A005325 Column of Motzkin triangle.

1, 6, 27, 104, 369, 1242, 4037, 12804, 39897, 122694, 373581, 1128816, 3390582, 10136556, 30192102, 89662216, 265640691, 785509362, 2319218869, 6839057544, 20147488020, 59306494520, 174466248840, 512987904000, 1507780192035, 4430417492826, 13015498076181

Offset: 5

Author

N. J. A. Sloane

Keywords

• nonn

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 5..1000
• R. Donaghey and L. W. Shapiro, Motzkin numbers, J. Combin. Theory, Series A, 23 (1977), 291-301.
• Nickolas Hein, Jia Huang, Variations of the Catalan numbers from some nonassociative binary operations, arXiv:1807.04623 [math.CO], 2018.
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, Une méthode pour obtenir la fonction génératrice d'une série, arXiv:0912.0072 [math.NT], 2009; FPSAC 1993, Florence. Formal Power Series and Algebraic Combinatorics.

Crossrefs

Cf. A026300, A026126, A001006.

A diagonal of triangle A020474.

Programs

Mathematica

RecurrenceTable[{3(-1+n)*n*a[-2+n]+n*(1+2n)*a[-1+n]-(-5+n)*(7+n)*a[n]==0, a[5]==1,a[6]==6}, a,{n,5,20}] (* Vaclav Kotesovec, Oct 05 2012 *)
a = DifferenceRoot[Function[{b, n}, {(-2n^2 - 25n - 78)b[n+1] - 3(n+5)(n+6) b[n] + (n+1)(n+13)b[n+2] == 0, b[1] == 1, b[2] == 6}]][# - 4]&;
Table[a[n], {n, 5, 31}] (* Jean-François Alcover, Jan 24 2019 *)

Maxima

a(n) := 6*sum(binomial(j,2*j-n+5)*binomial(n+1,j),j,ceiling((n-5)/2),(n+1))/(n+1);
/* Vladimir Kruchinin, Mar 18 2014 */

Formula

G.f.: z^5*M^6, where M=1+z*M+z^2*M^2 is the g.f. for the Motzkin numbers (A001006). - Emeric Deutsch, Aug 13 2004

a(n) = (sqrt(-3)/81)*((-1)^n*n*(4*n^3-15*n^2-55*n+102)/(n+7)/(n+3)/(n+2)*hypergeom([1/2, n+7],[3],4/3)-(-1)^n*(4*n^4-17*n^3+23*n^2+ 242*n-288)/(n+7)/(n+3)/(n+2)*hypergeom([1/2, n+6],[3],4/3)). - Mark van Hoeij, Oct 29 2011.

a(n) (n + 11) (n - 1) = (n + 4) (3 n + 9) a(n - 2) + (n + 4) (2 n + 9) a(n - 1). - Simon Plouffe, Feb 09 2012

a(n) ~ 3^(n+5/2)/(n^(3/2)*sqrt(Pi)). - Vaclav Kotesovec, Oct 05 2012

a(n) = 6*sum(j=ceiling((n-5)/2)..(n+1), C(j,2*j-n+5)*C(n+1,j))/(n+1). - Vladimir Kruchinin, Mar 17 2014

Extensions

More terms from Vincenzo Librandi, May 03 2013