%I #60 Aug 28 2024 02:34:35
%S 1,1,3,8,25,83,289,1041,3847,14504,55569,215727,846761,3354858,
%T 13398965,53888063,218053915,887107888,3626373205,14887942624,
%U 61358959587,253771944529,1052917272543,4381374717994,18280470530047,76459765772375
%N G.f. satisfies: A(x) = (1 + x*A(x)^2)/(1 - x^2).
%C Diagonal sums of A060693. - _Paul Barry_, Feb 11 2009
%C Starting with the second 1 and inserting a 2 between the 1 and 3: (1, 2, 3, 8, 25, 83, ...) the INVERT transform of that sequence deletes the 2, getting (1, 3, 8, 25, 83, ...). - _Gary W. Adamson_, Jun 24 2015
%H Vincenzo Librandi, <a href="/A143330/b143330.txt">Table of n, a(n) for n = 0..1000</a>
%H James East and Nicholas Ham, <a href="https://arxiv.org/abs/1811.05735">Lattice paths and submonoids of Z^2</a>, arXiv:1811.05735 [math.CO], 2018.
%H J. P. S. Kung and A. de Mier, <a href="http://dx.doi.org/10.1016/j.jcta.2012.08.010">Catalan lattice paths with rook, bishop and spider steps</a>, Journal of Combinatorial Theory, Series A 120 (2013) 379-389. - From _N. J. A. Sloane_, Dec 27 2012
%F G.f.: A(x) = (1-x^2 - sqrt(1 - 4*x - 2*x^2 + x^4))/(2*x).
%F From _Paul Barry_, Feb 11 2009: (Start)
%F G.f.: 1/(1-x^2-x/(1-x^2-x/(1-x^2-x/(1-x^2-x/(1-...))))) (continued fraction).
%F a(n) = Sum_{k=0..floor(n/2)} C(2n-3k,k)*A000108(n-2k).
%F (End)
%F D-finite with recurrence (n+1)*a(n) +(n+2)*a(n-1) +2*(17-11n)*a(n-2) +10*(3-n)*a(n-3) +(n-5)*a(n-4) +5*(n-6)*a(n-5)=0. - _R. J. Mathar_, Dec 11 2011
%F a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 4.439109106851354261627... is the root of the equation 1 - 2*d^2 - 4*d^3 + d^4 = 0 and c = 1/2*sqrt(d*(d^2+3)/(d^2-1)) = 1.16064231... - _Vaclav Kotesovec_, Feb 03 2014
%F G.f. satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*x^k*A(x)^(j-k). - _Ilya Gutkovskiy_, Apr 11 2019
%F G.f.: 1/G(x), with G(x) = 1-(x+x^2)/(1-x/G(x)) (continued fraction). - _Nikolaos Pantelidis_, Jan 11 2023
%F From _Peter Luschny_, Jan 25 2023: (Start)
%F a(n) = CatalanNumber(n)*hypergeom([-n/2, -n/2, -n/2 - 1/2, -n/2 + 1/2], [-(2*n)/3, -(2*n)/3 + 1/3, -(2*n)/3 + 2/3], -16/27).
%F a(n) = ((5 - n)*a(n - 4) + (2*n - 4)*a(n - 2) + (4*n - 2)*a(n - 1))/(n + 1) for n >= 4. (End)
%F G.f. A(x) = -x + (1/x)*series_reversion(x*G(-x)), where G(x) = 1 + 2*x + 5*x^2 + 18*x^3 + 70*x^4 + 293*x^5 + 1283*x^6 + ... is the g.f. of A073157. - _Peter Bala_, Aug 27 2024
%e G.f. = 1 + x + 3*x^2 + 8*x^3 + 25*x^4 + 83*x^5 + 289*x^6 + 1041*x^7 + ...
%p a := proc(n) option remember; if n <= 3 then return [1, 1, 3, 8][n + 1] fi;
%p ((5 - n)*a(n - 4) + (2*n - 4)*a(n - 2) + (4*n - 2)*a(n - 1))/(n + 1) end:
%p seq(a(n), n = 0..25); # _Peter Luschny_, Jan 25 2023
%t CoefficientList[Series[(1 - x^2 - Sqrt[1 - 4 x - 2 x^2 + x^4])/(2 x), {x, 0, 30}], x] (* _Vaclav Kotesovec_, Sep 17 2013 *)
%o (PARI) {a(n)=polcoeff((1-x^2-sqrt((1-x^2)^2-4*x+x^2*O(x^n)))/(2*x),n)}
%Y Cf. A000108, A060693, A073157.
%K nonn,easy
%O 0,3
%A _Paul D. Hanna_, Aug 08 2008
%E Minor edits by _Vaclav Kotesovec_, Mar 31 2014