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A190166 Number of (1,0)-steps at levels 0,2,4,... in all peakless Motzkin paths of length n.

%I A190166 #11 May 29 2022 04:26:38
%S A190166 0,1,2,3,6,14,34,83,202,495,1224,3046,7616,19115,48130,121527,307602,
%T A190166 780244,1982834,5047377,12867438,32847357,83952780,214806750,
%U A190166 550170300,1410412561,3618785462,9292203549,23877482490,61397367692,157972743178,406693829059,1047585820586,2699811117189
%N A190166 Number of (1,0)-steps at levels 0,2,4,... in all peakless Motzkin paths of length n.
%C A190166 a(n)=Sum(k*A190164(n,k),k>=0).
%C A190166 a(n)=A110236(n) - A190169(n).
%F A190166 G.f. = z/[(1-z+z^2)sqrt((1+z+z^2)(1-3z+z^2))].
%F A190166 Conjecture: (-n+1)*a(n) +(3*n-4)*a(n-1) +2*(-n+1)*a(n-2) +3*(n-2)*a(n-3) +2*(-n+3)*a(n-4) +(3*n-8)*a(n-5) +(-n+3)*a(n-6)=0. - _R. J. Mathar_, Apr 09 2019
%F A190166 a(n) ~ phi^(2*n+2) / (4 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, May 29 2022
%e A190166 a(4)=6 because in h'h'h'h', h'uhd, uhdh', and uhhd, where u=(1,1), h=(1,0), d=(1,-1), we have 4+1+1+0 h-steps at even levels (marked).
%p A190166 G := z/((1-z+z^2)*sqrt((1+z+z^2)*(1-3*z+z^2))): Gser := series(G,z=0,36): seq(coeff(Gser,z,n),n=0..33);
%Y A190166 Cf. A190164, A110236, A190169, A004148.
%K A190166 nonn
%O A190166 0,3
%A A190166 _Emeric Deutsch_, May 06 2011