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%I #17 May 22 2017 02:38:40
%S 1,2,5,14,39,110,312,890,2550,7334,21161,61226,177575,516114,1502867,
%T 4383462,12804429,37452870,109682319,321563658,943701141,2772060618,
%U 8149661730,23978203662,70600640796,208014215066,613266903927
%N Number of (s(0),s(1),...,s(n)) such that every s(i) is a nonnegative integer, s(0) = 1, |s(1) - s(0)| = 1, |s(i) - s(i-1)| <= 1 for i >= 2. Also sum of numbers in row n+1 of the array T defined in A026120.
%C a(n) is the total number of rows of consecutive peaks in all Motzkin (n+2)-paths. For example, with U=upstep, D=downstep, F=flatstep, the path FU(UD)FU(UDUDUD)DD(UD) contains 3 rows of peaks (in parentheses). The 9 Motzkin 4-paths are FFFF, FF(UD), F(UD)F, FUFD, (UD)FF, (UDUD), UFDF, UFFD, U(UD)D, containing a total of 5 rows of peaks and so a(2)=5. - _David Callan_, Aug 16 2006
%H G. C. Greubel, <a href="/A026135/b026135.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = Sum_{k=0..n} binomial(n-1, k-1)*binomial(k+1, floor((k+1)/2)). - _Vladeta Jovovic_, Sep 18 2003
%F G.f.: ((x-1)^2*((1+x)/(1-3x))^(1/2) + x^2 - 1)/(2*x^2). - _David Callan_, Aug 16 2006
%F G.f. = (1+z)*(1+z^2)/(1-z) where z=x*A001006(x). [From _R. J. Mathar_, Jul 07 2009]
%F Conjecture: (n+2)*a(n) +3*(-n-1)*a(n-1) +(-n-2)*a(n-2) +3*(n-3)*a(n-3)=0. - _R. J. Mathar_, Jun 23 2013
%t CoefficientList[Series[((x - 1)^2*((1 + x)/(1 - 3 x))^(1/2) + x^2 - 1)/(2*x^2), {x,0,50}], x] (* _G. C. Greubel_, May 22 2017 *)
%o (PARI) x='x+O('x^50); Vec(((x-1)^2*((1+x)/(1-3x))^(1/2) + x^2 - 1)/(2*x^2)) \\ _G. C. Greubel_, May 22 2017
%Y First differences are in A025566, second differences in A005773.
%Y Pairwise sums of A025179.
%K nonn
%O 0,2
%A _Clark Kimberling_
%E More terms from _David Callan_, Aug 16 2006
%E Typo in a(19) corrected by _R. J. Mathar_, Jul 07 2009
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