%I
%S 1,1,3,6,15,33,78,177,411,942,2175,5001,11526,26529,61107,140694,
%T 324015,746097,1718142,3956433,9110859,20980158,48312735,111253209,
%U 256191414,589951041,1358525283,3128378406,7203954255,16589089473,38200952238,87968220657
%N Number of compositions of n when each even part can be of two kinds.
%C Row sums of A105475.
%C Starting (1, 3, 6, 15,...) = sum of (n1)th row terms of triangle A140168.  _Gary W. Adamson_, May 10 2008
%C a(n) is also the number of compositions of n using 1's and 2's such that each run of like numbers can be grouped arbitrarily. For example, a(4) = 15 because 4 = (1)+(1)+(1)+(1) = (1+1)+(1)+(1) = (1)+(1+1)+(1) = (1)+(1)+(1+1) = (1+1)+(1+1) = (1+1+1)+(1) = (1)+(1+1+1) = (1+1+1+1) = (2)+(1)+(1) = (1)+(2)+(1) = (1)+(1)+(2) = (2)+(1+1) = (1+1)+(2) = (2)+(2) = (2+2).  Martin J. Erickson (erickson(AT)truman.edu), Dec 09 2008
%C An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 69, 261, 321 and 324, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A006138.  _Johannes W. Meijer_, Aug 15 2010
%C Inverse INVERT transform of the left shifted sequence gives A000034.
%C Eigensequence of the triangle
%C 1,
%C 2, 1,
%C 1, 2, 1,
%C 2, 1, 2, 1,
%C 1, 2, 1, 2, 1,
%C 2, 1, 2, 1, 2, 1,
%C 1, 2, 1, 2, 1, 2, 1,
%C 2, 1, 2, 1, 2, 1, 2, 1 ...  _Paul Barry_, Feb 10 2011
%C Pisano period lengths: 1, 3, 1, 6, 24, 3, 24, 6, 1, 24,120, 6,156, 24, 24, 12, 16, 3, 90, 24,...  _R. J. Mathar_, Aug 10 2012
%H Vincenzo Librandi, <a href="/A105476/b105476.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,3).
%F G.f.: (1x^2) / (1x3*x^2).
%F a(n) = a(n1) + 3*a(n2) for n>=3.
%F a(n) = 3*A006138(n2), n>=2.
%F a(n) = ((2+sqrt(13))*(1+sqrt(13))^n(2sqrt(13))*(1sqrt(13))^n)/(3*2^n*sqrt(13)) for n>0.  _Bruno Berselli_, May 24 2011
%F G.f.: 1/(1  sum(k>=1, x^k*(1+x^k))).  _Joerg Arndt_, Mar 09 2014
%F G.f.: 1/(1  (x/(1x))  x^2/(1x^2)) = 1/(1  (x+2*x^2+x^3+2*x^4+x^5+2*x^6+...) ); in general 1/(1  sum(j>=1, m(j)*x^j) ) is the g.f. for compositions with m(k) sorts of part k.  _Joerg Arndt_, Feb 16 2015
%e a(3)=6 because we have (3),(1,2),(1,2'),(2,1),(2',1) and (1,1,1).
%p G:=(1z^2)/(1z3*z^2): Gser:=series(G,z=0,35): 1,seq(coeff(Gser,z^n),n=1..33);
%t CoefficientList[Series[(x^2  1) / (3 x^2 + x  1), {x, 0, 100}], x] (* or *) Join[{1}, LinearRecurrence[{1, 3}, {1, 3}, 50]] (* _Vladimir Joseph Stephan Orlovsky_, Jul 17 2011 *)
%o (PARI) Vec((1x^2)/(1x3*x^2)+O(x^99)) \\ _Charles R Greathouse IV_, Jun 13 2013
%o (MAGMA) I:=[1,1,3]; [n le 3 select I[n] else Self(n1)+3*Self(n2): n in [1..35]]; // _Vincenzo Librandi_, Jul 21 2013
%Y Cf. A105475, A006130, A105963, A274977.
%K nonn,easy
%O 0,3
%A _Emeric Deutsch_, Apr 09 2005
%E Typo in Mathematica code fixed by _Vincenzo Librandi_, Jul 21 2013
