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%I #14 Jul 26 2023 17:39:22
%S 1,1,1,3,1,1,8,4,1,1,23,12,5,1,1,71,36,17,6,1,1,229,114,54,23,7,1,1,
%T 759,377,176,78,30,8,1,1,2566,1279,596,263,109,38,9,1,1,8817,4408,
%U 2070,912,382,148,47,10,1,1,30717,15375,7289,3240,1358,541,196,57,11,1,1,108278
%N Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n having k peaks of maximum height (1 <= k <= n).
%C Row sums are the Catalan numbers (A000108).
%C T(n,1) = A152880(n).
%C Sum_{k=1..n} k*T(n,k) = A152880(n+1).
%H Alois P. Heinz, <a href="/A152879/b152879.txt">Rows n = 1..150, flattened</a>
%F G.f. = G(t,z) = Sum_{j>=1}tz^j/(f(j)(f(j)-tzf(j-1))), where the f(j)'s are the Fibonacci polynomials (in z) defined by f(0)=f(1)=1, f(j) = f(j-1) - zf(j-2), j>=2 (Sergi Elizalde).
%e T(4,2)=4 because we have UU(UD)(UD)DD, U(UD)DU(UD)D, U(UD)(UD)DUD and UDU(UD)(UD)D, where U=(1,1), D=(1,-1), with the peaks of maximum height shown between parentheses.
%e Triangle starts:
%e 1;
%e 1, 1;
%e 3, 1, 1;
%e 8, 4, 1, 1;
%e 23, 12, 5, 1, 1;
%e 71, 36, 17, 6, 1, 1;
%e ...
%p f[0] := 1: f[1] := 1: for i from 2 to 20 do f[i] := sort(expand(f[i-1]-z*f[i-2])) end do: G := sum(t*z^j/(f[j]*(f[j]-t*z*f[j-1])), j = 1 .. 20): Gser := simplify(series(G, z = 0, 17)): for n to 12 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 12 do seq(coeff(P[n], t, k), k = 1 .. n) end do; # yields sequence in triangular form
%Y Cf. A000108, A152880.
%Y T(2n,n) gives A364030.
%K nonn,tabl
%O 1,4
%A _Emeric Deutsch_, Jan 02 2009
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