A187761 -id:A187761 - cp's OEIS frontend

A264430 Triangle read by rows, Bell transform of second order Bell numbers (A187761).

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 23, 50, 35, 10, 1, 0, 106, 268, 225, 85, 15, 1, 0, 568, 1645, 1603, 735, 175, 21, 1, 0, 3459, 11348, 12572, 6713, 1960, 322, 28, 1, 0, 23544, 86775, 107738, 65352, 22323, 4536, 546, 36, 1, 0, 176850, 727629, 1001895, 678980, 263865, 63021, 9450, 870, 45, 1

Offset: 0

Peter Luschny, Nov 13 2015

			Triangle starts:
[1]
[0,    1]
[0,    1,     1]
[0,    2,     3,     1]
[0,    6,    11,     6,    1]
[0,   23,    50,    35,   10,    1]
[0,  106,   268,   225,   85,   15,   1]
[0,  568,  1645,  1603,  735,  175,  21,  1]
[0, 3459, 11348, 12572, 6713, 1960, 322, 28, 1]

Peter Luschny, The Bell transform

Cf. A000110, A187761, A264428.

nmax = 10;
A187761[n_] := Sum[BellY[n, k, BellB /@ Range[0, n-1]], {k, 0, n}];
Table[BellY[n, k, A187761 /@ Range[0, nmax]], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 10 2019 *)

# uses[bell_transform from A264428]
def A264430_triangle(dim):
    uno = [1]*dim
    bell_numbers = [sum(bell_transform(n, uno)) for n in range(dim)]
    bell_number_2 = [sum(bell_transform(n, bell_numbers)) for n in range(dim)]
    for n in range(dim): print(bell_transform(n, bell_number_2))
A264430_triangle(10)

A264431 Triangle read by rows, inverse Bell transform of second order Bell numbers (A187761).

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 1, -3, 1, 0, -1, 7, -6, 1, 0, 2, -15, 25, -10, 1, 0, -8, 37, -90, 65, -15, 1, 0, 27, -133, 322, -350, 140, -21, 1, 0, -70, 587, -1330, 1757, -1050, 266, -28, 1, 0, 265, -2526, 6607, -9114, 7077, -2646, 462, -36, 1

Offset: 0

Peter Luschny, Nov 13 2015

			[ 1 ]
[ 0,     1 ]
[ 0,    -1,     1 ]
[ 0,     1,    -3,     1 ]
[ 0,    -1,     7,    -6,     1 ]
[ 0,     2,   -15,    25,   -10,     1 ]
[ 0,    -8,    37,   -90,    65,   -15,     1 ]
[ 0,    27,  -133,   322,  -350,   140,   -21,     1 ]
[ 0,   -70,   587, -1330,  1757, -1050,   266,   -28,     1 ]
[ 0,   265, -2526,  6607, -9114,  7077, -2646,   462,   -36,   1 ]

Peter Luschny, The Bell transform

Cf. A000110, A187761, A264428, A264429.

# uses[bell_transform from A264428, inverse_bell_transform from A264429]
def A264431_matrix(dim):
    uno = [1]*dim
    bell_numbers = [sum(bell_transform(n, uno)) for n in range(dim)]
    bell_number_2 = [sum(bell_transform(n, bell_numbers)) for n in range(dim)]
    return inverse_bell_transform(dim, bell_number_2)
A264431_matrix(10)

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

Original entry on oeis.org

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

Peter Luschny, Nov 13 2015

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

			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]

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

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

# 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

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 *)

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

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

# 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))

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)

A187824 a(n) is the largest m such that n is congruent to -1, 0 or 1 mod k for all k from 1 to m.

Original entry on oeis.org

3, 4, 5, 6, 3, 4, 4, 5, 3, 6, 4, 4, 3, 5, 5, 4, 3, 6, 5, 5, 3, 4, 6, 6, 3, 4, 4, 7, 3, 6, 4, 4, 3, 7, 7, 4, 3, 5, 5, 8, 3, 4, 5, 5, 3, 4, 4, 8, 3, 5, 4, 4, 3, 9, 5, 4, 3, 6, 6, 6, 3, 4, 5, 6, 3, 4, 4, 5, 3, 10, 4, 4, 3, 5, 5, 4, 3, 6, 5, 5, 3, 4, 7, 7, 3, 4, 4, 6, 3, 7, 4, 4, 3, 6, 6, 4, 3, 5, 5, 6, 3

Offset: 2

Kival Ngaokrajang, Dec 27 2012

This sequence and A187771 and A187761 are winners in the contest held at the 2013 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 14 2013
If n = t!-1 then a(n) >= t, so sequence is unbounded. - N. J. A. Sloane, Dec 30 2012
First occurrence of k = 3, 4, 5, ...:  2, 3, 4, 5, 29, 41, 55, 71, 881, 791, 9360, 10009, 1079, 30239, (17 unknown), 246960, (19 unknown), 636481, 1360800, 3160079, (23 unknown), 2162161, 266615999, 39412801 (27 unknown), 107881201, ... Searched up to 3*10^9. - Robert G. Wilson v, Dec 31 2012

			For n = 6, a(6) = 3 as follows.
m    Residue of 6 (mod m)
1             0
2             0
3             0
4             2
5             1
6             0
7            -1

N. J. A. Sloane, Table of n, a(n) for n = 2..10000
Don Reble, Division gets rough: OEIS A187824 and A220890

For values of n which set a new record see A220891.

For smallest inverse see A220890 and A056697.

A187824:= proc(n)
   local j,r;
   for j from 4 do
     r:= mods(n, j);
     if r <> r^3 then return j-1 end if
   end do
end proc; # Robert Israel, Dec 31 2012

f[n_] := Block[{k = 4, r}, While[r = Mod[n, k]; r < 2 || k - r < 2, k++]; k - 1]; Array[f, 101, 2] (* Robert G. Wilson v, Dec 31 2012 *)

A187824(n)={n++>2 && for(k=4,oo, n%k>2 && return(k-1))} \\ M. F. Hasler, Dec 31 2012, minor edits Aug 20 2020

a(n)=my(k=3);n++;while(n%k++<3,);k-1 \\ Charles R Greathouse IV, Jan 02 2013

from gmpy2 import t_mod
def A187824(n):
    k = 1
    while t_mod(n+1,k) < 3:
        k += 1
    return k-1 # Chai Wah Wu, Aug 31 2014

def a(n):
   m=1
   while abs(n%m) < 2:
      m += 1
   return m
[a(n) for n in range(1,100)]
# Derek Orr, Aug 31 2014, corrected & edited by M. F. Hasler, Aug 20 2020

If n == 0 (mod 20), then a(n-2) = a(n+2) = 3, while a(n) = 5,5,6, 5,5,8, 5,5,6, 5,5,6, 5,5,7, 5,5,6, 5,5,7, ... with records a(20) = 5, a(60) = 6, a(120) = 8, a(720) = 10, a(2520) = 12, a(9360) = 13, ... If n == 0 (mod 5), but is not a multiple of 20, then always a(n-2) = a(n+2) = 4, while a(n) = 6,3,5, 6,3,7, 5,3,9, 6,3,5, 7,3,6, 5,3,6, 7,3,5, ... - Vladimir Shevelev, Dec 31 2012

a(n)=3 iff n == 2 (mod 4). a(n)=4 iff n == 3, 7, 8, 12, 13, 17 (mod 20), i.e., n == 2 or 3 (mod 5) but not n == 2 (mod 4). In the same way one can obtain a covering set for any value taken by a(n), this is actually nothing else than the definition. For example, n == 2, 3 or 4 (mod 6) but not 2 or 3 (mod 5) nor 2 (mod 4) yields a(n)=5 iff n == 4, 9, 15, 16, 20, 21, 39, 40, 44, 45, 51 or 56 (mod 60), etc. - M. F. Hasler, Dec 31 2012

Corrected m = 100 by Kival Ngaokrajang, Dec 30 2012
Definition & example corrected by Kival Ngaokrajang, Dec 30 2012
More terms from N. J. A. Sloane, Dec 30 2012

A179455 Triangle read by rows: number of permutation trees of power n and height <= k + 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 5, 6, 1, 15, 23, 24, 1, 52, 106, 119, 120, 1, 203, 568, 700, 719, 720, 1, 877, 3459, 4748, 5013, 5039, 5040, 1, 4140, 23544, 36403, 39812, 40285, 40319, 40320, 1, 21147, 176850, 310851, 354391, 362057, 362836, 362879, 362880

Offset: 0

Peter Luschny, Aug 11 2010

Partial row sums of A179454. Special cases: A179455(n,1) = BellNumber(n) = A000110(n) for n > 1; A179455(n,n-1) = n! for n > 1 and A179455(n,n-2) = A033312(n) for n > 1. Column 3 is A187761(n) for n >= 3.

See the interpretation of Joerg Arndt in A187761: Maps such that f^[k](x) = f^[k-1](x) correspond to column k of A179455 (for n >= k). - Peter Luschny, Jan 08 2013

			As a (0,0)-based triangle with an additional column [1,0,0,0,...] at the left hand side:
1;
0, 1;
0, 1, 2;
0, 1, 5, 6;
0, 1, 15, 23, 24;
0, 1, 52, 106, 119, 120;
0, 1, 203, 568, 700, 719, 720;
0, 1, 877, 3459, 4748, 5013, 5039, 5040;
0, 1, 4140, 23544, 36403, 39812, 40285, 40319, 40320;
0, 1, 21147, 176850, 310851, 354391, 362057, 362836, 362879, 362880;

Alois P. Heinz, Rows n = 0..141, flattened
Swapnil Garg, Alan Peng, Classical and consecutive pattern avoidance in rooted forests, arXiv:2005.08889 [math.CO], May 2020.
Peter Luschny, Permutation Trees.

Row sums are A264151.

Cf. A000110, A179454, A179456, A187761, A264428.

b[n_, t_, h_] := b[n, t, h] = If[n == 0 || h == 0, 1, Sum[Binomial[n - 1, j - 1]*b[j - 1, 0, h - 1]*b[n - j, t, h], {j, 1, n}]];
T[0, 0] = 1; T[n_, k_] := b[n, 1, k];
Table[T[n, k], {n, 0, 9}, {k, 0, If[n == 0, 0, n-1]}] // Flatten (* Jean-François Alcover, Jul 10 2019, after Alois P. Heinz in A179454 *)

# Generating algorithm from Joerg Arndt.
def A179455row(n):
    def generate(n, k):
        if n == 0 or k == 0: return 0
        for j in range(n-1, 0, -1):
            f = a[j] + 1
            while f <= j:
                a[j] = f1 = fl = f
                for i in range(k):
                    fl = f1
                    f1 = a[fl]
                if f1 == fl: return j
                f += 1
            a[j] = 0
        return 0
    count = [1 for j in range(n)] if n > 0 else [1]
    for k in range(n):
        a = [0 for j in range(n)]
        while generate(n, k) != 0:
            count[k] += 1
    return count
for n in range(9): A179455row(n) # Peter Luschny, Jan 08 2013

# uses[bell_transform from A264428]
# Adds the column (1,0,0,0,..) to the left hand side and starts at n=0.
def A179455_matrix(dim):
    b = [1]+[0]*(dim-1); L = [b]
    for k in range(dim):
        b = [sum(bell_transform(n, b)) for n in range(dim)]
        L.append(b)
    return matrix(ZZ, dim, lambda n, k: L[k][n] if k<=n else 0)
print(A179455_matrix(10)) # Peter Luschny, Dec 06 2015

A187771 Numbers whose sum of divisors is the cube of the sum of its prime divisors.

Original entry on oeis.org

245180, 612408, 639198, 1698862, 1721182, 5154168, 7824284, 15817596, 20441848, 25969788, 27688078, 28404862, 35860609, 67149432, 77378782, 91397838, 96462862, 179302264, 191550135, 289772221, 306901244, 311657084, 392802179, 441839706, 572673855, 652117774, 988918364

Offset: 1

Manuel Valdivia, Jan 04 2013

This sequence and A187824 and A187761 are winners in the contest held at the 2013 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 14 2013
The identity sigma(k) = (sopf(k))^m only occurs for m = 3 (this sequence) in the given range, however it is likely that it also occurs for other powers m in higher numbers.
The smallest k such that sigma(k) = sopf(k)^m, for m=4,5,6 are 1056331752 (A221262), 213556659624 (A221263) and 45770980141656, respectively. - Giovanni Resta, Jan 07 2013
Prime divisors are taken without multiplicity. - Harvey P. Dale, Dec 17 2016

			a(13) = 35860609 = 41 * 71 * 97 * 127, then sigma(35860609) = 37933056 = (41 + 71 + 97 + 127)^3.

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38.

Donovan Johnson and Robert Gerbicz, Table of n, a(n) for n = 1..1105 (first 100 terms from _Donovan Johnson_)
W. Sierpinski, Number Of Divisors And Their Sum, Elementary theory of numbers, Warszawa, 1964.

Cf. A000203, A008472, A020477, A070222.

Cf. A221262 (sigma(k)=sopf(k)^4), A221263 (sigma(k)=sopf(k)^5).

d[n_]:= If[Plus@@Divisors[n]-Power[Plus@@Select[Divisors[n], PrimeQ], 3]==0, n]; Select[Range[2,10^9], #==d[#]&]
Select[Range[2, 10^9],DivisorSigma[1,#]==Total[FactorInteger[#][[All, 1]]]^3&] (* Harvey P. Dale, Dec 17 2016 *)

is(n)=my(f=factor(n));sum(i=1,#f~,f[i,1])^3==sigma(n) \\ Charles R Greathouse IV, Jun 29 2013

a(n) = k if sigma(k) = (sopf(k))^3, where sigma(k) = A000203(k) and sopf(k) = A008472(k).

A265312 Square array read by ascending antidiagonals, Bell numbers iterated by the Bell transform.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 5, 1, 1, 1, 2, 6, 15, 1, 1, 1, 2, 6, 23, 52, 1, 1, 1, 2, 6, 24, 106, 203, 1, 1, 1, 2, 6, 24, 119, 568, 877, 1, 1, 1, 2, 6, 24, 120, 700, 3459, 4140, 1, 1, 1, 2, 6, 24, 120, 719, 4748, 23544, 21147, 1, 1, 1, 2, 6, 24, 120, 720, 5013, 36403, 176850, 115975, 1

Offset: 0

Peter Luschny, Dec 06 2015

			[1, 1, 1, 1,  1,   1,   1,    1,     1, ...] A000012
[1, 1, 2, 5, 15,  52, 203,  877,  4140, ...] A000110
[1, 1, 2, 6, 23, 106, 568, 3459, 23544, ...] A187761
[1, 1, 2, 6, 24, 119, 700, 4748, 36403, ...] A264432
[1, 1, 2, 6, 24, 120, 719, 5013, 39812, ...]
[1, 1, 2, 6, 24, 120, 720, 5039, 40285, ...]
[...                                    ...]
[1, 1, 2, 6, 24, 120, 720, 5040, 40320, ...] A000142 = main diagonal.

Alois P. Heinz, Antidiagonals n = 0..140, flattened
Peter Luschny, The Bell transform

Cf. A000012, A000110, A000142, A187761, A264428, A264432, A265313.

A:= proc(n, h) option remember; `if`(min(n, h)=0, 1, add(
      binomial(n-1, j-1)*A(j-1, h-1)*A(n-j, h), j=1..n))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Aug 21 2017

A[n_, h_]:=A[n, h]=If[Min[n, h]==0, 1, Sum[Binomial[n - 1, j - 1] A[j - 1, h - 1] A[n - j, h] , {j, n}]]; Table[A[n, d - n], {d, 0, 12}, {n, 0, d}]//Flatten (* Indranil Ghosh, Aug 21 2017, after maple code *)

from sympy.core.cache import cacheit
from sympy import binomial
@cacheit
def A(n, h): return 1 if min(n, h)==0 else sum([binomial(n - 1, j - 1)*A(j - 1, h - 1)*A(n - j, h) for j in range(1, n + 1)])
for d in range(13): print([A(n, d - n) for n in range(d + 1)]) # Indranil Ghosh, Aug 21 2017, after Maple code

# uses[bell_transform from A264428]
def bell_number_matrix(ord, len):
    b = [1]*len; L = [b]
    for k in (1..ord-1):
        b = [sum(bell_transform(n, b)) for n in range(len)]
        L.append(b)
    return matrix(ZZ, L)
print(bell_number_matrix(6, 9))

A264432 Third-order Bell numbers.

Original entry on oeis.org

1, 1, 2, 6, 24, 119, 700, 4748, 36403, 310851, 2922606, 29977587, 332929492, 3978258079, 50872884285, 692985674373, 10015172966221, 153021613683924, 2464031776132958, 41698912656882644, 739771703127828419, 13727160292457369098, 265876635231121617716

Offset: 0

Peter Luschny, Dec 02 2015

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..469
Peter Luschny, The Bell transform

Cf. A000110, A187761, A264428, A265312.

b:= proc(n, h) option remember; `if`(min(n, h)=0, 1, add(
      binomial(n-1, j-1)*b(j-1, h-1)*b(n-j, h), j=1..n))
    end:
a:= n-> b(n, 3):
seq(a(n), n=0..22);  # Alois P. Heinz, Aug 21 2017

b[n_, h_]:=b[n, h]=If[Min[n, h]==0, 1, Sum[Binomial[n - 1, j - 1] b[j - 1, h - 1] b[n - j, h] , {j, n}]]; Table[b[n, 3], {n, 0, 30}] (* Indranil Ghosh, Aug 21 2017, after Maple code *)

\\ For n>23 precision has to be adapted as needed!
A = exp('x + O('x^33) );
B = exp( intformal(A) );
C = exp( intformal(B) );
D = exp( intformal(C) );
Vec( serlaplace(D) )

from sympy.core.cache import cacheit
from sympy import binomial
@cacheit
def b(n, h): return 1 if min(n, h)==0 else sum(binomial(n - 1, j - 1)*b(j - 1, h - 1)*b(n - j, h) for j in range(1, n + 1))
def a(n): return b(n, 3)
print([a(n) for n in range(31)]) # Indranil Ghosh, Aug 21 2017, after Maple code

# uses[bell_transform from A264428]
def A264432_list(dim):
    uno = [1]*dim
    bell_number = [sum(bell_transform(n, uno)) for n in range(dim)]
    bell_number_2 = [sum(bell_transform(n, bell_number)) for n in range(dim)]
    return [sum(bell_transform(n, bell_number_2)) for n in range(dim)]
print(A264432_list(23))

cp's OEIS Frontend

A264430 Triangle read by rows, Bell transform of second order Bell numbers (A187761).

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A264431 Triangle read by rows, inverse Bell transform of second order Bell numbers (A187761).

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

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A187824 a(n) is the largest m such that n is congruent to -1, 0 or 1 mod k for all k from 1 to m.

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A179455 Triangle read by rows: number of permutation trees of power n and height <= k + 1.

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A187771 Numbers whose sum of divisors is the cube of the sum of its prime divisors.

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A265312 Square array read by ascending antidiagonals, Bell numbers iterated by the Bell transform.

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