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.

A101785 G.f. satisfies: A(x) = 1 + x*A(x)/(1 - x^2*A(x)^2).

Original entry on oeis.org

1, 1, 1, 2, 5, 12, 30, 79, 213, 584, 1628, 4600, 13138, 37871, 110043, 321978, 947813, 2805104, 8341608, 24912004, 74686460, 224694128, 678143656, 2052640752, 6229616730, 18952875247, 57792705415, 176596786934, 540679385663
Offset: 0

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Author

Paul D. Hanna, Dec 16 2004

Keywords

Comments

Formula may be derived using the Lagrange Inversion theorem (cf. A049124).
a(n) = number of Dyck n-paths (A000108) all of whose descents have odd length. For example, a(3) counts UUUDDD, UDUDUD. - David Callan, Jul 25 2005
The number of noncrossing partitions of [n] with all blocks of odd size. E.g.: a(4)=5 with the five partitions being 123/4, 124/3, 134/2,1/234 and 1/2/3/4. - Louis Shapiro, Jan 07 2006
Number of ordered trees with n edges in which every non-leaf vertex has an odd number of children. - David Callan, Mar 30 2007
Number of valid hook configurations of permutations of [n] that avoid the patterns 312 and 321. - Colin Defant, Apr 28 2019

Examples

			Generated from Fibonacci polynomials (A011973) and
coefficients of odd powers of 1/(1-x):
a(1) = 1*1/1
a(2) = 1*1/1 + 0*1/3
a(3) = 1*1/1 + 1*3/3
a(4) = 1*1/1 + 2*6/3 + 0*1/5
a(5) = 1*1/1 + 3*10/3 + 1*5/5
a(6) = 1*1/1 + 4*15/3 + 3*15/5 + 0*1/7
a(7) = 1*1/1 + 5*21/3 + 6*35/5 + 1*7/7
a(8) = 1*1/1 + 6*28/3 + 10*70/5 + 4*28/7 + 0*1/9
This process is equivalent to the formula:
a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1,k)*C(n,2*k)/(2*k+1).
		

Crossrefs

Column k=2 of A212382.

Programs

  • Magma
    [n eq 0 select 1 else (&+[Binomial(n-k-1,k)*Binomial(n, 2*k)/(2*k+1): k in [0..Floor((n-1)/2)]]): n in [0..30]]; // G. C. Greubel, May 03 2019
    
  • Mathematica
    Flatten[{1,Table[Sum[Binomial[n-k-1,k]*Binomial[n,2*k]/(2*k+1),{k,0,Floor[(n-1)/2]}],{n,1,20}]}] (* Vaclav Kotesovec, Sep 17 2013 *)
    CoefficientList[InverseSeries[Series[x*(1-x^2)/(1+x-x^2), {x, 0, 30}], x]/x, x] (* G. C. Greubel, May 03 2019 *)
  • PARI
    {a(n)=if(n==0,1,sum(k=0,(n-1)\2,binomial(n-k-1,k)*binomial(n,2*k)/(2*k+1)))}
    for(n=1, 40, print1(a(n), ", "))
    
  • PARI
    N=66; Vec(serreverse(x/(1+sum(k=1,N,x^(2*k-1)))+O(x^N))/x) /* Joerg Arndt, Aug 19 2012 */
    
  • Sage
    [1]+[sum(binomial(n-k-1, k)*binomial(n, 2*k)/(2*k+1) for k in (0..floor((n-1)/2))) for n in (1..30)] # G. C. Greubel, May 03 2019

Formula

a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1, k)*C(n, 2*k)/(2*k+1) for n>0, with a(0)=1.
G.f.: (1/x) * Series_Reversion( x*(1-x^2)/(1+x-x^2) ).
Recurrence: 4*n*(n+1)*(91*n^2 - 379*n + 360)*a(n) = 6*n*(182*n^3 - 849*n^2 + 1075*n - 264)*a(n-1) - 2*(182*n^4 - 1122*n^3 + 2011*n^2 - 603*n - 648)*a(n-2) + 6*(n-3)*(364*n^3 - 1698*n^2 + 2267*n - 696)*a(n-3) - 5*(n-4)*(n-3)*(91*n^2 - 197*n + 72)*a(n-4). - Vaclav Kotesovec, Sep 17 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 3/4 + 1/(4*sqrt(3/(19 - 304/(4103 + 273*sqrt(273))^(1/3) + 2*(4103 + 273*sqrt(273))^(1/3)))) + 1/2*sqrt(19/6 + 76/(3*(4103 + 273*sqrt(273))^(1/3)) - 1/6*(4103 + 273*sqrt(273))^(1/3) + 63/2*sqrt(3/(19 - 304/(4103 + 273*sqrt(273))^(1/3) + 2*(4103 + 273*sqrt(273))^(1/3)))) = 3.228704951094501729... is the root of the equation 5 - 24*d + 4*d^2 - 12*d^3 + 4*d^4 = 0 and c = 0.82499074317860885542266460957609663272... is the root of the equation -125 - 3376*c^2 - 22080*c^4 - 23296*c^6 + 93184*c^8 = 0. - Vaclav Kotesovec, added Sep 17 2013, updated Jan 04 2014
G.f.: 1/(9*(3-3*x+x^2))*(x^2+27- x^2*(2*x+3)^3*(x-6)^3/(9*(3-3*x+x^2)^3*S(0) - x^2*(2*x+3)^2*(x-6)^2 )), where S(k) = 4*k+3 - x^2*(2*x^2-9*x-18)^2*(3*k+4)*(6*k+5)/( 18*(4*k+5)*(3-3*x+x^2)^3 - x^2*(2*x^2-9*x-18)^2*(3*k+5)*(6*k+7)/S(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 26 2013