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A190169 Number of (1,0)-steps at levels 1,3,5,... in all peakless Motzkin paths of length n.

%I A190169 #6 Apr 09 2019 12:13:41
%S A190169 0,0,0,1,4,10,24,60,152,386,980,2488,6324,16098,41032,104711,267512,
%T A190169 684138,1751316,4487217,11506792,29530524,75841152,194910254,
%U A190169 501234960,1289755668,3320603016,8553723949,22044934324,56841474482,146626826376,378392593206,976884539336,2522936490418
%N A190169 Number of (1,0)-steps at levels 1,3,5,... in all peakless Motzkin paths of length n.
%C A190169 a(n)=Sum(k*A190167(n,k),k>=0).
%C A190169 a(n)=A110236(n) - A190166(n).
%F A190169 G.f. = (1-2z+z^2-2z^3+z^4)/[2z(1-z+z^2)sqrt((1+z+z^2)(1-3z+z^2))]-1/(2z).
%F A190169 Conjecture: -(n-1)*(n+1)*a(n) -n*(n-19)*a(n-1) +2*(n-1)*(7*n-40)*a(n-2) -(n-2)*(17*n-97)*a(n-3) +2*(9*n^2-64*n+119)*a(n-4) -17*(n-4)*(n-5)*a(n-5) +(19*n-59)*(n-5)*a(n-6) -2*(8*n-21)*(n-6)*a(n-7) +2*(2*n-5)*(n-7)*a(n-8)=0. - _R. J. Mathar_, Apr 09 2019
%e A190169 a(4)=4 because in hhhh, huh'd, uh'dh, and uh'h'd, where u=(1,1), h=(1,0), d=(1,-1), we have 0+1+1+2 h-steps at odd levels (marked).
%p A190169 G := ((1-2*z+z^2-2*z^3+z^4)*1/2)/(z*(1-z+z^2)*sqrt((1+z+z^2)*(1-3*z+z^2)))-(1/2)/z: Gser:=series(G,z=0,36): seq(coeff(Gser,z,n),n=0..33);
%Y A190169 Cf. A190167, A110236, A190166
%K A190169 nonn
%O A190169 0,5
%A A190169 _Emeric Deutsch_, May 06 2011