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
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
Programs
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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
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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)
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