A178519 - OEIS

%I #6 Nov 12 2019 04:16:38

%S 1,1,1,1,4,1,11,3,32,10,98,33,1,309,114,6,998,402,30,3285,1439,137,1,

%T 10981,5205,600,10,37178,18976,2562,70,127227,69610,10758,416,1,

%U 439369,256626,44640,2250,15,1529280,949974,183594,11452,140,5359314,3528725

%N Triangle read by rows: T(n,k) is the number of ordered trees with n edges and having k vertices of outdegree 2 that have (two) leaves as their (two) children.

%C Row n has 1 + floor(n/3) entries (n >= 3).

%C Sum of entries in row n is A000108(n) (Catalan numbers).

%C T(n,0) = A178520(n).

%C Sum_{k>=0} k*T(n,k) = binomial(2n-5, n-2) = A001700(n-3).

%C Statistic suggested by Lou Shapiro.

%F G.f. G=G(t,z) satisfies z*G^2 - (1 - z^3 + tz^3)*G + 1 - z^2 + tz^2 = 0.

%e T(2,1)=1 because we have \/ .

%e Triangle starts:

%e    1;

%e    1;

%e    1,  1;

%e    4,  1;

%e   11,  3;

%e   32, 10;

%e   98, 33,  1;

%p eq := z*G^2-(1-z^3+t*z^3)*G+1-z^2+t*z^2: G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 18)): for n from 0 to 15 do P[n] := sort(coeff(Gser, z, n)) end do: 1; 1; 1, 1; for n from 3 to 15 do seq(coeff(P[n], t, j), j = 0 .. floor((1/3)*n)) end do; # yields sequence in triangular form

%Y Cf. A000108, A178520, A001700.

%K nonn,tabf

%O 0,5

%A _Emeric Deutsch_, May 31 2010