|
%I #15 Feb 10 2014 07:13:11
%S 1,1,2,3,7,12,29,53,130,247,611,1192,2965,5897,14726,29723,74443,
%T 152020,381617,786733,1978582,4111295,10355303,21661168,54628201,
%U 114925697,290148890,613442227,1550177791,3291704108,8324934533,17745496453
%N Number of lattice paths from (0,0) to the line x+y=n that use the step set {(0,1),(1,0),(2,0),(3,0),...} and never pass below y=x.
%C a(n) = Sum(A011117(i,n-i), i=0..floor(n/2)), i.e. diagonal sums in A011117 formatted as an upper right triangle.
%C Hankel transform is A060656. - _Paul Barry_, Mar 01 2010
%H Vincenzo Librandi, <a href="/A089324/b089324.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: 2/[(1-z)^2+sqrt(1-6z^2+z^4)].
%F G.f.: 1/(1-x-x^2/(1-2x^2/(1-x^2/(1-2x^2/(1-x^2/(1-2x^2/(1-... (continued fraction). - _Paul Barry_, Mar 01 2010
%F Conjecture: (n+1)*a(n) +3*(-n-1)*a(n-1) +(-5*n+13)*a(n-2) +18*(n-2)*a(n-3) +(-5*n+7)*a(n-4) +3*(-n+5)*a(n-5) +(n-5)*a(n-6)=0. - _R. J. Mathar_, Nov 24 2012
%F a(n) ~ sqrt(6*sqrt(2)-8) * (1 - (12*sqrt(2)-17)*(-1)^n) * (sqrt(2)+1)^(n+4) / (2 * sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Feb 09 2014
%e a(4) = 7 because we have VVVV, VVVh, VVhV, VhVV, VVH, VVhh and VhVh, where V=(0,1), h=(1,0) and H=(2,0).
%t CoefficientList[Series[2/((1-x)^2+Sqrt[1-6*x^2+x^4]), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 09 2014 *)
%Y Cf. A011117.
%K nonn
%O 0,3
%A _Emeric Deutsch_, Dec 25 2003
|
