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 A126931 a(n)=A127359(n+1)/2-A127359(n). 7

%I

%S 1,3,10,33,110,366,1220,4065,13550,45162,150540,501786,1672620,

%T 5575356,18584520,61948257,206494190,688313490,2294378300,7647926046,

%U 25493086820,84976950468,283256501560,944188318938,3147294396460

%N a(n)=A127359(n+1)/2-A127359(n).

%C Hankel transform is A000012=[1,1,1,1,1,1,1,...].

%C a(n) is the number of Motzkin paths of length n in which the (1,0)-steps at level 0 come in 3 colors and there are no (1,0)-steps at a higher level. Example: a(3)=33 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 3^3 = 27 paths of shape HHH, 3 paths of shape HUD, and 3 paths of shape UDH. - Emeric Deutsch, May 02 2011

%F G.f.: 1/(1-3x-x^2/(1-x^2/(1-x^2/(1-x^2/(1-... (continued fraction). [From _Paul Barry_, Mar 10 2009]

%F G.f. = 2/[1-6z+sqrt(1-4z^2)]. - Emeric Deutsch, May 02 2011

%F Conjecture: 3*(n+1)*a(n) +10*(-n-1)*a(n-1) +12*(-n+2)*a(n-2) +40*(n-2)*a(n-3)=0. - _R. J. Mathar_, Nov 26 2012

%p A127359 := proc(n) add(binomial(n,floor(k/2))*3^(n-k), k=0..n) ; end proc: A126931 := proc(n) A127359(n+1)/2-A127359(n) ; end proc: seq(A126931(n),n=0..50) ; [From _R. J. Mathar_, Mar 25 2010]

%K nonn

%O 0,2

%A _Philippe DELEHAM_, Mar 17 2007

%E More terms from _R. J. Mathar_, Mar 25 2010

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