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%I #21 Jul 23 2017 12:50:36
%S 1,2,6,20,73,281,1125,4635,19525,83710,364070,1602327,7123041,
%T 31937010,144255802,655804649,2998354717,13777825186,63596593430,
%U 294743653360,1371017707245,6398580086645,29952930770185,140604572777250,661708404611603,3121439743413256,14756658303857332
%N 1/n times A104631(n), the coefficient of x^(2n+1) in the expansion of (1+x+x^2+x^3+x^4)^n.
%C This sequence may be viewed as a higher-order form of the Motzkin numbers, A001006, which are 1/n times the coefficient of x^(n+1) in the expansion of (1+x+x^2)^n. According to Superseeker, this sequence is the INVERT transform of A104184, which is related to Motzkin numbers also. See A104631 for additional comments.
%C Alternatively, this sequence corresponds to the number of positive walks with n steps {-2,-1,0,1,2} starting at the origin, ending at altitude 1, and staying strictly above the x-axis. - _David Nguyen_, Dec 01 2016
%H C. Banderier, C. Krattenthaler, A. Krinik, D. Kruchinin, V. Kruchinin, D. Nguyen, and M. Wallner, <a href="https://arxiv.org/abs/1609.06473">Explicit formulas for enumeration of lattice paths: basketball and the kernel method</a>, arXiv preprint arXiv:1609.06473 [math.CO], 2016.
%F a(n) = Sum_{i=0..(2*n+1)/5}((-1)^i*binomial(n,i)*binomial(3*n-5*i,n-1))/n. - _Vladimir Kruchinin_, Apr 06 2017
%F Conjecture: 2*n*(2*n+1)*(n-1)*a(n) -(n-1)*(19*n^2-19*n+2)*a(n-1) -5*(n-2)*(2*n^2-3*n-1)*a(n-2) +25*n*(n-2)*(n-3)*a(n-3)=0. - _R. J. Mathar_, Jul 23 2017
%t f=1; Table[f=Expand[f(x^4+x^3+x^2+x+1)]; Coefficient[f, x, 2n+1]/n, {n, 30}]
%t a[ n_] := If[ n < 1, 0, Coefficient[ (1 + x + x^2 + x^3 + x^4)^n, x, 2 n + 1] / n]; (* _Michael Somos_, Dec 01 2016 *)
%o (PARI) a(n) = polcoeff((1+x+x^2+x^3+x^4)^n, 2*n+1)/n \\ _Michel Marcus_, Sep 24 2016
%o (Maxima)
%o a(n):=sum((-1)^i*binomial(n,i)*binomial(3*n-5*i,n-1),i,0,(2*n+1)/5)/n; /* _Vladimir Kruchinin_, Apr 06 2017 */
%Y Cf. A005717 (coefficient of x^(n+1) in the expansion of (1+x+x^2)^n).
%K easy,nonn
%O 1,2
%A _T. D. Noe_, Mar 17 2005
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