G.f.: x*(c(x))^3 = (-1+(1-x)*c(x))/x, c(x) = g.f. for Catalan numbers. Also a(n) = 3*n*Catalan(n)/(n+2). -
Wolfdieter Lang
G.f. is A(x) = C(x)*(1-x)/x-1/x = x(1+x*C(x)^2)*C(x)^2 where C(x) is g.f. for Catalan numbers,
A000108.
G.f. satisfies x^2*A(x)^2 + (3*x-1)*A(x) + x = 0.
Series reversion of g.f. A(x) is -A(-x). -
Michael Somos, Jan 21 2004
a(n+1) = Sum_{i+j+k=n} C(i)C(j)C(k) with i, j, k >= 0 and where C(k) denotes the k-th Catalan number. -
Benoit Cloitre, Nov 09 2003
An inverse Chebyshev transform of x^2. -
Paul Barry, Oct 13 2004
The sequence is 0, 0, 1, 0, 3, 0, 9, 0, ... with zeros restored. Second binomial transform of (-1)^n*
A005322(n). The g.f. is transformed to x^2 under the Chebyshev transformation A(x)->(1/(1+x^2))A(x/(1+x^2)). For a sequence b(n), this corresponds to taking Sum_{k=0..floor(n/2)} C(n-k, k)(-1)^k*b(n-2k), or Sum_{k=0..n} C((n+k)/2, k)*b(k)*(-1)^((n-k)/2)*(1+(-1)^(n-k))/2. -
Paul Barry, Oct 13 2004
G.f.: (c(x^2)*(1-x^2)-1)/x^2, c(x) the g.f. of
A000108; a(n) = Sum_{k=0..n} (k+1)*C(n, (n-k)/2)*(-1)^k*(C(2,k)-2*C(1,k)+C(0, k))*(1+(-1)^(n-k))/(n+k+2). -
Paul Barry, Oct 13 2004
a(n) = Sum_{k=0..n} binomial(n,k)*2^(n-k)*(-1)^(k+1)*binomial(k, floor((k-1)/2)). -
Paul Barry, Feb 16 2006
E.g.f.: exp(2*x)*(Bessel_I(1,2x) - Bessel_I(2,2*x)). -
Paul Barry, Jun 04 2007
a(n) = (1/Pi)*Integral_{x=0..4} x^n*(x-1)*sqrt(x*(4-x))/(2*x). -
Paul Barry, Feb 08 2008
D-finite with recurrence: For n > 1, a(n+1) = 2*(2n+1)*(n+1)*a(n)/((n+3)*n). -
Sean A. Irvine, Dec 09 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j] = Catalan(j-i), (i<=j), and A[i,j] = 0, otherwise. Then, for n >= 2, a(n-1) = (-1)^(n-2)*coeff(charpoly(A,x),x^2). -
Milan Janjic, Jul 08 2010
a(n) = sum of top row terms of M^(n-1), M = an infinite square production matrix as follows:
2, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, 1, ...
...
E.g.f.: exp(2*x)*(BesselI(2,2*x)) = Q(0) - 1 where Q(k) = 1 - 2*x/(k + 1 - 3*((k+1)^2)/((k^2) + 8*k + 9 - (k+2)*((k+3)^2)*(2*k+3)/((k+3)*(2*k+3) - 3*(k+1)/Q(k+1)))); (continued fraction). -
Sergei N. Gladkovskii, Dec 05 2011
a(n) = -binomial(2*n,n)/(n+1)*hypergeom([-1,n+1/2],[n+2],4). -
Peter Luschny, Aug 15 2012
a(n) = Sum_{i=0..n-1} C(i)*C(n-i), where C(i) denotes the i-th Catalan number. -
Dmitry Kruchinin, Mar 02 2013
a(n) = ((-1)^n/(n+1))*Sum_{i=0..n-1} (-1)^(i+1)*(n+1-i)*binomial(2*n+2,i), n>=0. -
Taras Goy, Aug 09 2018
Sum_{n>=1} 1/a(n) = 14*Pi/(27*sqrt(3)) + 5/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 164*log(phi)/(75*sqrt(5)) + 7/25, where phi is the golden ratio (
A001622). (End)
a(n) = 3*Sum_{k = 0..n-2} (-1)^k * binomial(2*n-k-1, n+1)*binomial(n+1, k)/(k + 1) for n >= 2. -
Peter Bala, Sep 02 2024
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