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%I #7 Apr 06 2019 09:29:45
%S 0,0,1,1,0,1,3,2,0,1,10,5,2,0,1,33,16,5,2,0,1,111,51,16,5,2,0,1,379,
%T 168,51,16,5,2,0,1,1312,565,168,51,16,5,2,0,1,4596,1934,565,168,51,16,
%U 5,2,0,1,16266,6716,1934,565,168,51,16,5,2,0,1,58082,23604,6716,1934,565,168
%N Triangle read by rows: T(n,k) is the number of hill-free Dyck paths of semilength n and having length of first descent equal to k (1<=k<=n; n>=1). A hill in a Dyck path is a peak at level 1.
%C Row sums are the Fine numbers (A000957). T(n,1)=A001558(n-3) for n>=3. T(n,k)=A118973(n-k) for n>=k>=2. Sum(k*T(n,k),k=1..n)=A118974(n)
%H E. Deutsch and L. Shapiro, <a href="https://doi.org/10.1016/S0012-365X(01)00121-2">A survey of the Fine numbers</a>, Discrete Math., 241 (2001), 241-265.
%F G:=tz^2*CF[C-(1-t)/(1-tz)], where F=[1-sqrt(1-4z)]/[z(3-sqrt(1-4z)] and C=[1-sqrt(1-4z)]/(2z) is the Catalan function.
%e T(5,2)=5 because we have uu(dd)uududd, uu(dd)uuuddd,uuu(dd)uuddd,uuu(dd)ududd and uuuu(dd)uddd, where u=(1,1), d=(1,-1) (the first descents are shown between parentheses).
%e Triangle starts:
%e 0;
%e 0,1;
%e 1,0,1;
%e 3,2,0,1;
%e 10,5,2,0,1;
%e 33,16,5,2,0,1;
%e ...
%p F:=(1-sqrt(1-4*z))/z/(3-sqrt(1-4*z)): C:=(1-sqrt(1-4*z))/2/z: G:=t*z^2*C*F*(C-(1-t)/(1-t*z)): Gser:=simplify(series(G,z=0,15)): for n from 1 to 12 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 1 to 12 do seq(coeff(P[n],t,j),j=1..n) od; # yields sequence in triangular form
%Y Cf. A000957, A001558, A118973, A118974.
%K nonn,tabl
%O 1,7
%A _Emeric Deutsch_, May 08 2006
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