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%I #13 Feb 13 2014 03:32:47
%S 1,2,8,28,103,382,1432,5408,20546,78436,300636,1156188,4459267,
%T 17241526,66807856,259361920,1008598126,3928120924,15319329472,
%U 59817190552,233826979750,914962032172,3583556424208,14047386554368,55108441878868
%N Number of peaks at even levels in all hill-free Dyck paths of semilength n+2 (a hill in a Dyck path is a peak at level 1).
%H Vincenzo Librandi, <a href="/A114590/b114590.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: (1+2*z^2-(1+2*z)*sqrt(1-4*z))/(2*z^2*(2+z)^2*sqrt(1-4*z)).
%F a(n) = sum(k*A114588(n+2,k),k=0..n+1).
%F a(n)=sum{k=0..n, sum{j=0..n-k, C(n-j,k-j)*C(n-j,k)*(j+1)}}; - _Paul Barry_, Nov 03 2006
%F Conjecture: 2*(n+2)*a(n) +(-7*n-9)*a(n-1) -18*a(n-2) +2*(-7*n+19)*a(n-3) +4*(-2*n+3)*a(n-4)=0. - _R. J. Mathar_, Nov 15 2012
%F Recurrence: 2*n*(n+2)*(3*n+1)*a(n) = (21*n^3 + 34*n^2 + n - 8)*a(n-1) + 2*(n+1)*(2*n+1)*(3*n+4)*a(n-2). - _Vaclav Kotesovec_, Feb 12 2014
%F a(n) ~ 4^(n+2) / (9*sqrt(Pi*n)). - _Vaclav Kotesovec_, Feb 12 2014
%e a(1)=2 because in the 2 (=A000957(4)) hill-free Dyck paths of semilength 3, namely UUUDDD and U(UD)(UD)D (U=(1,1), D=(1,-1)) we have altogether 2 peaks at even level (shown between parentheses).
%p G:=(1+2*z^2-(1+2*z)*sqrt(1-4*z))/2/z^2/(2+z)^2/sqrt(1-4*z): Gser:=series(G,z=0,30): 1, seq(coeff(Gser,z^n),n=1..25);
%t CoefficientList[Series[(1+2*x^2-(1+2*x)*Sqrt[1-4*x])/2/x^2/(2+x)^2/Sqrt[1-4*x], {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 12 2014 *)
%Y Cf. A114588, A114587, A114515.
%K nonn
%O 0,2
%A _Emeric Deutsch_, Dec 11 2005
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