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

Author

Peter Luschny, Nov 13 2015

Keywords

• nonn
• tabl

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

Examples

			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]
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1325
• Peter Luschny, The Bell transform
• Peter Luschny, Permutation Trees

Crossrefs

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

Programs

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 *)

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

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))

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)