cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A115178 Expansion of c(x^2+x^3), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 1, 2, 4, 7, 15, 29, 61, 126, 266, 566, 1212, 2619, 5685, 12419, 27247, 60049, 132847, 294931, 656877, 1467258, 3286218, 7378240, 16603458, 37441990, 84599854, 191501532, 434224404, 986161959, 2243009869, 5108859821
Offset: 0

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Author

Paul Barry, Mar 14 2006

Keywords

Comments

Diagonal sums of number triangle A117434.
a(n) = number of Motzkin n-paths (A001006) in which every flatstep (F) is followed by a downstep (D). For example, a(5)=4 counts UDUFD, UFDUD, UUDFD, UUFDD. - David Callan, Jun 07 2006
a(n) = number of lattice paths in the first quadrant from (0,0) to (n,0) using only steps U1=(1,1), U2=(2,1) and D=(1,-1). E.g. a(6)=7 because we have U1DU1DU1D, U1U1U1DDD, U1U1DU1DD, U1DU1U1DD, U1U1DDU1D, U2DU2D and U2U2DD. - José Luis Ramírez Ramírez, May 27 2013

Examples

			1 + x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 15*x^7 + 29*x^8 + 61*x^9 + ...
		

Crossrefs

Cf. A007477.

Programs

  • Mathematica
    Table[Sum[Binomial[k, n - 2*k]*CatalanNumber[k], {k, 0, Floor[n/2]}], {n, 0, 50}] (* G. C. Greubel, Feb 03 2017 *)
  • PARI
    {a(n) = local(A); A = O(x^0); for( k=0, n\5, A = 1 / (1 - x^2 / (1 - x / (1 - x^2 * A)))); polcoeff( A, n)} /* Michael Somos, May 12 2012 */

Formula

a(n) = Sum_{k=0..floor(n/2)} C(k)*C(k,n-2k).
D-finite with recurrence (n+2)*a(n) +(n+2)*a(n-1) +4*(1-n)*a(n-2) +2*(7-4*n)*a(n-3) +2*(5-2*n)*a(n-4)=0. - R. J. Mathar, Nov 15 2011
G.f. A(x) satisfies A(x) = 1 / (1 - x^2 / (1 - x / (1 - x^2 * A(x)))). - Michael Somos, May 12 2012
G.f.: (1-sqrt(1-4*z^2*(1+z)))/(2*z^2*(1+z)). - José Luis Ramírez Ramírez, May 27 2013
a(n) ~ sqrt(3 - 1/9*(-2 + (19-3*sqrt(33))^(1/3) + (19+3*sqrt(33))^(1/3))^2) * (((-2 + (19-3*sqrt(33))^(1/3) + (19+3*sqrt(33))^(1/3)) * (4 + (19-3*sqrt(33))^(1/3) + (19+3*sqrt(33))^(1/3)))/9)^n /(n^(3/2)*sqrt(Pi)). - Vaclav Kotesovec, Sep 16 2013