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A060899 Number of walks of length n on square lattice, starting at origin, staying on points with x+y >= 0.

%I A060899 #53 May 14 2022 04:53:25
%S A060899 1,2,8,24,96,320,1280,4480,17920,64512,258048,946176,3784704,14057472,
%T A060899 56229888,210862080,843448320,3186360320,12745441280,48432676864,
%U A060899 193730707456,739699064832,2958796259328,11342052327424
%N A060899 Number of walks of length n on square lattice, starting at origin, staying on points with x+y >= 0.
%C A060899 The number of lattice paths consisting of 2*n steps either (1,1) or (1,-1) that return to the x-axis only at times that are a multiple of 4. - _Peter Bala_, Jan 02 2020
%H A060899 Harry J. Smith, <a href="/A060899/b060899.txt">Table of n, a(n) for n=0..200</a>
%H A060899 Paul Barry, <a href="https://www.emis.de/journals/JIS/VOL22/Barry1/barry411.html">The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths</a>, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.
%H A060899 Alin Bostan, <a href="https://www-apr.lip6.fr/sem-comb-slides/IHP-bostan.pdf">Computer Algebra for Lattice Path Combinatorics</a>, Séminaire de Combinatoire Ph. Flajolet, March 28 2013.
%H A060899 Alin Bostan, Andrew Elvey Price, Anthony John Guttmann, and Jean-Marie Maillard, <a href="https://arxiv.org/abs/2001.00393">Stieltjes moment sequences for pattern-avoiding permutations</a>, arXiv:2001.00393 [math.CO], 2020.
%H A060899 Math Overflow, <a href="https://mathoverflow.net/questions/86924/combinatorial-proof-for-the-number-of-lattice-paths-that-return-to-the-axis-only">Combinatorial proof for the number of lattice paths that return to the axis only at times that are a multiple of 4</a>, 2012.
%F A060899 a(n) = 2^n*binomial(n, [n/2]);
%F A060899 G.f.: (sqrt((1+4*x)/(1-4*x))-1)/4/x. - _Vladeta Jovovic_, Apr 28 2003
%F A060899 E.g.f.: BesselI(0, 4*x)+BesselI(1, 4*x). - _Vladeta Jovovic_, Apr 28 2003
%F A060899 a(n) = 4^n*sum{k=0..n, C(n,k)C(k)/(-2)^k}, with C(n)=A000108(n). - _Paul Barry_, Dec 28 2006
%F A060899 (n+1)*a(n) -4*a(n-1) +16*(-n+1)*a(n-2)=0. - _R. J. Mathar_, Nov 24 2012
%F A060899 a(n) = (-4)^n*hypergeom([3/2,-n],[2],2). - _Peter Luschny_, Apr 26 2016
%F A060899 Sum_{n>=0} a(n)/6^n = 3/phi = A134973. - _Peter McNair_, Apr 30 2022
%F A060899 In general, for k>4, Sum_{n>=0} a(n)/k^n = (sqrt((k+4)/(k-4)) - 1) * k/4. - _Vaclav Kotesovec_, May 13 2022
%F A060899 From _Amiram Eldar_, May 14 2022: (Start)
%F A060899 Sum_{n>=0} 1/a(n) = 16*asin(1/4)/(3*sqrt(15)) + 4/3.
%F A060899 Sum_{n>=0} (-1)^n/a(n) = 4/5 - 16*asin(1/4)/(5*sqrt(15)). (End)
%t A060899 Table[2^n Binomial[n,Floor[n/2]],{n,0,30}] (* _Harvey P. Dale_, Oct 15 2017 *)
%o A060899 (PARI) { for (n=0, 200, write("b060899.txt", n, " ", 2^n*binomial(n, n\2)); ) } \\ _Harry J. Smith_, Jul 14 2009
%Y A060899 Cf. A005566, A001700, A060897, A060898, A060900.
%Y A060899 Cf. A001405.
%K A060899 nonn,easy
%O A060899 0,2
%A A060899 _David W. Wilson_, May 05 2001