OFFSET
1,3
COMMENTS
Compare with triangle A019538, whose entries are given by
... Sum multinomial(n; n_1,n_2,...,n_k), where the sum extends over all compositions (n_1,n_2,...,n_k) of n into exactly k nonnegative parts.
Let P be the poset of all ordered pairs (S,T) of subsets of [n] with |S|=|T|, ordered componentwise by inclusion. T(n,k) is the number of length k chains in P from ({},{}) to ([n],[n]). - Geoffrey Critzer, Apr 15 2020
LINKS
Alois P. Heinz, Rows n = 1..100, flattened
FORMULA
Generating function: Let J(z) = Sum_{n>=0} z^n/n!^2. Then
1 + Sum_{n>=1} (Sum_{k = 1..n} T(n,k)*x^k)*z^n/n!^2) = 1/(1 - x*(J(z) - 1))
= 1 + x*z + (x + 4*x^2)*z^2/2!^2 + (x + 18*x^2 + 36*x^3)* z^3/3!^2 + ....
Relations with other sequences:
The change of variable z -> z/x followed by x -> 1/(x - 1) transforms the above bivariate generating function 1/(1 - x*(J(z) - 1)) into (1 - x)/(-x + J(z*(x-1))), which is the generating function for A192721.
1/k!*T(n,k) = A061691(n,k).
T(n,n) = n!^2 = A001044(n).
Row sums = A102221.
For n>=1, Sum_{k = 1..n} (-1)^(n+k)*T(n,k)/k = A002190(n).
EXAMPLE
The triangle begins
n/k|..1.....2.......3........4........5........6
================================================
.1.|..1
.2.|..1.....4
.3.|..1....18.....36
.4.|..1....68.....432......576
.5.|..1...250....3900....14400....14400
.6.|..1...922...32400...252000...648000...518400
...
T(4,2) = 68:
There are 3 compositions of 4 into 2 parts, namely, 4 = 2 + 2 = 1 + 3 = 3 + 1; hence
T(4,2) = (4!/(2!*2!))^2 + (4!/(1!*3!))^2 + (4!/(3!*1!))^2
= 36 + 16 + 16 = 68.
/...1................\ /..1..............\
|...3.....1...........||..1....1..........|
|..19....16.....5.....||..1....2....1.....|
|.211...299....65....1||..1....3....3....1|
|.....................||..................|
=
/...1...................\
|...4......1.............|
|..36.....18......1......|
|.576....432.....68.....1|
|........................|
MAPLE
J := unapply(BesselJ(0, 2*sqrt(-1)*sqrt(z)), z):
G := 1/(1-x*(J(z)-1)):
Gser := simplify(series(G, z = 0, 15)):
for n from 1 to 14 do
P[n] := n!^2*sort(coeff(Gser, z, n)) od:
for n from 1 to 14 do seq(coeff(P[n], x, k), k = 1..n) od;
# yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; expand(
`if`(n=0, 1, add(x*b(n-i)/i!^2, i=1..n)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n)*n!^2):
seq(T(n), n=1..14); # Alois P. Heinz, Sep 10 2019
MATHEMATICA
b[n_] := b[n] = Expand[If[n == 0, 1, Sum[x b[n-i]/i!^2, {i, 1, n}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n] n!^2];
Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, Dec 07 2019, after Alois P. Heinz *)
CROSSREFS
KEYWORD
AUTHOR
Peter Bala, Jul 11 2011
STATUS
approved