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Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n such that the difference between the sum of the valley abscissae and number of valleys is k (0 <= k <= (n-1)^2).
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%I #14 Feb 16 2021 02:06:32

%S 1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,1,1,1,1,1,1,2,3,3,4,4,4,4,4,3,3,2,1,

%T 1,1,1,1,1,2,3,4,5,6,7,8,9,9,10,10,9,9,8,7,6,5,4,3,2,1,1,1,1,1,1,2,3,

%U 4,6,7,9,11,13,15,18,19,21,23,24,24,25,24,24,23,21,19,18,15,13,11,9,7,6,4,3

%N Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n such that the difference between the sum of the valley abscissae and number of valleys is k (0 <= k <= (n-1)^2).

%C Row n contains 1+(n-1)^2 entries (n >= 1).

%C Row sums are the Catalan numbers (A000108).

%C Sum_{k=0..(n-1)^2} k*T(n,k) = A145885(n).

%C In the R. Stanley reference one has the equivalent statistic (maj(w) - des(w)) on Dyck words w.

%D R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see p. 236, Exercise 6.34 d.

%H Alois P. Heinz, <a href="/A145884/b145884.txt">Rows n = 0..32, flattened</a>

%F The generating polynomial for row n is P[n](t) = ((1+t)/(1+t^n))*binomial[2n,n]/[n+1], where [n+1]=1+t+t^2+...+t^n and binomial[2n,n] is a Gaussian polynomial.

%e T(4,5)=2 because we have UD.UUD.UDD (2+5-2=5) and UUUDDD.UD (6-1=5); here U=(1,1), D=(1,-1) and each valley is shown by a dot.

%e Triangle starts:

%e 1;

%e 1;

%e 1,1;

%e 1,1,1,1,1;

%e 1,1,1,2,2,2,2,1,1,1;

%e 1,1,1,2,3,3,4,4,4,4,4,3,3,2,1,1,1;

%p br:=proc(n) options operator, arrow: sum(q^i,i=0..n-1) end proc: f:=proc(n) options operator, arrow: product(br(j),j=1..n) end proc: cbr:=proc(n,k) options operator, arrow: f(n)/(f(k)*f(n-k)) end proc: P:=proc(n) options operator, arrow: sort(expand(simplify((q+1)*cbr(2*n, n)/(br(n+1)*(1+q^n))))) end proc: 1; for n to 7 do seq(coeff(P(n),q,k),k=0..(n-1)^2) end do; # yields sequence in triangular form

%t g[k_] := (1 - t^k)/(1 - t);

%t gpol[n_, k_] := If[0 <= k <= n, Product[g[n - j + 1]/g[j], {j, 1, k}], 0];

%t P[n_] := ((1 + t)/(1 + t^n)) gpol[2n, n]/Sum[t^k, {k, 0, n}];

%t T[n_] := CoefficientList[P[n] + O[t]^(n^2), t]; T[0] = {1};

%t T /@ Range[0, 7] // Flatten (* _Jean-François Alcover_, Feb 16 2021 *)

%Y Cf. A000108, A145885.

%K nonn,tabf

%O 0,13

%A _Emeric Deutsch_, Nov 06 2008