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A190169
Number of (1,0)-steps at levels 1,3,5,... in all peakless Motzkin paths of length n.
2
0, 0, 0, 1, 4, 10, 24, 60, 152, 386, 980, 2488, 6324, 16098, 41032, 104711, 267512, 684138, 1751316, 4487217, 11506792, 29530524, 75841152, 194910254, 501234960, 1289755668, 3320603016, 8553723949, 22044934324, 56841474482, 146626826376, 378392593206, 976884539336, 2522936490418
OFFSET
0,5
COMMENTS
a(n)=Sum(k*A190167(n,k),k>=0).
a(n)=A110236(n) - A190166(n).
FORMULA
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).
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
EXAMPLE
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).
MAPLE
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);
CROSSREFS
KEYWORD
nonn
AUTHOR
Emeric Deutsch, May 06 2011
STATUS
approved