A292668 -id:A292668 - cp's OEIS frontend

A275431 Triangle read by rows: T(n,k) = number of ways to insert n pairs of parentheses in k words.

1, 2, 1, 5, 2, 1, 14, 8, 2, 1, 42, 24, 8, 2, 1, 132, 85, 28, 8, 2, 1, 429, 286, 100, 28, 8, 2, 1, 1430, 1008, 358, 105, 28, 8, 2, 1, 4862, 3536, 1309, 378, 105, 28, 8, 2, 1, 16796, 12618, 4772, 1410, 384, 105, 28, 8, 2, 1, 58786, 45220, 17556, 5220, 1435, 384, 105, 28, 8, 2, 1

Offset: 1

R. J. Mathar, Jul 27 2016

Multiset transformation of A000108. Each word is dissected by a number of parentheses associated to its length.

Also the number of forests of exactly k (unlabeled) ordered rooted trees with a total of n non-root nodes where each tree has at least 1 non-root node. - Alois P. Heinz, Sep 20 2017

			       1
       2     1
       5     2     1
      14     8     2     1
      42    24     8     2     1
     132    85    28     8     2     1
     429   286   100    28     8     2     1
    1430  1008   358   105    28     8     2     1
    4862  3536  1309   378   105    28     8     2     1
   16796 12618  4772  1410   384   105    28     8     2     1
   58786 45220 17556  5220  1435   384   105    28     8     2     1

Alois P. Heinz, Rows n = 1..141, flattened
Index entries for triangles generated by the Multiset Transformation

Cf. A000108 (1st column), A007223 (2nd column), A056711 (3rd column), A088327 (row sums).

T(2n,n) gives A292668.

C:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
b:= proc(n, i, p) option remember; `if`(p>n, 0, `if`(n=0, 1,
      `if`(min(i, p)<1, 0, add(b(n-i*j, i-1, p-j)*
       binomial(C(i)+j-1, j), j=0..min(n/i, p)))))
    end:
T:= (n, k)-> b(n$2, k):
seq(seq(T(n, k), k=1..n), n=1..12);  # Alois P. Heinz, Apr 13 2017

c[n_] := c[n] = Binomial[2*n, n]/(n + 1);
b[n_, i_, p_] := b[n, i, p] = If[p > n, 0, If[n == 0, 1, If[Min[i, p] < 1, 0, Sum[b[n - i*j, i - 1, p - j]*Binomial[c[i] + j - 1, j], {j, 0, Min[n/i, p]}]]]];
T[n_, k_] := b[n, n, k];
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 18 2018, after Alois P. Heinz *)

T(n,1) = A000108(n).

T(n,k) = Sum_{c_i*N_i=n,i=1..k} binomial(T(N_i,1)+c_i-1,c_i) for 1

G.f.: Product_{j>=1} 1/(1-y*x^j)^A000108(j). - Alois P. Heinz, Apr 13 2017

A304787 Expansion of Product_{k>=1} (1 + x^k)^(binomial(2*k,k)/(k+1)).

Original entry on oeis.org

1, 1, 2, 7, 20, 67, 222, 758, 2617, 9189, 32554, 116494, 420046, 1525221, 5571065, 20457808, 75476447, 279636977, 1039965746, 3880891892, 14527657602, 54537434161, 205270200229, 774460385687, 2928429307876, 11095878177649, 42122749335654, 160192845018335, 610224764470011

Offset: 0

Ilya Gutkovskiy, May 18 2018

nonn

N. J. A. Sloane, Transforms
Eric Weisstein's World of Mathematics, Catalan Number

Cf. A000108, A052805, A052854, A088327, A292668.

nmax = 28; CoefficientList[Series[Product[(1 + x^k)^CatalanNumber[k], {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[(-1)^(k/d + 1) d CatalanNumber[d], {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 28}]

G.f.: Product_{k>=1} (1 + x^k)^A000108(k).

a(n) ~ c * 4^n / n^(3/2), where c = exp( Sum_{k>=1} (-1)^k * (2 - 4^k + 4^k*sqrt(1 - 4^(1-k)))/(2*k) ) / sqrt(Pi) = 1.4863036894111457491052224706533674748514957... - Vaclav Kotesovec, Mar 21 2021

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cp's OEIS Frontend

A275431 Triangle read by rows: T(n,k) = number of ways to insert n pairs of parentheses in k words.

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