A000296 -id:A000296 - cp's OEIS frontend

A337043 a(0) = 1; thereafter a(n) = exp(-1/n) * Sum_{k>=0} (nk - 1)^n / (n^k k!).

1, 0, 2, 9, 112, 1875, 43416, 1310946, 49778688, 2313362673, 128894500000, 8469572721533, 647341071298560, 56871349337125648, 5684260661585401728, 640631299771142578125, 80788871646072851660800, 11323828537291632967145015, 1753760620207362607774290432

Offset: 0

Ilya Gutkovskiy, Aug 12 2020

nonn

Cf. A000296, A301419, A330605, A334162, A337038, A337039, A337040, A337041, A337042.

Table[SeriesCoefficient[1/(1 + x) Sum[(x/(1 + x))^k/Product[(1 - n j x/(1 + x)), {j, 1, k}], {k, 0, n}], {x, 0, n}], {n, 0, 18}]
Join[{1}, Table[n! SeriesCoefficient[Exp[(Exp[n x] - 1)/n - x], {x, 0, n}], {n, 1, 18}]]
Join[{1}, Table[Sum[(-1)^(n - k) Binomial[n,k] n^k BellB[k, 1/n], {k, 0, n}], {n, 1, 18}]]

a(n) = [x^n] (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - n*j*x/(1 + x)).
a(n) = n! * [x^n] exp((exp(n*x) - 1) / n - x), for n > 0.
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * n^k * BellPolynomial_k(1/n), for n > 0.

A344489 a(n) = 1 + Sum_{k=0..n-2} binomial(n-1,k) * a(k).

Original entry on oeis.org

1, 1, 2, 4, 11, 34, 122, 487, 2144, 10276, 53165, 294760, 1740950, 10899841, 72033470, 500664496, 3648211139, 27792215302, 220802394110, 1825428024367, 15672798590804, 139499676115312, 1285109772354941, 12235037442987028, 120220980122266010, 1217655627762149857

Offset: 0

Ilya Gutkovskiy, May 21 2021

nonn

Cf. A000296, A000629, A007476, A186021, A337186, A344490, A344491, A344492, A344493.

a[n_] := a[n] = 1 + Sum[Binomial[n - 1, k] a[k] , {k, 0, n - 2}]; Table[a[n], {n, 0, 25}]
nmax = 25; A[] = 0; Do[A[x] = (1 + x A[x/(1 - x)])/(1 - x^2) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = (1 + x * A(x/(1 - x))) / (1 - x^2).

A355254 Expansion of e.g.f. exp(3*(exp(x) - 1) - x).

Original entry on oeis.org

1, 2, 7, 29, 142, 785, 4813, 32240, 233449, 1812161, 14980768, 131174939, 1211111629, 11745451658, 119255234371, 1264050651953, 13952113296766, 160006824960725, 1902825936046105, 23423342243273696, 297982102750214605, 3911917977005948453, 52926119656555824520

Offset: 0

Vaclav Kotesovec, Jun 26 2022

nonn

Inverse binomial transform of A027710.
In general, if m >= 1 and e.g.f. = exp(m*exp(x) + r*x + s) then
a(n) ~ n^(n+r) * exp(n/LambertW(n/m) - n + s) / (m^r * sqrt(1 + LambertW(n/m)) * LambertW(n/m)^(n+r)).
Equivalently, a(n) ~ n! * (n/m)^r * exp(n/LambertW(n/m) + s) / (sqrt(2*Pi*n * (1 + LambertW(n/m))) * LambertW(n/m)^(n+r)).

Vaclav Kotesovec, Table of n, a(n) for n = 0..545

Cf. A000296, A027710, A078940, A217924.

nmax = 25; CoefficientList[Series[Exp[3*Exp[x]-3-x], {x, 0, nmax}], x] * Range[0, nmax]!

my(x='x+O('x^30)); Vec(serlaplace(exp(3*(exp(x) - 1) - x))) \\ Michel Marcus, Dec 04 2023

a(n) ~ 3 * n^(n-1) * exp(n/LambertW(n/3) - n - 3) / (sqrt(1 + LambertW(n/3)) * LambertW(n/3)^(n-1)).

a(0) = 1; a(n) = -a(n-1) + 3 * Sum_{k=1..n} binomial(n-1,k-1) * a(n-k). - Ilya Gutkovskiy, Dec 04 2023

A050372 Number of ways to factor n into distinct composite factors.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 0, 1, 0, 2, 1, 1, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 1, 0, 3, 1, 1, 1, 1, 0, 2, 1, 2, 1, 1, 0, 3, 0, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 4, 0, 1, 1, 1, 1, 1, 0, 3, 1, 1, 0, 3, 1, 1, 1, 2, 0, 3, 1, 1, 1, 1, 1, 4, 0, 1, 1, 2, 0, 1

Offset: 1

Christian G. Bower, Nov 15 1999

nonn

a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3,1).

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A002808, A045778, A050370-A050375. a(p^k)=A025147. a(A002110)=A000296.

with(numtheory):
b:= proc(n, k) option remember;
     `if`(isprime(n), 0, `if`(n>k, 0, 1)+
      add(`if`(d>k or isprime(d), 0, b(n/d, d-1))
          , d=divisors(n) minus {1, n}))
    end:
a:= n-> b(n$2):
seq(a(n), n=1..120);  # Alois P. Heinz, May 26 2013

b[n_, k_] := b[n, k] = If[PrimeQ[n], 0, If[n>k, 0, 1] + Sum[If[d>k || PrimeQ[d], 0, b[n/d, d-1]], {d, Divisors[n] ~Complement~ {1, n}}]];
a[n_] := b[n, n];
Array[a, 120] (* Jean-François Alcover, Mar 21 2017, after Alois P. Heinz *)

Dirichlet g.f.: Product_{n is composite}(1+1/n^s).

A086659 T(n,k) counts the set partitions of n containing k-1 blocks of length 1.

Original entry on oeis.org

1, 1, 3, 4, 4, 6, 11, 20, 10, 10, 41, 66, 60, 20, 15, 162, 287, 231, 140, 35, 21, 715, 1296, 1148, 616, 280, 56, 28, 3425, 6435, 5832, 3444, 1386, 504, 84, 36, 17722, 34250, 32175, 19440, 8610, 2772, 840, 120, 45, 98253, 194942, 188375, 117975, 53460, 18942, 5082, 1320, 165, 55

Offset: 2

Wouter Meeussen, Jul 27 2003

			The 15 set partitions of {1,2,3,4} consist of 4 partitions with 0 blocks of length 1 : {{1,2,3,4}},{{1,2},{3,4}},{{1,3},{2,4}},{{1,4},{2,3}},
4 partitions with 1 block of length 1 : {{1},{2,3,4}},{{1,2,3},{4}},{{1,2,4},{3}},{{1,3,4},{2}}
6 partitions with 2 blocks of length 1 : {{1},{2},{3,4}},{{1},{2,3},{4}},{{1},{2,4},{3}},{{1,2},{3},{4}},{{1,3},{2},{4}},{{1,4},{2},{3}}.
(There are no partitions with n-1 blocks of length 1 and 1 with n of them)
    1;
    1,   3;
    4,   4,   6;
   11,  20,  10,  10;
   41,  66,  60,  20, 15;
  162, 287, 231, 140, 35, 21;
  ...

Alois P. Heinz, Rows n = 2..142, flattened

Row sums = Bell[n]-1 (A058692), first column=A000296, main diagonal = triangular numbers A000217.

with(combinat):
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
      `if`(i<1, 0, add(multinomial(n, n-i*j, i$j)/j!*
      b(n-i*j, i-1)*`if`(i=1, x^j, 1), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n-2))(b(n$2)):
seq(T(n), n=2..16);  # Alois P. Heinz, Mar 08 2015

Table[Count[Count[ #, {_Integer}]&/@SetPartitions[n], # ]&/@Range[0, n-2], {n, 2, 10}]

E.g.f.: exp(x*y)*(exp(exp(x)-1-x)-1). - Vladeta Jovovic, Jul 28 2003

More terms from Vladeta Jovovic, Jul 28 2003

A098742 Number of indecomposable set partitions of [1..n] without singletons.

Original entry on oeis.org

0, 0, 1, 1, 3, 9, 33, 135, 609, 2985, 15747, 88761, 531561, 3366567, 22462017, 157363329, 1154257683, 8841865833, 70573741857, 585753925047, 5046128460801, 45044554041897, 416005748766771, 3969321053484921, 39077616720410409

Offset: 0

Don Knuth, Oct 01 2004

After a(3) = 1, always divisible by 3. a(n) is 3 times a prime (A001748) when n = 5, 6, 11, 14, 15, 16, 19. - Jonathan Vos Post, Jun 22 2008

			a(5)=9 because of the set partitions 135|24, 134|25, 125|34, 145|23, 15|234, 13|245, 124|35, 12345, 14|235. [Puttenham missed the last of these.]

D. E. Knuth, TAOCP, Vol. 4, Section 7.2.1.7, Problem 26.
George Puttenham, The Arte of English Poesie (1589), page 72, can be said to have stated the problem; but he omitted one case for n=5 and 22 cases for n=6, so he must have had other constraints in mind!

Alois P. Heinz, Table of n, a(n) for n = 0..500

Cf. A000296, A074664, A001748, A008585.

F:= proc(n) option remember; convert(series(1 -1/add(coeff(series(exp(exp(x)-1), x,n+1), x,j)*j!*x^j, j=0..n), x,n+1), polynom) end: a:= n-> coeff(series(x*F(n)/(1+x-F(n)), x,n+1), x,n): seq(a(n), n=0..24); # Alois P. Heinz, Sep 05 2008

f[n_] := f[n] = Normal[ Series[ 1-1/Sum[ SeriesCoefficient[ Series[ Exp[Exp[x] - 1], {x, 0, n + 1}], {x, 0, j}]*j!*x^j, {j, 0, n}], {x, 0, n + 1}]]; a[0] = 0; a[n_] := SeriesCoefficient[ Series[ x*(f[n]/(1 + x - f[n])), {x, 0, n + 1}], {x, 0, n}]; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, translated from Maple *)

def A098742_list(dim):
    T = matrix(ZZ,dim,dim)
    for n in range(dim): T[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            T[n,k] = T[n-1,k-1]+(k+1)*T[n-1,k]+(k+2)*T[n-1,k+1]
    return [0,0]+list(T.column(0))
A098742_list(23) # - Peter Luschny, Sep 20 2012

If f(z) is the generating function for A074664, then a(z)=zf(z)/(1+z-f(z)).
Also, if g(z) is the generating function for A000296, then a(z) = 1-1/g(z).
O.g.f.: x^2/(1-x-2*x^2/(1-2*x-3*x^2/(1-3*x-4*x^2/(1-... -n*x-(n+1)*x^2/(1- ...)))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
From Sergei N. Gladkovskii, Sep 20 2012, Nov 04 2012, Feb 04 2013, Feb 23 2013, Apr 18 2013, May 12 2013: (Start) Continued fractions:
G.f.: -x + 2*x/E(0) where E(k)= 1 + 1/(1 + 2*x/(1 - 2*(k+2)*x/E(k+1))).
G.f.: 1 - x*U(0,1/x) where U(k,x)= x - k - (k+1)/U(k+1,x).
G.f.: (1+x)*x/G(0) - x where G(k) = 1 + x - x*(k+1)/(1 - x/G(k+1)).
G.f.: x/Q(0) - x where Q(k)= 1 + x/(x*k-x-1)/Q(k+1).
G.f.: 1 - Q(0) where Q(k)= 1 + x - x/(1 - x*(k+1)/Q(k+1)).
G.f.: 1-x-1/Q(0) where Q(k)= 1 + x/(1 - x - x*(k+1)/(x + 1/Q(k+1))). (End)

More terms from Vladeta Jovovic, Oct 21 2004

A184175 Number of set partitions of {1,2,...,n} having no blocks of the form {i, i+1}.

Original entry on oeis.org

1, 1, 1, 3, 10, 35, 139, 611, 2925, 15128, 83903, 495929, 3108129, 20565721, 143134606, 1044489265, 7968879387, 63407648443, 525016067171, 4514661402304, 40245681692885, 371319303282381, 3540506731807277, 34840411462506887, 353394158240095874, 3690577066014598575

Offset: 0

Emeric Deutsch, Feb 09 2011

nonn

a(n) = A184174(n,0).

			a(3)=3 because we have 1-2-3, 13-2, and 123. a(4)=10 because among the 15 (=bell(4)) partitions of {1,2,3,4} only 12-34, 14-23, 12-3-4, 1-23-4, and 1-2-34, have adjacent blocks of size 2.
Contribution from _Paul D. Hanna_, Sep 03 2017: (Start)
G.f.: A(x) = 1 + x + x^2 + 3*x^3 + 10*x^4 + 35*x^5 + 139*x^6 + 611*x^7 + 2925*x^8 + 15128*x^9 + 83903*x^10 + 495929*x^11 + 3108129*x^12 +...
where
G.f.: A(x) = 1/(1+x^2) + x/((1+x^2)*(1-x+x^2)) + x^2/((1+x^2)*(1-x+x^2)*(1-2*x+x^2)) + x^3/((1+x^2)*(1-x+x^2)*(1-2*x+x^2)*(1-3*x+x^2)) +... (End)

Alois P. Heinz, Table of n, a(n) for n = 0..550

Cf. A184174, A184175, A184176, A000296.

with(combinat): seq(add((-1)^j*binomial(n-j, j)*bell(n-2*j), j = 0 .. floor((1/2)*n)), n = 0 .. 25);

Table[Sum[(-1)^j*Binomial[n-j, j]*BellB[n-2*j], {j, 0, Floor[n/2]}], {n, 0, 25}] (* Jean-François Alcover, Feb 22 2017, translated from Maple *)

{a(n) = my(A = sum(m=0,n, x^m/prod(k=0,m,1-k*x+x^2 +x*O(x^n)))); polcoeff(A,n)}
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Sep 03 2017

a(n) = Sum((-1)^j*binomial(n-j,j)*bell(n-2j), j=0..floor(n/2)).

G.f.: Sum_{n>=0} x^n / Product_{k=0..n} (1 - k*x + x^2). - Paul D. Hanna, Sep 03 2017

A250104 Triangle read by rows: T(n,k) = number of partitions of n with k circular successions (n>=0, 0 <= k <= n).

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 1, 3, 0, 1, 4, 4, 6, 0, 1, 11, 20, 10, 10, 0, 1, 41, 66, 60, 20, 15, 0, 1, 162, 287, 231, 140, 35, 21, 0, 1, 715, 1296, 1148, 616, 280, 56, 28, 0, 1, 3425, 6435, 5832, 3444, 1386, 504, 84, 36, 0, 1, 17722, 34250, 32175, 19440, 8610, 2772, 840, 120, 45, 0, 1

Offset: 0

N. J. A. Sloane, Nov 16 2014

			Triangle begins:
0
1, 0,
1, 0, 1,
1, 3, 0, 1,
4, 4, 6, 0, 1,
11, 20, 10, 10, 0, 1,
41, 66, 60, 20, 15, 0, 1,
162, 287, 231, 140, 35, 21, 0, 1,
715, 1296, 1148, 616, 280, 56, 28, 0, 1,
3425, 6435, 5832, 3444, 1386, 504, 84, 36, 0, 1,
17722, 34250, 32175, 19440, 8610, 2772, 840, 120, 45, 0, 1
...

T. Mansour, A. O. Munagi, Set partitions with circular successions, European Journal of Combinatorics, 42 (2014), 207-216.

A124323 is an essentially identical triangle, differing only in row 0 and 1.

For columns see A000296, A250105 - A250107.

t[n_, k_] := Binomial[n, k]*((-1)^(n-k)+Sum[(-1)^(j-1)*BellB[n-k-j], {j, 1, n-k}]); t[0, 0]=0; t[1, 0]=1; t[1, 1]=0; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 09 2014 *)

A367786 Expansion of e.g.f. exp(exp(4*x) - x - 1).

Original entry on oeis.org

1, 3, 25, 235, 2737, 36947, 563657, 9542715, 176920417, 3555369635, 76820077945, 1772943290763, 43469116126737, 1127040956393203, 30779951676185385, 882453651485815003, 26480355971228530369, 829522636694530362691, 27064267045022876869337, 917751849133986186857003

Offset: 0

Ilya Gutkovskiy, Nov 30 2023

nonn

Cf. A000110, A000296, A124311, A285064, A285065, A330605, A367785.

nmax = 19; CoefficientList[Series[Exp[Exp[4 x] - x - 1], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = -a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 4^k BellB[k], {k, 0, n}], {n, 0, 19}]

my(x='x+O('x^30)); Vec(serlaplace(exp(exp(4*x) - x - 1))) \\ Michel Marcus, Nov 30 2023

a(n) = exp(-1) * Sum_{k>=0} (4*k-1)^n / k!.
a(0) = 1; a(n) = -a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k).
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * 4^k * Bell(k).

A367888 Expansion of e.g.f. exp(3(exp(x) - 1) - 2x).

Original entry on oeis.org

1, 1, 4, 13, 61, 304, 1747, 10945, 74830, 550687, 4335109, 36272086, 320980645, 2991373597, 29253607780, 299258487553, 3193634980753, 35469069928792, 409082335024591, 4890313138089133, 60489400453642822, 772967507343358171, 10189818916331129017, 138398721137005215526

Offset: 0

Ilya Gutkovskiy, Dec 04 2023

nonn

Cf. A000296, A027710, A126617, A194689, A355254, A367889.

b:= proc(n, k, m) option remember; `if`(n=0, 3^m, `if`(k>0,
      b(n-1, k-1, m+1)*k, 0)+m*b(n-1, k, m)+b(n-1, k+1, m))
    end:
a:= n-> b(n, 0$2):
seq(a(n), n=0..23);  # Alois P. Heinz, Apr 29 2025

nmax = 23; CoefficientList[Series[Exp[3 (Exp[x] - 1) - 2 x], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = -2 a[n - 1] + 3 Sum[Binomial[n - 1, k - 1] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 23}]
Table[Sum[Binomial[n, k] (-2)^(n - k) BellB[k, 3], {k, 0, n}], {n, 0, 23}]

my(x='x+O('x^30)); Vec(serlaplace(exp(3*(exp(x) - 1) - 2*x))) \\ Michel Marcus, Dec 04 2023

G.f. A(x) satisfies: A(x) = 1 - x * ( 2 * A(x) - 3 * A(x/(1 - x)) / (1 - x) ).
a(n) = exp(-3) * Sum_{k>=0} 3^k * (k-2)^n / k!.
a(0) = 1; a(n) = -2 * a(n-1) + 3 * Sum_{k=1..n} binomial(n-1,k-1) * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * (-2)^(n-k) * A027710(k).

cp's OEIS Frontend

A337043 a(0) = 1; thereafter a(n) = exp(-1/n) * Sum_{k>=0} (n*k - 1)^n / (n^k * k!).

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A344489 a(n) = 1 + Sum_{k=0..n-2} binomial(n-1,k) * a(k).

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A355254 Expansion of e.g.f. exp(3*(exp(x) - 1) - x).

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PARI

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A050372 Number of ways to factor n into distinct composite factors.

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Maple

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A086659 T(n,k) counts the set partitions of n containing k-1 blocks of length 1.

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Maple

Mathematica

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Extensions

A098742 Number of indecomposable set partitions of [1..n] without singletons.

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Extensions

A184175 Number of set partitions of {1,2,...,n} having no blocks of the form {i, i+1}.

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Maple

Mathematica

PARI

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A250104 Triangle read by rows: T(n,k) = number of partitions of n with k circular successions (n>=0, 0 <= k <= n).

A337043 a(0) = 1; thereafter a(n) = exp(-1/n) * Sum_{k>=0} (nk - 1)^n / (n^k k!).

A367888 Expansion of e.g.f. exp(3(exp(x) - 1) - 2x).