A264428 Triangle read by rows, Bell transform of Bell numbers.

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 5, 11, 6, 1, 0, 15, 45, 35, 10, 1, 0, 52, 205, 210, 85, 15, 1, 0, 203, 1029, 1330, 700, 175, 21, 1, 0, 877, 5635, 8946, 5845, 1890, 322, 28, 1, 0, 4140, 33387, 63917, 50358, 20055, 4410, 546, 36, 1, 0, 21147, 212535, 484140, 450905, 214515, 57855, 9240, 870, 45, 1

Offset

0,8

Comments

Consider the sequence S0 -> T0 -> S1 -> T1 -> S2 -> T2 -> ... Here Sn -> Tn indicates the Bell transform mapping a sequence Sn to a triangle Tn as defined in the link and Tn -> S{n+1} the operator associating a triangle with the sequence of its row sums. If

S0 = A000012 = <1,1,1,...> then

T0 = A048993 # Stirling subset numbers,

S1 = A000110 # Bell numbers,

T1 = A264428 # Bell transform of Bell numbers,

S2 = A187761 # second-order Bell numbers,

T2 = A264430 # Bell transform of second-order Bell numbers,

S3 = A264432 # third-order Bell numbers.

This construction is closely related to permutations trees and A179455. Sn is A179455_col(n+1) prepended by A179455_diag(k) = k! for k <= n. In other words, Sn 'converges' to n! for n -> oo.

Given a sequence (s(n))n>=0 with s(0) = 0 and with e.g.f. B(x) = Sum_{n >= 1} s(n)*x^n/n!, then the Bell matrix associated with s(n) equals the exponential Riordan array [1, B(x)] belonging to the Lagrange subgroup of the exponential Riordan group. Omitting the first row and column from the Bell matrix produces the exponential Riordan array [d/dx(B(x)), B(x)] belonging to the Derivative subgroup of the exponential Riordan group. - Peter Bala, Jun 07 2016

Links

G. C. Greubel, Table of n, a(n) for n = 0..1325

Peter Luschny, The Bell transform

Peter Luschny, Permutation Trees

Formula

From Peter Bala, Jun 07 2016: (Start)

E.g.f.: exp(t*B(x)), where B(x) = Integral_{u = 0..x} exp(exp(u) - 1) du = x + x^2/2! + 2*x^3/3! + 5*x^4/4! + 15*x^5/5! + 52*x^6/6! + ....

Row polynomial recurrence: R(n+1,t) = t*Sum_{k = 0 ..n} binomial(n,k)*Bell(k)* R(n-k,t) with R(0,t) = 1. (End)

Example

Triangle starts:
[1]
[0,   1]
[0,   1,    1]
[0,   2,    3,    1]
[0,   5,   11,    6,    1]
[0,  15,   45,   35,   10,    1]
[0,  52,  205,  210,   85,   15,   1]
[0, 203, 1029, 1330,  700,  175,  21,  1]
[0, 877, 5635, 8946, 5845, 1890, 322, 28, 1]

Maple

# Computes sequence in matrix form.
BellMatrix := proc(f, len) local T, A; A := [seq(f(n), n=0..len-2)];
T := proc(n, k) option remember; if k=0 then k^n else
add(binomial(n-1, j-1)*T(n-j, k-1)*A[j], j=1..n-k+1) fi end;
Matrix(len, (n, k)->T(n-1, k-1), shape=triangular[lower]) end:
BellMatrix(n -> combinat:-bell(n), 9); # Peter Luschny, Jan 21 2016
# Alternative, using the recurrence of Peter Bala:
R := proc(n) option remember; if n = 0 then 1 else
t*add(binomial(n-1, k)*combinat:-bell(k)*R(n-k-1, t), k=0..n-1) fi end:
T_row := n-> seq(coeff(R(n), t, k), k=0..n):
seq(print(T_row(n)), n=0..8); # Peter Luschny, Jun 09 2016

Mathematica

BellMatrix[f_Function|f_Symbol, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
rows = 11;
M = BellMatrix[BellB, rows];
Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 21 2016, updated Jul 14 2018 *)
With[{r = 8}, Flatten[Table[BellY[n, k, BellB[Range[0, r]]], {n, 0, r}, {k, 0, n}]]] (* Jan Mangaldan, May 22 2016 *)

Prog

(Sage)
# The functions below are referenced in various other sequences.
def bell_transform(n, a): # partition_based
    row = []
    fn = factorial(n)
    for k in (0..n):
        result = 0
        for p in Partitions(n, length=k):
            factorial_product = 1
            power_factorial_product = 1
            for part, count in p.to_exp_dict().items():
                factorial_product *= factorial(count)
                power_factorial_product *= factorial(part)**count
            coefficient = fn//(factorial_product*power_factorial_product)
            result += coefficient*prod([a[i-1] for i in p])
        row.append(result)
    return row
def bell_matrix(generator, dim):
    G = [generator(k) for k in srange(dim)]
    row = lambda n: bell_transform(n, G)
    return matrix(ZZ, [row(n)+[0]*(dim-n-1) for n in srange(dim)])
def inverse_bell_matrix(generator, dim):
    G = [generator(k) for k in srange(dim)]
    row = lambda n: bell_transform(n, G)
    M = matrix(ZZ, [row(n)+[0]*(dim-n-1) for n in srange(dim)]).inverse()
    return matrix(ZZ, dim, lambda n, k: (-1)^(n-k)*M[n, k])
bell_numbers = [sum(bell_transform(n, [1]*10)) for n in range(11)]
for n in range(11): print(bell_transform(n, bell_numbers))
(PARI)
bell_matrix(f, len) = { my( m = matrix(len, len) );  m[1, 1] = 1;
  for( n = 1, len-1, m[n+1, 2] = f(n-1) );
  for( n = 0, len-1, for( k = 1, n,
     m[n+1, k+1] = sum(j = 1, n-k+1, binomial(n-1, j-1)*m[n-j+1, k]*m[j+1, 2]) ));
  return( m )
}
f(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1 - i*x)), x^n * O(x)), n);
bell_matrix(f, 9) \\ Peter Luschny, Jan 24 2016
(Python)
from functools import cache
from math import comb as binomial
def BellMatrix(f, size):
    A = [f(n) for n in range(size - 1)]
    @cache
    def T(n, k):
        if k == 0: return k ** n
        return sum(
            binomial(n - 1, j) * T(n - j - 1, k - 1) * A[j]
            for j in range(n - k + 1) )
    return [[T(n, k) for k in range(n + 1)] for n in range(size)]
@cache
def b(n, k=0): return n < 1 or k*b(n-1, k) + b(n-1, k+1)
print(BellMatrix(b, 9))  # Peter Luschny, Jun 14 2022

Crossrefs

Cf. A000012, A000110, A000217, A000914, A027801, A048993, A051836, A179455, A187761 (row sums), A264430, A264432, A265312.

Sequence in context: A172380 A144633 A352366 * A256550 A005210 A352363

Adjacent sequences: A264425 A264426 A264427 * A264429 A264430 A264431

Keyword

nonn,tabl

Author

Peter Luschny, Nov 13 2015

Status

approved