Also a(0)=1, a(n) = Sum_{k=0..n-1} M(k)*a(n-k-1), where M(n) are the Motzkin numbers (
A001006).
D-finite with recurrence n*a(n) = 2*n*a(n-1) + 3*(n-2)*a(n-2), a(0)=a(1)=1. -
Michael Somos, Feb 02 2002
G.f.: 1/2+(1/2)*((1+x)/(1-3*x))^(1/2). Related to Motzkin numbers
A001006 by a(n+1) = 3*a(n) -
A001006(n-1) [see Yaqubi Lemma 2.6].
a(n) = Sum_{q=0..n} binomial(q, floor(q/2))*binomial(n-1, q) for n > 0. -
Emeric Deutsch, Aug 15 2002
a(n+1) = Sum_{k=0..n} (-1)^(n+k)*C(n, k)*C(2*k+1, k+1).
a(n) = 0^n + Sum_{k=0..n-1} (-1)^(n+k-1)*C(n-1, k)*C(2*k+1, k+1). (End)
a(n+1) = Sum_{k=0..n} (-1)^k*3^(n-k)*binomial(n, k)*
A000108(k). -
Paul Barry, Jan 27 2005
G.f.: 1/(1-x/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). -
Paul Barry, Jan 19 2009
G.f.: 1+x/(1-2*x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-.... (continued fraction). -
Paul Barry, Jan 19 2009
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}...Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any (l_i - l_(i+1))^2 >= 2 for i=1..n-1 and delta(l_1,l_2,..., l_i,...,l_n) = 1 otherwise. -
Thomas Wieder, Feb 25 2009
a(n) = Sum_{k=1..n} (k/n * Sum_{j=0..n} binomial(n,j)*binomial(j,2*j-n-k)). -
Vladimir Kruchinin, Sep 06 2010
a(0) = 1; a(n+1) = Sum_{t=0..n} n!/((n-t)!*ceiling(t/2)!*floor(t/2)!). -
Andrew S. Hays, Feb 02 2011
a(n) = leftmost column term of M^n*V, where M = an infinite quadradiagonal matrix with all 1's in the main, super and subdiagonals, [1,0,0,0,...] in the diagonal starting at position (2,0); and rest zeros. V = vector [1,0,0,0,...]. -
Gary W. Adamson, Jun 16 2011
a(n) = upper left term of M^n, a(n+1) = sum of top row terms of M^n; M = an infinite square production matrix in which the main diagonal is (1,1,0,0,0,...) as follows:
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 0, 1, 0, 0, ...
1, 1, 1, 0, 1, 0, ...
1, 1, 1, 1, 0, 1, ...
1, 1, 1, 1, 1, 0, ... (End)
Limit_{n->oo} a(n+1)/a(n) = 3.0 = lim_{n->oo} (1 + 2*cos(Pi/n)). -
Gary W. Adamson, Feb 10 2012
With first term deleted: E.g.f.: a(n) = n! * [x^n] exp(x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). -
Peter Luschny, Aug 25 2012
G.f.: G(0)/2 + 1/2, where G(k) = 1 + 2*x*(4*k+1)/( (2*k+1)*(1+x) - x*(1+x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1+x)*(k+1)/G(k+1))); (continued fraction). -
Sergei N. Gladkovskii, Jun 24 2013
a(n) = GegenbauerC(n-2,-n+1,-1/2) + GegenbauerC(n-1,-n+1,-1/2) for n >= 1. -
Peter Luschny, May 12 2016
0 = a(n)*(+9*a(n+1) + 18*a(n+2) - 9*a(n+3)) + a(n+1)*(-6*a(n+1) + 7*a(n+2) - 2*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for n >= 0. -
Michael Somos, Dec 01 2016
a(n) = (-1)^(n + 1)*2*JacobiP(n - 1, 3, -n - 1/2, -7)/(n^2 + n). -
Peter Luschny, May 25 2021
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