A000110 -id:A000110 - cp's OEIS frontend

A135084 a(n) = A000110(2^n-1).

1, 5, 877, 1382958545, 10293358946226376485095653, 8250771700405624889912456724304738028450190134337110943817172961

Offset: 1

Thomas Wieder, Nov 18 2007

nonn

Number of set partitions of all nonempty subsets of a set, Bell(2^n-1).

			Let S={1,2,3,...,n} be a set of n elements and let
SU be the set of all nonempty subsets of S. The number of elements of SU is |SU| = 2^n-1. Now form all possible set partitions from SU where the empty set is excluded. This gives a set W and its number of elements is |W| = Sum_{k=1..2^n-1} Stirling2(2^n-1,k).
For S={1,2} we have SU = { {1}, {2}, {1,2} } and W =
{
{{1}, {2}, {1, 2}},
{{1, 2}, {{1}, {2}}},
{{2}, {{1}, {1, 2}}},
{{1}, {{2}, {1, 2}}},
{{{1}, {2}, {1, 2}}}
}
and |W| = 5.

Amiram Eldar, Table of n, a(n) for n = 1..9

Cf. A000079, A000110, A008277, A077585, A135085.

ZahlDerMengenAusMengeDerZerlegungenEinerMenge:=proc() local n,nend,arg,k,w; nend:=5; for n from 1 to nend do arg:=2^n-1; w[n]:=sum((stirling2(arg,k)), k=1..arg); od; print(w[1],w[2],w[3],w[4],w[5],w[6],w[7],w[8],w[9],w[10]); end proc;

BellB[2^Range[6]-1] (* Harvey P. Dale, Jul 22 2012 *)

from sympy import bell
def A135084(n): return bell(2**n-1) # Chai Wah Wu, Jun 22 2022

a(n) = Sum_{k=1..2^n-1} Stirling2(2^n-1,k) = Bell(2^n-1), where Stirling2(n, k) is the Stirling number of the second kind and Bell(n) is the Bell number.

A146094 Bell numbers (A000110) read mod 4.

Original entry on oeis.org

1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0, 3, 3, 2, 1, 1, 2, 1, 3, 0, 3, 1, 0

Offset: 0

N. J. A. Sloane, Feb 07 2009

G. C. Greubel, Table of n, a(n) for n = 0..2500
P. Pleasants, W. Lunnon, and N. Stephens, Arithmetic properties of Bell numbers to a composite modulus I, Acta Arithmetica 35 (1979), pp. 1-16.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,1).

Cf. A000110, A146093-A146122 (Bell numbers read mod 3 to mod 32).

Cf. A054767 (periods).

[Bell(n) mod 4: n in [0..100]]; // Vincenzo Librandi, Jan 30 2016

Mod[BellB[Range[0, 50]], 4] (* G. C. Greubel, Jan 30 2016 *)

a(n) = a(n-12). - Charles R Greathouse IV, Jul 06 2011 [See A054767. - Jianing Song, Jun 20 2025]

A152431 Eigentriangle, row sums = A000110, the Bell numbers.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 6, 2, 2, 5, 22, 6, 4, 5, 15, 92, 22, 12, 10, 15, 52, 426, 92, 44, 30, 30, 52, 203, 2146, 426, 184, 110, 90, 104, 203, 877, 11624, 2146, 852, 460, 330, 312, 406, 877, 4140, 67146, 11624, 4292, 2130, 1380, 1144, 1218, 1754, 4140, 21147

Offset: 1

Gary W. Adamson, Dec 04 2008

Row sums = the Bell numbers, A000110, starting with offset 1: (1, 2, 5, 15, 52,...).
Left border = A074664 (1, 1, 2, 6, 22 92, 426,...), the INVERTi transform of (1, 2, 5, 15, 52,...).
Sum of n-th row terms = rightmost term of next row.

			First few rows of the triangle =
1;
1, 1;
2, 1, 2;
6, 2, 2, 5;
22, 6, 4, 5, 15;
92, 22, 12, 10, 15, 52;
426, 92, 44, 30, 30, 52, 203;
2146, 426, 184, 110, 90, 104, 203, 877;
11624, 2146, 852, 460, 330, 312, 406, 877, 4140;
67146, 11624, 4292, 2130, 1380, 1144, 1218, 1754, 4140, 21147;
411142, 67146, 23248, 10730, 6390, 4784, 4466, 5262, 8280, 21147, 115975;
...
Row 4 = (6, 2, 2, 5) = termwise products of (6, 2, 1, 1) and (1, 1, 2, 5).

A000110, A074664

Triangle read by rows, M*Q. M = an infinite lower triangular matrix with A074664 in every column: (1, 1, 2, 6, 22, 92, 426,...). Q = a matrix with the Bell numbers (1, 1, 2, 5, 15,...) as the main diagonal and the rest zeros.

A154107 A000110 / A014182: (A154107 convolved with A014182 = Bell numbers).

Original entry on oeis.org

1, 1, 3, 5, 15, 61, 207, 881, 4491, 21493, 117543, 710021, 4266279, 28107745, 196120515, 1397747525, 10648637151, 84304440685, 688868927151, 5913133211249, 52348170504555, 479326416322933, 4557380168574135, 44560107679838549, 449806788855058407, 4680686977970550721

Offset: 0

Gary W. Adamson, Jan 04 2009

nonn

A000110 / A014182 = (the eigensequence of Pascal's triangle) /
(eigensequence of the inverse of Pascal's triangle).
A014182 = expansion of exp(1-x-exp(-x)).

			A000110 = 52 = (1, 1, 3, 5, 15, 61) convolved with (1, 0, -1, 1, 2, -9)
= (61 - 5 + 3 + 2 - 9)

Cf. A000110, A014182.

nmax = 30; Table[a[j]/.SolveAlways[Table[Sum[a[k]*Sum[(-1)^(n-k-m)*StirlingS2[n-k+1, m+1], {m, 0, n-k}], {k, 0, n}]==BellB[n], {n, 0, nmax}], a][[1]], {j, 0, nmax}] (* Vaclav Kotesovec, Jul 26 2021 *)

A000110 / A014182 = (1, 1, 2, 5, 15, 52, 203,...) / (1, 0, -1, 1, 2, -9, 9, 50,...).

a(12) corrected and more terms added from Vaclav Kotesovec, Jul 26 2021

A165194 Triangle of 2^n terms by rows, left half of (n+1)-th row = row n; right half = "reverse and increment" row n; using terms in A000110.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 1, 1, 2, 1, 2, 5, 2, 1, 1, 2, 5, 15, 5, 2, 5, 2, 1

Offset: 1

Gary W. Adamson, Sep 06 2009

Row sums = A000110, the Bell sequence starting with offset 1; (1, 2, 5, 15,...).

Rows tend to A165195.

			First few rows of the triangle =
1;
1, 1;
1, 1, 2, 1;
1, 1, 2, 1, 2, 5, 2, 1;
1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1;
...
For example: row 4, left half = (1, 1, 2, 1); right half = (1, 2, 1, 1)
replaced with the next higher Bell numbers: (2, 5, 2, 1). Appending the two \kQ halves, we obtain row 4: (1, 1, 2, 1, 2, 5, 2, 1), sum = 15 = A000110(4).

A000110, A165195, A165196

Given the Bell sequence, A000110: (1, 1, 2, 5, 15,...); row 1 = 1, row 2 =
(1, 1);...where left half of row (n+1) = row n. Right half of row (n+1)
= reversal of row n, replacing terms with the next Bell number.

A165195 Rows of triangle A165194 tend to this sequence; generated from A000110.

Original entry on oeis.org

1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1, 2, 5, 15, 5, 15, 52, 15, 5, 2, 5, 15, 5, 2, 5, 2, 1

Offset: 1

Gary W. Adamson, Sep 06 2009

nonn

			Given terms in the Bell sequence, A000110; A165195 begins (1, 1, 2, 1,... then to obtain the first 2^3 terms, the first 2^2 terms = (1, 1, 2, 1,... then append to the latter the reversal of (1, 1, 2, 1) = (1, 2, 1, 1) but incremented with the next higher Bell number = (2, 5, 2, 1). The first 2^3 terms are thus (1, 1, 2, 1, 2, 5, 2, 1). Repeat with analogous operations to obtain 2^4 terms, and so on..

Cf. A000110, A165194, A165196.

The sequence can be generated from strings of 2^n terms starting (1, 1,... then the next string of 2^(n+1) terms is obtained by appending a "reverse and increment" substring to the previous substring.

A173108 Triangle, A000110 in every column > 0, shifted down twice.

Original entry on oeis.org

1, 1, 2, 1, 5, 1, 15, 2, 1, 52, 5, 1, 203, 15, 2, 1, 877, 52, 5, 1, 4140, 203, 15, 2, 1, 21147, 877, 52, 5, 1, 115975, 4140, 203, 15, 2, 1, 678570, 21147, 877, 52, 5, 1, 4213597, 115975, 4140, 203, 15, 2, 1, 27644437, 678570, 21147, 877, 52, 5, 1

Offset: 0

Gary W. Adamson, Feb 09 2010

Row sums = A173109: (1, 1, 3, 6, 18, 58, 221, 935, ...).

Let the triangle = M. Then lim_{n->oo} M^n = A173110: (1, 1, 3, 6, 20, 60, ...).

			First few rows of the triangle:
       1;
       1;
       2,    1;
       5,    1;
      15,    2,   1;
      52,    5,   1;
     203,   15,   2,  1;
     877,   52,   5,  1;
    4140,  203,  15,  2, 1;
   21147,  877,  52,  5, 1;
  115975, 4140, 203, 15, 2, 1;
  ...

Cf. A000110, A173109, A173110, A173111.

T[n_, k_] := BellB[n - 2 k];
Table[T[n, k], {n, 0, 10}, {k, 0, Quotient[n, 2]}] // Flatten (* Jean-François Alcover, Apr 22 2022 *)

B(n) = sum(k=0, n, stirling(n, k, 2)); \\ A000110
tabf(nn) = for (n=0, nn, for(k=0, n\2, print1(B(n-2*k), ", "));); \\ Michel Marcus, Nov 19 2022

Bell sequence in every column, for columns > 0, shifted down twice.

Keyword tabf and more terms from Michel Marcus, Nov 19 2022

A179508 a(n) is the unique integer such that Sum_{k=0..p-1} b(k)/(-n)^k == a(n) (mod p) for any prime p not dividing n, where b(0), b(1), b(2), ... are Bell numbers given by A000110.

Original entry on oeis.org

2, 1, 2, -1, 10, -43, 266, -1853, 14834, -133495, 1334962, -14684569, 176214842, -2290792931, 32071101050, -481066515733, 7697064251746, -130850092279663, 2355301661033954, -44750731559645105, 895014631192902122, -18795307255050944539

Offset: 1

Zhi-Wei Sun, Jul 17 2010

sign

On July 17, 2010 Zhi-Wei Sun conjectured that a(n) exists for every n=1,2,3,... He noted that a(1)=2 since Sum_{k=0..p-1} (-1)^k * b(k) == b(p) (mod p), and conjectured that a(2)=1, a(3)=2, a(4)=-1, a(5)=10, a(6)=-43, a(7)=266, a(8)=-1853, a(9)=14834, a(10)=-133495. It seems that (-1)^(n-1)*a(n) > 0 for all n=3,4,5,...
I guess that a(2n) == (-1)^(n-1) (mod 4) and a(2n-1) == 2 (mod 4) for all n=1,2,3,... Perhaps a(2n-1) == 2 (mod 8) for every positive integer n. - Zhi-Wei Sun, Jul 18 2010
On August 5, 2010 Zhi-Wei Sun and Don Zagier proved that a(n) actually equals (-1)^(n-1)*D(n-1)+1, where D(0), D(1), D(2), ... are derangement numbers given by A000166. - Zhi-Wei Sun, Aug 07 2010

Seiichi Manyama, Table of n, a(n) for n = 1..451
Zhi-Wei Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2010.
Zhi-Wei Sun, On Apery numbers and generalized central trinomial coefficients, preprint, arXiv:1006.2776 [math.NT], 2010-2011.
Zhi-Wei Sun, A conjecture on Bell numbers (a message to Number Theory List on July 17, 2010) [From _Zhi-Wei Sun_, Jul 18 2010]
Zhi-Wei Sun and Don Zagier, On a curious property of Bell numbers, Bulletin of the Australian Mathematical Society, Volume 84, Issue 1, August 2011. [_Zhi-Wei Sun_, Aug 07 2010]

Cf. A000110, A000166.

A179508:= n-> (-1)^n*(n!*add((-1)^(k)/k!, k=0..n))+1 : seq(A179508(n), n=0..21);
# second program:
G(x):=(2-x)*exp(-x)/(1-x): f[0]:=G(x): for n from 1 to 21 do f[n]:=diff(f[n-1], x) od: x:=0; seq((-1)^n*f[n], n=0..21); # Mélika Tebni, Jul 10 2021

a[1] = 2;
a[n_]:=a[n]=a[n-1]*(1-n)+n+1;
Array[a, 30] (* Jon Maiga, Jul 10 2021 *)

a(n) = a(n-1)*(1-n)+n+1. - Jon Maiga, Jul 10 2021

A193274 a(n) = binomial(Bell(n), 2) where B(n) = Bell numbers A000110(n).

Original entry on oeis.org

0, 0, 1, 10, 105, 1326, 20503, 384126, 8567730, 223587231, 6725042325, 230228283165, 8877197732406, 382107434701266, 18221275474580181, 956287167902779240, 54916689705422813731, 3433293323775503064306, 232614384749689991763561, 17010440815323680947084096

Offset: 0

N. J. A. Sloane, Aug 26 2011

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..300
Frank Ruskey and Jennifer Woodcock, The Rand and block distances of pairs of set partitions, in International Workshop on Combinatorial Algorithms, Victoria, 2011. LNCS.
Frank Ruskey, Jennifer Woodcock and Yuji Yamauchi, Counting and computing the Rand and block distances of pairs of set partitions, Journal of Discrete Algorithms, Volume 16, October 2012, Pages 236-248. - From _N. J. A. Sloane_, Oct 03 2012

Row sums of A193297.

[Binomial(Bell(n),2): n in [0..20]]; // Vincenzo Librandi, Feb 17 2018

a:= n-> binomial(combinat[bell](n), 2):
seq(a(n), n=0..20);  # Alois P. Heinz, Aug 28 2011

a[n_] := With[{b = BellB[n]}, b*(b-1)/2]; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Mar 18 2014 *)

from itertools import accumulate, islice
def A193274_gen(): # generator of terms
    yield 0
    blist, b = (1,), 1
    while True:
        blist = list(accumulate(blist, initial=(b:=blist[-1])))
        yield b*(b-1)//2
A193274_list = list(islice(A193274_gen(),30)) # Chai Wah Wu, Jun 22 2022

A205543 Logarithmic derivative of the Bell numbers (A000110).

Original entry on oeis.org

1, 3, 10, 39, 171, 822, 4271, 23759, 140518, 878883, 5789015, 40019058, 289513303, 2186421919, 17199606090, 140662816543, 1193865048363, 10499107480518, 95528651305671, 898071593401559, 8712429618413678, 87118795125708283, 896925422648691735

Offset: 1

Paul D. Hanna, Jan 28 2012

nonn

a(n) = number of indecomposable partitions (A074664) of [n+3] in which n+3 lies in a doubleton block (see Link). - David Callan, Oct 08 2014

			L.g.f.: L(x) = x + 3*x^2/2 + 10*x^3/3 + 39*x^4/4 + 171*x^5/5 + 822*x^6/6 +...
where exponentiation yields the o.g.f. of the Bell numbers:
exp(L(x)) = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 203*x^6 + 877*x^7 +...
which equals the series:
exp(L(x)) = 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) + x^3/((1-x)*(1-2*x)*(1-3*x)) +...

David Callan, A combinatorial interpretation for this sequence

Cf. A000110.

{a(n)=n*polcoeff(log(sum(m=0,n, x^m/prod(k=1,m, 1-k*x +x*O(x^n)))),n)}

L.g.f.: log( Sum_{n>=0} x^n / Product_{k=1..n} (1 - k*x) ).

cp's OEIS Frontend

A135084 a(n) = A000110(2^n-1).

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A146094 Bell numbers (A000110) read mod 4.

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A152431 Eigentriangle, row sums = A000110, the Bell numbers.

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A154107 A000110 / A014182: (A154107 convolved with A014182 = Bell numbers).

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A165194 Triangle of 2^n terms by rows, left half of (n+1)-th row = row n; right half = "reverse and increment" row n; using terms in A000110.

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A165195 Rows of triangle A165194 tend to this sequence; generated from A000110.

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A173108 Triangle, A000110 in every column > 0, shifted down twice.

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A179508 a(n) is the unique integer such that Sum_{k=0..p-1} b(k)/(-n)^k == a(n) (mod p) for any prime p not dividing n, where b(0), b(1), b(2), ... are Bell numbers given by A000110.

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