A138155 - OEIS

%I #28 Aug 10 2018 09:16:49

%S 1,1,1,1,2,1,1,4,3,1,1,7,8,4,1,1,12,19,13,5,1,1,20,42,37,19,6,1,1,33,

%T 89,97,62,26,7,1,1,54,183,240,184,95,34,8,1,1,88,368,570,511,312,137,

%U 43,9,1,1,143,728,1312,1351,951,491,189,53,10,1

%N Triangle read by rows: T(n,k) is the number of Dyck paths with nondecreasing peaks having semilength n and with height of last peak equal to k (1 <= k <= n).

%C Row sums yield A048285.

%C T(n,2) = A000071(n+1) (Fibonacci numbers - 1).

%C T(n,3) = A095681(n-3).

%H Alois P. Heinz, <a href="/A138155/b138155.txt">Rows n = 1..141, flattened</a>

%H J. G. Penaud and O. Roques, <a href="https://dx.doi.org/10.1016/S0012-365X(01)00261-8">Génération de chemins de Dyck à pics croissants</a>, Discrete Mathematics, Vol. 246, no. 1-3 (2002), 255-267.

%F G.f.: Sum_{n >= 0} {(-1)^n tz^{2n+1}(1-z)}/ {Product_{i=1...n+1}((1-z)(1-tz^i)-z)}.

%F Conjectural g.f.: Sum_{n>=1} (t*x*(1 - x))^n/( Product_{i=2..n+1} (1 - 2*x + x^i) ) = t*x + (t + t^2)*x^2 + (t + 2*t^2 + t^3)*x^3 + ... (checked up to x^12). - _Peter Bala_, Mar 31 2017

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

%e T(5,4)=4 because we have UDUUUUDDDD, UUDUUUDDDD, UUUDUUDDDD and UUUUDUDDDD, where U=(1,1) and D=(1,-1).

%e Triangle T(n,k) begins:

%e 1;

%e 1,  1;

%e 1,  2,   1;

%e 1,  4,   3,   1;

%e 1,  7,   8,   4,   1;

%e 1, 12,  19,  13,   5,  1;

%e 1, 20,  42,  37,  19,  6,  1;

%e 1, 33,  89,  97,  62, 26,  7, 1;

%e 1, 54, 183, 240, 184, 95, 34, 8, 1;

%p g:=sum((-1)^n*t*z^(2*n+1)*(1-z)/(product((1-z)*(1-t*z^i)-z,i=1..n+1)), n=0.. 30): gser:=simplify(series(g,z=0,15)): for n to 11 do P[n]:=sort(coeff(gser, z,n)) end do: for n to 11 do seq(coeff(P[n],t,j),j=1..n) end do; # yields sequence in triangular form

%p # second Maple program:

%p b:= proc(x, y, k, t) option remember; `if`(x=0, z^k,

%p       `if`(t and y<k or y<1, 0, b(x-1, y-1, max(y, k),

%p        false))+ `if`(y+2>x, 0, b(x-1, y+1, k, true)))

%p     end:

%p T:= n-> (p-> seq(coeff(p, z, i), i=1..n))(b(2*n, 0$2, true)):

%p seq(T(n), n=1..12);  # _Alois P. Heinz_, Apr 02 2017

%t b[x_, y_, k_, t_] := b[x, y, k, t] = If[x==0, z^k, If[t && y<k || y<1, 0, b[x-1, y-1, Max[y, k], False]] + If[y+2 > x, 0, b[x-1, y+1, k, True]]];

%t T[n_] := Function[p, Table[Coefficient[p, z, i], {i, 1, n}]][b[2*n, 0, 0, True]];

%t Array[T, 12] // Flatten (* _Jean-François Alcover_, Jun 19 2018, from _Alois P. Heinz_'s 2nd Maple program *)

%Y Cf. A000071, A048285, A095681.

%K nonn,tabl

%O 1,5

%A _Emeric Deutsch_, Mar 04 2008