A097331 Expansion of 1 + 2x/(1 + sqrt(1 - 4x^2)).

1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, 0, 132, 0, 429, 0, 1430, 0, 4862, 0, 16796, 0, 58786, 0, 208012, 0, 742900, 0, 2674440, 0, 9694845, 0, 35357670, 0, 129644790, 0, 477638700, 0, 1767263190, 0, 6564120420, 0, 24466267020, 0, 91482563640, 0, 343059613650, 0

Offset: 0

Author

Paul Barry, Aug 05 2004

Keywords

• easy
• nonn

Comments

Binomial transform is A097332. Second binomial transform is A014318.

Essentially the same as A126120. - R. J. Mathar, Jun 15 2008

Hankel transform is A087960(n) = (-1)^binomial(n+1,2). - Paul Barry, Aug 10 2009

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..3340
• Jean-Luc Baril, Sergey Kirgizov, Armen Petrossian, Motzkin paths with a restricted first return decomposition, Integers (2019) Vol. 19, A46.

Programs

Maple

A097331_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w]:=a[w-1]-(-1)^w*add(a[j]*a[w-j-1],j=1..w-1) od; convert(a,list)end: A097331_list(48); # Peter Luschny, May 19 2011

Mathematica

a[0] = 1; a[n_?OddQ] := CatalanNumber[(n-1)/2]; a[] = 0; Table[a[n], {n, 0, 48}] (* _Jean-François Alcover, Jul 24 2013 *)

Sage

def A097331_list(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1; R = []
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] -= D[k-1]
            h += 1; R.append(abs(D[1]))
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
    return R
A097331_list(49) # Peter Luschny, Jun 03 2012

Formula

a(n) = 0^n + Catalan((n-1)/2)(1-(-1)^n)/2.

Unsigned version of A090192, A105523. - Philippe Deléham, Sep 29 2006

From Paul Barry, Aug 10 2009: (Start)

G.f.: 1+xc(x^2), c(x) the g.f. of A000108;

G.f.: 1/(1-x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+... (continued fraction);

G.f.: 1+x/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-... (continued fraction). (End)

G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+2*x) (continued fraction); more generally g.f. C(x/(1+2*x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011

Conjecture: (n+1)*a(n) + n*a(n-1) + 4*(-n+2)*a(n-2) + 4*(-n+3)*a(n-3)=0. - R. J. Mathar, Dec 02 2012

Recurrence: (n+3)*a(n+2) = 4*n*a(n), a(0)=a(1)=1. For nonzero terms, a(n) ~ 2^(n+1)/((n+1)^(3/2)*sqrt(2*Pi)). - Fung Lam, Mar 17 2014