C This sequence has a(0) = 1 and for all n > 0, a(n) = 2^(n-1)+1. Consequently 2*a(n) >= a(n+1) for all n > 0 and the sequence is complete. - _Frank M Jackson_, Jan 29 2012
%C Row lengths of the triangle in A198069. - _Reinhard Zumkeller_, May 26 2013
%C Take A007843 and count the repeated values. The result is 1,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,.... Build a third sequence, where a(1) = 1 and a(n) equals the length (greater than 1) of the shortest palindromic subsequence of consecutive terms of the second sequence starting with a(n) of the second sequence. The third sequence starts 1,3,5,3,9,3,5,3,17,3,5,3,9,3,5,3,33,.... Conjecturally, in the third sequence: (1) the indices of the first occurrence of each value form the present sequence and (2) for n>1, a(n) is in the a(n-1)-th position. - _Ivan N. Ianakiev_, Aug 20 2019
%H Vincenzo Librandi, <a href="/A094373/b094373.txt">Table of n, a(n) for n = 0..1000</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/CompleteSequence.html">Complete Sequence</a>.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2).
%F a(n) = (2^n - 0^n)/2 + 1.
%F a(n) = 3*a(n-1) - 2*a(n-2).
%F a(2*n) = 2*a(2*n-1) - 1, n>0.
%F Row sums of triangle A135225. - _Gary W. Adamson_, Nov 23 2007
%F a(n) = A131577(n) + 1. - _Paul Curtz_, Aug 07 2008
%F a(n) = 2*a(n-1) - 1 for n>1, a(0)=1, a(1)=2. - _Philippe Deléham_, Sep 25 2009
%F E.g.f.: exp(x)*(1 + sinh(x)). - _Arkadiusz Wesolowski_, Aug 13 2012
%F G.f.: G(0), where G(k)= 1 + 2^k*x/(1 - x/(x + 2^k*x/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Jul 26 2013
%F a(n) = 2^(n-1) +1 = A000051(n-1) for n>0. - _M. F. Hasler_, Sep 22 2013
%e G.f. = 1 + 2*x + 3*x^2 + 5*x^3 + 9*x^4 + 17*x^5 + 33*x^6 + 65*x^7 + ...
%p 1, seq((2^n - 0^n)/2 +1, n=1..40); # _G. C. Greubel_, Nov 06 2019
%t CoefficientList[Series[(1-x-x^2)/((1-x)*(1-2*x)), {x, 0, 40}], x] (* or *) Join[{1}, LinearRecurrence[{3, -2}, {2, 3}, 40]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 22 2012 *)
%t a[ n_]:= If[n<0, 0, 1 + Quotient[2^n, 2]]; (* _Michael Somos_, May 26 2014 *)
%t a[ n_]:= SeriesCoefficient[(1-x-x^2)/((1-x)(1-2x)), {x, 0, n}]; (* _Michael Somos_, May 26 2014 *)
%t LinearRecurrence[{3,-2},{1,2,3},40] (* _Harvey P. Dale_, Aug 09 2015 *)
%o (Magma) [(2^n-0^n)/2+1: n in [0..40]]; // _Vincenzo Librandi_, Jun 10 2011
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 35); Coefficients(R!( (1-x-x^2)/((1-x)*(1-2*x)))); // _Marius A. Burtea_, Oct 25 2019
%o (PARI) a(n)=2^n\2+1 \\ _Charles R Greathouse IV_, Apr 05 2013
%o (PARI) Vec((1-x-x^2)/((1-x)*(1-2*x))+O(x^40)) \\ _Charles R Greathouse IV_, Apr 05 2013
%o (Sage) [(2^n - 0^n)/2 + 1 for n in (0..40)] # _G. C. Greubel_, Nov 06 2019
%o (GAP) a:=[2,3];; for n in [3..40] do a[n]:=3*a[n-1]-2*a[n-2]; od; Concatenation([1], a); # _G. C. Greubel_, Nov 06 2019
%Y Apart from the initial 1, identical to A000051.
%Y Cf. A135225.
%Y Column k=1 of A152977.
%Y Row n=2 of A238016.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Apr 28 2004
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