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G. C. Greubel, <a href="/A026135/b026135.txt">Table of n, a(n) for n = 0..1000</a>
G.f. : ((x-1)^2*((1+x)/(1-3x))^(1/2) + x^2 - 1)/(2*x^2). - David Callan, Aug 16 2006
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 *)
(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
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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
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a(n) = Sum_{k=0..n} binomial(n-1, k-1)*binomial(k+1, floor((k+1)/2)). - _Vladeta Jovovic (vladeta(AT)eunet.rs), _, Sep 18 2003
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 (callan(AT)stat.wisc.edu), _, Aug 16 2006
G.f. ((x-1)^2*((1+x)/(1-3x))^(1/2) + x^2 - 1)/(2*x^2). - _David Callan (callan(AT)stat.wisc.edu), _, Aug 16 2006
More terms from _David Callan (callan(AT)stat.wisc.edu), _, Aug 16 2006
Clark Kimberling (ck6(AT)evansville.edu)
G.f. = (1+z)*(1+z^2)/(1-z) where z=x*A001006(x). [From _R. J. Mathar (mathar(AT)strw.leidenuniv.nl), _, Jul 07 2009]
Typo in a(19) corrected by _R. J. Mathar (mathar(AT)strw.leidenuniv.nl), _, Jul 07 2009
1, 2, 5, 14, 39, 110, 312, 890, 2550, 7334, 21161, 61226, 177575, 516114, 1502867, 4383462, 12804429, 37452870, 109682319, 321564658, 321563658, 943701141, 2772060618, 8149661730, 23978203662, 70600640796, 208014215066, 613266903927
a(n) = Sum_{k=0..n} binomial(n-1, k-1)*binomial(k+1, floor((k+1)/2)). - Vladeta Jovovic (vladeta(AT)Euneteunet.yurs), Sep 18 2003
G.f. = (1+z)*(1+z^2)/(1-z) where z=x*A001006(x). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 07 2009]
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Typo in a(19) corrected by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 07 2009