%I #3 Mar 30 2012 17:36:07
%S 0,1,1,0,1,2,2,1,1,6,6,3,2,1,19,17,12,5,3,1,61,56,36,20,8,4,1,202,185,
%T 120,66,31,12,5,1,683,624,409,224,110,46,17,6,1,2348,2144,1408,784,
%U 385,172,66,23,7,1,8184,7468,4920,2760,1380,624,257,92,30,8,1,28855,26317
%N Triangle read by rows: T(n,k) is the number of hillfree Dyck paths of semilength n and having k peaks at level 2; 0<= k<=n1, n>=2 (a Dyck path is said to be hillfree if it has no peaks at level 1).
%C Row n has n terms (n>=2). Row sums yield the Fine numbers (A000957). T(n,0)=A114627(n3). Sum(kT(n,k),k=0..n1)=A114495(n).
%F G.f.=(1+ztzzC)/[1+z+z^2tztz^2z(1+z)C], where C=[1sqrt(14z)]/(2z) is the Catalan function.
%e T(5,2)=3 because we have U(UD)(UD)UUDDD, UUUDD(UD)(UD)D and U(UD)UUDD(UD)D, where U=(1,1), D=(1,1) (the peaks at level 2 are shown between parentheses).
%e Triangle begins:
%e 0,1;
%e 1,0,1;
%e 2,2,1,1;
%e 6,6,3,2,1;
%e 19,17,12,5,3,1;
%p C:=(1sqrt(14*z))/2/z: G:=(1+zt*zz*C)/(1+z+z^2t*zt*z^2z*(1+z)*C): Gser:=simplify(series(G,z=0,15)): for n from 2 to 12 do P[n]:=coeff(Gser,z^n) od: for n from 2 to 12 do seq(coeff(t*P[n],t^j),j=1..n) od; # yields sequence in triangular form
%Y Cf. A000957, A114627, A114495.
%K nonn,tabf
%O 2,6
%A _Emeric Deutsch_, Dec 18 2005
