OFFSET
0,2
COMMENTS
Partial sum of 1,1,1,2,4,8,...
Binomial transform of abs(A073097).
Binomial transform is A094374.
Partial sums are in A006127. - Paul Barry, Aug 05 2004
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
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
Row lengths of the triangle in A198069. - Reinhard Zumkeller, May 26 2013
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
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein, Complete Sequence.
Index entries for linear recurrences with constant coefficients, signature (3,-2).
FORMULA
a(n) = (2^n - 0^n)/2 + 1.
a(n) = 3*a(n-1) - 2*a(n-2).
a(2*n) = 2*a(2*n-1) - 1, n>0.
Row sums of triangle A135225. - Gary W. Adamson, Nov 23 2007
a(n) = A131577(n) + 1. - Paul Curtz, Aug 07 2008
a(n) = 2*a(n-1) - 1 for n>1, a(0)=1, a(1)=2. - Philippe Deléham, Sep 25 2009
E.g.f.: exp(x)*(1 + sinh(x)). - Arkadiusz Wesolowski, Aug 13 2012
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
a(n) = 2^(n-1) +1 = A000051(n-1) for n>0. - M. F. Hasler, Sep 22 2013
EXAMPLE
G.f. = 1 + 2*x + 3*x^2 + 5*x^3 + 9*x^4 + 17*x^5 + 33*x^6 + 65*x^7 + ...
MAPLE
1, seq((2^n - 0^n)/2 +1, n=1..40); # G. C. Greubel, Nov 06 2019
MATHEMATICA
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 *)
a[ n_]:= If[n<0, 0, 1 + Quotient[2^n, 2]]; (* Michael Somos, May 26 2014 *)
a[ n_]:= SeriesCoefficient[(1-x-x^2)/((1-x)(1-2x)), {x, 0, n}]; (* Michael Somos, May 26 2014 *)
LinearRecurrence[{3, -2}, {1, 2, 3}, 40] (* Harvey P. Dale, Aug 09 2015 *)
PROG
(Magma) [(2^n-0^n)/2+1: n in [0..40]]; // Vincenzo Librandi, Jun 10 2011
(Magma) R<x>:=PowerSeriesRing(Integers(), 35); Coefficients(R!( (1-x-x^2)/((1-x)*(1-2*x)))); // Marius A. Burtea, Oct 25 2019
(PARI) a(n)=2^n\2+1 \\ Charles R Greathouse IV, Apr 05 2013
(PARI) Vec((1-x-x^2)/((1-x)*(1-2*x))+O(x^40)) \\ Charles R Greathouse IV, Apr 05 2013
(Sage) [(2^n - 0^n)/2 + 1 for n in (0..40)] # G. C. Greubel, Nov 06 2019
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Apr 28 2004
STATUS
approved