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A104632
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1/n times A104631(n), the coefficient of x^(2n+1) in the expansion of (1+x+x^2+x^3+x^4)^n.
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2
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1, 2, 6, 20, 73, 281, 1125, 4635, 19525, 83710, 364070, 1602327, 7123041, 31937010, 144255802, 655804649, 2998354717, 13777825186, 63596593430, 294743653360, 1371017707245, 6398580086645, 29952930770185, 140604572777250, 661708404611603, 3121439743413256, 14756658303857332
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OFFSET
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1,2
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COMMENTS
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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.
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
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LINKS
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FORMULA
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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
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
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MATHEMATICA
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f=1; Table[f=Expand[f(x^4+x^3+x^2+x+1)]; Coefficient[f, x, 2n+1]/n, {n, 30}]
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 *)
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PROG
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(PARI) a(n) = polcoeff((1+x+x^2+x^3+x^4)^n, 2*n+1)/n \\ Michel Marcus, Sep 24 2016
(Maxima)
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 */
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CROSSREFS
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Cf. A005717 (coefficient of x^(n+1) in the expansion of (1+x+x^2)^n).
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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