A050352 -id:A050352 - cp's OEIS frontend

A032033 Stirling transform of A032031.

1, 3, 21, 219, 3045, 52923, 1103781, 26857659, 746870565, 23365498683, 812198635941, 31055758599099, 1295419975298085, 58538439796931643, 2848763394161128101, 148537065755389540539, 8261178848690959117605, 488177936257344615487803, 30544839926043868901604261

Offset: 0

Christian G. Bower

Also "AIJ" (ordered, indistinct, labeled) transform of 3,3,3,3...

Third row of array A094416 (generalized ordered Bell numbers).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
P. Blasiak, K. A. Penson and A. I. Solomon, Dobinski-type relations and the log-normal distribution, arXiv:quant-ph/0303030, J. Phys. A.: Math. Gen 36 (2003) L273.
Christian G. Bower, Transforms (2)
Jacob Sprittulla, On Colored Factorizations, arXiv:2008.09984 [math.CO], 2020.

Cf. A032031, A094416, A094418, A094419.

b:= proc(n, m) option remember;
     `if`(n=0, 3^m*m!, m*b(n-1, m)+b(n-1, m+1))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..20);  # Alois P. Heinz, Aug 04 2021

a[n_] := PolyLog[-n, 3/4]/4; a[0] = 1; Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Nov 14 2011 *)
t = 30; Range[0, t]! CoefficientList[Series[1/(4 - 3 Exp[x]), {x, 0, t}], x] (* Vincenzo Librandi, Mar 16 2014 *)

a(n)=ceil(polylog(-n,3/4)/4) \\ Charles R Greathouse IV, Jul 14 2014

my(N=25,x='x+O('x^N)); Vec(serlaplace(1/(4 - 3*exp(x)))) \\ Joerg Arndt, Jan 15 2024

E.g.f.: 1/(4-3*exp(x)).
a(n) = 3 * A050352(n), n > 0.
a(n) = Sum_{k=0..n} Stirling2(n, k) * (3^k) * k!.
a(n) = (1/4) * Sum_{k>=0} k^n * (3/4)^k. - Karol A. Penson, Jan 25 2002
a(n) = Sum_{k=0..n} A131689(n,k)*3^k. - Philippe Deléham, Nov 03 2008
G.f. A(x)=B(x)/x, where B(x)=x+3*x^2+21*x^3+... = Sum_{n>=1} b(n)*x^n satisfies 4*B(x)-x = 3*B(x/(1-x)), and b(n)=3*Sum_{k=1..n-1} binomial(n-1,k-1)*b(k), b(1)=1. - Vladimir Kruchinin, Jan 27 2011
a(n) = log(4/3)*Integral_{x = 0..inf} (floor(x))^n * (4/3)^(-x) dx. - Peter Bala, Feb 14 2015
a(0) = 1; a(n) = 3 * Sum_{k=1..n} binomial(n,k) * a(n-k). - Ilya Gutkovskiy, Jan 17 2020
a(0) = 1; a(n) = 3 * a(n-1) - 4 * Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023
a(n) = (3/4) * Sum_{k=0..n} 4^k * (-1)^(n-k) * k! * Stirling2(n,k) for n > 0. - Seiichi Manyama, Jun 01 2025

A050351 Number of 3-level labeled linear rooted trees with n leaves.

Original entry on oeis.org

1, 1, 5, 37, 365, 4501, 66605, 1149877, 22687565, 503589781, 12420052205, 336947795317, 9972186170765, 319727684645461, 11039636939221805, 408406422098722357, 16116066766061589965, 675700891505466507541

Offset: 0

Christian G. Bower, Oct 15 1999

nonn

Lists of lists of sets.

			G.f. = 1 + x + 5*x^2 + 37*x^3 + 365*x^4 + 4501*x^5 + 66605*x^6 + ...

T. S. Motzkin, Sorting numbers ...: for a link to an annotated scanned version of this paper see A000262.
T. S. Motzkin, Sorting numbers for cylinders and other classification numbers, in Combinatorics, Proc. Symp. Pure Math. 19, AMS, 1971, pp. 167-176.

G. C. Greubel, Table of n, a(n) for n = 0..390
Robert Gill, The number of elements in a generalized partition semilattice, Discrete mathematics 186.1-3 (1998): 125-134. See Example 1.
S. Giraudo, Combinatorial operads from monoids, arXiv preprint arXiv:1306.6938 [math.CO], 2013.
Marian Muresan, A concrete approach to classical analysis, CMS Books in Mathematics (2009) Table 10.2
Norihiro Nakashima, Shuhei Tsujie, Enumeration of Flats of the Extended Catalan and Shi Arrangements with Species, arXiv:1904.09748 [math.CO], 2019.
N. J. A. Sloane and Thomas Wieder, The Number of Hierarchical Orderings, Order 21 (2004), 83-89.
Index entries for sequences related to rooted trees

Cf. A000670, A050352-A050359.

Equals 1/2 * A004123(n) for n>0.

with(combstruct); SeqSeqSetL := [T, {T=Sequence(S), S=Sequence(U,card >= 1), U=Set(Z,card >=1)},labeled];

With[{nn=20},CoefficientList[Series[(2-E^x)/(3-2*E^x),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Feb 29 2012 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1/(2 - 1/(2 - Exp[x])), {x, 0, n}]]; (* Michael Somos, Nov 28 2014 *)

{a(n) = if( n<0, 0, n! * polcoeff( 1/(2 - 1/(2 - exp(x + x * O(x^n)))), n))};

{a(n)=if(n==0, 1, (1/6)*round(suminf(k=1, k^n * (2/3)^k *1.)))} \\ Paul D. Hanna, Nov 28 2014

A050351 = lambda n: sum(stirling_number2(n,k)*(2^(k-1))*factorial(k) for k in (0..n)) if n>0 else 1
[A050351(n) for n in (0..17)] # Peter Luschny, Jan 18 2016

E.g.f.: (2-exp(x))/(3-2*exp(x)).
a(n) is asymptotic to (1/6)*n!/log(3/2)^(n+1). - Benoit Cloitre, Jan 30 2003
For m-level trees (m>1), e.g.f. is (m-1-(m-2)*e^x)/(m-(m-1)*e^x) and number of trees is 1/(m*(m-1))*sum(k>=0, (1-1/m)^k*k^n). Here m=3, so a(n)=(1/6)*sum(k>=0, (2/3)^k*k^n) (for n>0). - Benoit Cloitre, Jan 30 2003
a(n) = Sum_{k=1..n} Stirling2(n, k)*k!*2^(k-1). - Vladeta Jovovic, Sep 28 2003
Recurrence: a(n+1) = 1 + 2*Sum_{j=1..n} binomial(n+1, j)*a(j). - Jon Perry, Apr 25 2005
With p(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, sum_{i=1}^{p(n)} = sum over i and prod_{j=1}^{d(i)} = product over j one has: a(n)=sum_{i=1}^{p(n)}(n!/(prod_{j=1}^{p(i)}p(i, j)!))*(p(i)!/(prod_{j=1}^{d(i)} m(i, j)!))*2^(p(i)-1). - Thomas Wieder, May 18 2005
Let f(x) = (1+x)*(1+2*x). Let D be the operator g(x) -> d/dx(f(x)*g(x)). Then for n>=1, a(n) = D^(n-1)(1) evaluated at x = 1/2. Compare with the result A000670(n) = D^(n-1)(1) at x = 0. See also A194649. - Peter Bala, Sep 05 2011
E.g.f.: 1 + x/(G(0)-3*x) where G(k)= x + k + 1 - x*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jul 11 2012
a(n) = (1/6) * Sum_{k>=1} k^n * (2/3)^k for n>0. - Paul D. Hanna, Nov 28 2014
E.g.f. A(x) satisfies 0 = 2 - A'(x) - 7*A(x) + 6*A(x)^2. - Michael Somos, Nov 28 2014

A201354 Expansion of e.g.f. exp(x) / (4 - 3*exp(x)).

Original entry on oeis.org

1, 4, 28, 292, 4060, 70564, 1471708, 35810212, 995827420, 31153998244, 1082931514588, 41407678132132, 1727226633730780, 78051253062575524, 3798351192214837468, 198049421007186054052, 11014905131587945490140, 650903915009792820650404, 40726453234725158535472348

Offset: 0

Paul D. Hanna, Nov 30 2011

			E.g.f.: E(x) = 1 + 4*x + 28*x^2/2! + 292*x^3/3! + 4060*x^4/4! + 70564*x^5/5! + ...
O.g.f.: A(x) = 1 + 4*x + 28*x^2 + 292*x^3 + 4060*x^4 + 70564*x^5 + ...
where A(x) = 1 + 4*x/(1+x) + 2!*4^2*x^2/((1+x)*(1+2*x)) + 3!*4^3*x^3/((1+x)*(1+2*x)*(1+3*x)) + 4!*4^4*x^4/((1+x)*(1+2*x)*(1+3*x)*(1+4*x)) + ...

G. C. Greubel, Table of n, a(n) for n = 0..370

Cf. A000629, A032033, A050352, A123227, A136728.

R:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( 1/(4*Exp(-x) -3) ))); // G. C. Greubel, Jun 08 2020

seq(coeff(series(1/(4*exp(-x) -3), x, n+1)*n!, x, n), n = 0..20); # G. C. Greubel, Jun 08 2020

Table[Sum[(-1)^(n-k)*4^k*StirlingS2[n,k]*k!,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 13 2013 *)

{a(n)=n!*polcoeff(exp(x+x*O(x^n))/(4 - 3*exp(x+x*O(x^n))), n)}

{a(n)=polcoeff(sum(m=0, n, 4^m*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)}

{Stirling2(n, k)=if(k<0||k>n, 0, sum(i=0, k, (-1)^i*binomial(k, i)/k!*(k-i)^n))}
{a(n)=sum(k=0, n, (-1)^(n-k)*4^k*Stirling2(n, k)*k!)}

[sum( (-1)^(n-j)*4^j*factorial(j)*stirling_number2(n,j) for j in (0..n)) for n in (0..20)] # G. C. Greubel, Jun 08 2020

O.g.f.: A(x) = Sum_{n>=0} n! * 4^n*x^n / Product_{k=0..n} (1+k*x).
O.g.f.: A(x) = 1/(1 - 4*x/(1-3*x/(1 - 8*x/(1-6*x/(1 - 12*x/(1-9*x/(1 - 16*x/(1-12*x/(1 - 20*x/(1-15*x/(1 - ...))))))))))), a continued fraction.
a(n) = Sum_{k=0..n} (-1)^(n-k) * 4^k * Stirling2(n,k) * k!.
a(n) = 4*A050352(n) for n>0.
a(n) = Sum_{k=0..n} A123125(n,k)*4^k*3^(n-k). - Philippe Deléham, Nov 30 2011
a(n) = log(4/3) * Integral_{x = 0..oo} (ceiling(x))^n * (4/3)^(-x) dx. - Peter Bala, Feb 06 2015
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 8*x*(k+1)/(8*x*(k+1) - 1 + 6*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
a(n) ~ n! / (3*(log(4/3))^(n+1)). - Vaclav Kotesovec, Jun 13 2013
a(n) = 1 + 3 * Sum_{k=0..n-1} binomial(n,k) * a(k). - Ilya Gutkovskiy, Jun 08 2020
From Seiichi Manyama, Nov 15 2023: (Start)
a(0) = 1; a(n) = -4*Sum_{k=1..n} (-1)^k * binomial(n,k) * a(n-k).
a(0) = 1; a(n) = 4*a(n-1) + 3*Sum_{k=1..n-1} binomial(n-1,k) * a(n-k). (End)
a(n) = (4/3)*A032033(n) - (1/3)*0^n. - Seiichi Manyama, Dec 21 2023

A090353 G.f. satisfies A^4 = BINOMIAL(A^3).

Original entry on oeis.org

1, 1, 4, 28, 286, 3886, 66260, 1361972, 32784353, 904412593, 28124223808, 973106096392, 37073604836768, 1541948625066176, 69513081435903392, 3376138396206853792, 175739519606046355540, 9760024269508314079444

Offset: 0

Paul D. Hanna, Nov 26 2003

In general, if A^n = BINOMIAL(A^(n-1)), then for all integer m>0 there exists an integer sequence B such that B^d = BINOMIAL(A^m) where d=gcd(m+1,n). Also, coefficients of A(k*x)^n = k-th binomial transform of coefficients in A(k*x)^(n-1) for all k>0.

			A^4 = BINOMIAL(A090355), since A090355=A^3. Also, BINOMIAL(A) = A090354^2.

Vaclav Kotesovec, Table of n, a(n) for n = 0..320

Cf. A032033, A050352, A084784, A090351, A090354, A090355, A090356, A090358.

m:=40;
f:= func< n,x | Exp((&+[(&+[3^(j-1)*Factorial(j)* StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1); // A090353
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 09 2023

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^4 - A[x/(1 - x)]^3/(1 - x) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)
With[{m=40}, CoefficientList[Series[Exp[Sum[Sum[3^(j-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Jun 09 2023 *)

{a(n) = my(A); if(n<0,0,A=1+x +x*O(x^n); for(k=1,n, B = subst(A^3,x,x/(1-x))/(1-x)+x*O(x^n); A = A - A^4 + B); polcoef(A,n,x))}
for(n=0,20,print1(a(n),", "))

m=50
def f(n, x): return exp(sum(sum(3^(j-1)*factorial(j)* stirling_number2(k,j)*x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
def A090353_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090353_list(m-9) # G. C. Greubel, Jun 09 2023

G.f. satisfies: A(x)^4 = A(x/(1-x))^3/(1-x).
a(n) ~ (n-1)! / (12 * (log(4/3))^(n+1)). - Vaclav Kotesovec, Nov 19 2014
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*3^(k-1) = A050352(n) = 1/3*A032033(n) for n >= 1. - Peter Bala, May 26 2015
G.f. satisfies [x^n] 1/A(x)^(3*n-3) = [x^n] 1/A(x)^(4*n-4) for n >= 0. - Paul D. Hanna, Apr 28 2025
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/12) * (3/4)^k). - Seiichi Manyama, May 26 2025

A050354 Number of ordered factorizations of n with one level of parentheses.

Original entry on oeis.org

1, 1, 1, 3, 1, 5, 1, 9, 3, 5, 1, 21, 1, 5, 5, 27, 1, 21, 1, 21, 5, 5, 1, 81, 3, 5, 9, 21, 1, 37, 1, 81, 5, 5, 5, 111, 1, 5, 5, 81, 1, 37, 1, 21, 21, 5, 1, 297, 3, 21, 5, 21, 1, 81, 5, 81, 5, 5, 1, 201, 1, 5, 21, 243, 5, 37, 1, 21, 5, 37, 1, 513, 1, 5, 21, 21, 5, 37, 1, 297, 27, 5, 1, 201

Offset: 1

Christian G. Bower, Oct 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).

Dirichlet inverse of (A074206*A153881). - Mats Granvik, Jan 12 2009

			For n=6, we have (6) = (3*2) = (2*3) = (3)*(2) = (2)*(3), thus a(6) = 5.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A002033, A050351, A050352, A050353, A050355, A050356, A050357, A050358, A050359.

A[n_]:=If[n==1, n/2, 2*Sum[If[dIndranil Ghosh, May 19 2017 *)

A050354aux(n) = if(1==n,n/2, 2*sumdiv(n,d, if(dA050354aux(d), 0)));
A050354(n) = if(1==n,n,A050354aux(n)); \\ Antti Karttunen, May 19 2017, after Jovovic's general recurrence.

def A(n): return 1/2 if n==1 else 2*sum(A(d) for d in divisors(n) if dIndranil Ghosh, May 19 2017, after Antti Karttunen's PARI program

Dirichlet g.f.: (2-zeta(s))/(3-2*zeta(s)).
Recurrence for number of ordered factorizations of n with k-1 levels of parentheses is a(n) = k*Sum_{d|n, d1, a(1)= 1/k. - Vladeta Jovovic, May 25 2005
a(p^k) = 3^(k-1).
a(A002110(n)) = A050351(n).
Sum_{k=1..n} a(k) ~ -n^r / (4*r*Zeta'(r)), where r = 2.185285451787482231198145140899733642292971552057774261555354324536... is the root of the equation Zeta(r) = 3/2. - Vaclav Kotesovec, Feb 02 2019

Duplicate comment removed by R. J. Mathar, Jul 15 2010

A090355 G.f. satisfies A^4 = BINOMIAL(A)^3.

Original entry on oeis.org

1, 3, 15, 109, 1086, 14178, 232906, 4647006, 109376595, 2967406345, 91130074437, 3123199831983, 118106517900868, 4883161763750820, 219076867059030300, 10597531747143624820, 549768536732090716371, 30443800514118532762329

Offset: 0

Paul D. Hanna, Nov 26 2003

See comments in A090353.

Cf. A090353, A090354; A019538, A032033, A050352, A201354.

Cf. A090352, A090357, A090362.

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^4 - A[x/(1 - x)]^3/(1 - x)^3 + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)

{a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^4+B^3);polcoeff(A,n,x))}

G.f.: A(x)^4 = A(x/(1-x))^3/(1-x)^3.
From Peter Bala, May 26 2015: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*3^k = A032033(n) = 3*A050352(n).
BINOMIAL(A(x)) = exp( Sum_{n >= 1} c(n)*x^n/n ) where c(n) = (-1)^n*Sum_{k = 1..n} k!*Stirling2(n,k)*4^k = A201354(n) = 4*A050352(n) for n >= 1. A(x) = B(x)^3 and BINOMIAL(A(x)) = B(x)^4 where B(x) = 1 + x + 4*x^2 + 28*x^3 + 286*x^4 + ... is the o.g.f. for A090353. See also A019538. (End)
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/4) * (3/4)^k). - Seiichi Manyama, May 26 2025
a(n) ~ (n-1)! / (4 * log(4/3)^(n+1)). - Vaclav Kotesovec, May 28 2025

A257565 Generalized Fubini numbers. Square array read by ascending antidiagonals, A(n,k) = 1 + k(Sum_{j=1..n-1} C(n,j)A(j,k)); n>=0 and k>=0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 13, 5, 1, 1, 1, 75, 37, 7, 1, 1, 1, 541, 365, 73, 9, 1, 1, 1, 4683, 4501, 1015, 121, 11, 1, 1, 1, 47293, 66605, 17641, 2169, 181, 13, 1, 1, 1, 545835, 1149877, 367927, 48601, 3971, 253, 15, 1, 1, 1, 7087261, 22687565, 8952553, 1306809, 108901, 6565, 337, 17, 1, 1

Offset: 0

Peter Luschny, May 08 2015

M. Mureşan defined the generalized Fubini numbers as the enumerators of the k-labeled ordered p partitions of an n-set.

			      1,       1,       1,       1,        1,         1, ...  A000012
      1,       1,       1,       1,        1,         1, ...  A000012
      1,       3,       5,       7,        9,        11, ...  A005408
      1,      13,      37,      73,      121,       181, ...  A003154
      1,      75,     365,    1015,     2169,      3971, ...  A193252
      1,     541,    4501,   17641,    48601,    108901, ...
      1,    4683,   66605,  367927,  1306809,   3583811, ...
      1,   47293, 1149877, 8952553, 40994521, 137595781, ...
A000012, A000670, A050351, A050352,  A050353,

M. Mureşan, On the generalized Fubini numbers. (Romanian) Stud. Cercet. Mat. 37, 70-76 (1985).

Robert Gill, The number of elements in a generalized partition semilattice, Discrete mathematics 186.1-3 (1998): 125-134. See Example 1.
N. Kilar and Y. Simsek, A new family of Fubini type numbers and polynomials associated with Apostol-Bernoulli numbers and polynomials, J. Korean Math. Soc. 54 (5) (2017) 1605-1621.

Rows: A005408, A003154, A193252.

Columns: A000670, A050351, A050352, A050353.

F := proc(n,k) option remember; 1+k*add(binomial(n,j)*F(j,k),j=1..n-1) end:
seq(print(seq(F(n-k,k),k=0..n)), n=0..7); # triangular form
egf := k -> 1+1/(1/(exp(z)-1)-k): # egf of column k
for k from 0 to 4 do seq(j!*coeff(series(egf(k),z,10),z,j),j=0..8) od;
A := (n,k) -> `if`(n=0,1,add(k^(n-j-1)*(k+1)^j*combinat:-eulerian1(n,j),j=0..n-1)): seq(print(seq(A(n,k),k=0..5)),n=0..7);

A[n_, k_] := A[n, k] = 1 + k Sum[Binomial[n, j] A[j, k], {j, 1, n - 1}]; Table[A[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 30 2016 *)

E.g.f. of column k: 1+1/(1/(exp(z)-1)-k).
A(n,k) = Sum_{j=0..n-1} k^j*j!*{n,j+1} for n>0, else 1; {n,j} denotes the Stirling subset numbers.
A(n,k) = Sum_{j=0..n-1} k^(n-j-1)*(k+1)^j* for n>0, else 1;  denotes the Eulerian numbers.

A250914 E.g.f.: (18 - 17cosh(x)) / (25 - 24cosh(x)).

Original entry on oeis.org

1, 7, 1015, 367927, 248956855, 270732878647, 431806658432695, 949587798053709367, 2753726282896986372535, 10181613308681289633868087, 46749244630988859672950920375, 260970234691672017384493753162807, 1740621952318191255997909826897420215, 13670746044282245244660044262911331401527

Offset: 0

Paul D. Hanna, Nov 28 2014

nonn

The number of 4-level labeled linear rooted trees with 2*n leaves.
A bisection of A050352.
a(n) == 7 (mod 1008) for n>0.

			E.g.f.: E(x) = 1 + 7*x^2/2! + 1015*x^4/4! + 367927*x^6/6! + 248956855*x^8/8! +...
where E(x) = (18 - 17*cosh(x)) / (25 - 24*cosh(x)).
ALTERNATE GENERATING FUNCTION.
E.g.f.: A(x) = 1 + 7*x + 1015*x^2/2! + 367927*x^3/3! + 248956855*x^4/4! +...
where
12*A(x) = 9 + exp(x)*(3/4) + exp(4*x)*(3/4)^2 + exp(9*x)*(3/4)^3 + exp(16*x)*(3/4)^4 + exp(25*x)*(3/4)^5 + exp(36*x)*(3/4)^6 +...

Seiichi Manyama, Table of n, a(n) for n = 0..186

Cf. A249938, A249939, A247082, A250915, A050352.

nmax=20; Table[(CoefficientList[Series[(18-17*Cosh[x]) / (25-24*Cosh[x]), {x, 0, 2*nmax}], x] * Range[0, 2*nmax]!)[[n]],{n,1,2*nmax+2,2}] (* Vaclav Kotesovec, Nov 29 2014 *)

/* E.g.f.: (18 - 17*cosh(x)) / (25 - 24*cosh(x)): */
{a(n) = local(X=x+O(x^(2*n+1))); (2*n)!*polcoeff( (18 - 17*cosh(X)) / (25 - 24*cosh(X)) , 2*n)}
for(n=0, 20, print1(a(n), ", "))

/* Formula for a(n): */
{Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
{a(n) = if(n==0, 1, sum(k=0, 2*n, 3^(k-1) * k! * Stirling2(2*n, k) ))}
for(n=0, 20, print1(a(n), ", "))

/* As the Sum of an Infinite Series: */
\p60 \\ set precision
Vec(serlaplace(2/3 + 1/12*sum(n=0,2000,exp(n^2*x)*(3/4)^n*1.)))

E.g.f.: 2/3 + (1/12)*Sum_{n>=0} exp(n^2*x) * (3/4)^n  =  Sum_{n>=0} a(n)*x^n/n!.
a(n) = Sum_{k=0..2*n} 3^(k-1) * k! * Stirling2(2*n, k) for n>0 with a(0)=1.
a(n) ~ (2*n)! / (12 * (log(4/3))^(2*n+1)). - Vaclav Kotesovec, Nov 29 2014

A321189 a(n) = n! * [x^n] 1 - 1/(n - 1/(exp(x) - 1)).

Original entry on oeis.org

1, 1, 5, 73, 2169, 108901, 8288293, 890380177, 128364028145, 23918924529901, 5595490598128221, 1605718043992482553, 554663179293965398825, 227038711419826844827381, 108674023653792712066606229, 60142879347501714200454327841, 38108071228342727619600464659425

Offset: 0

Ilya Gutkovskiy, Oct 29 2018

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..200
Robert Gill, The number of elements in a generalized partition semilattice, Discrete mathematics 186.1-3 (1998): 125-134. See Example 1.

Main diagonal of A257565.

Cf. A000670, A050351, A050352, A050353, A094420.

Concatenation([1],List([1..16],n->Sum([1..n],k->Stirling2(n,k)*Factorial(k)*n^(k-1)))); # Muniru A Asiru, Oct 29 2018

seq(coeff(series(factorial(n)*(1-1/(n-1/(exp(x)-1))),x,n+1), x, n), n = 0 .. 15); # Muniru A Asiru, Oct 29 2018
# Or, using the recurrence of the Fubini polynomials:
F := proc(n) option remember; if n = 0 then return 1 fi;
expand(add(binomial(n, k)*F(n-k)*x, k = 1..n)) end:
a := n -> `if`(n=0, 1, subs(x = n, F(n)) / n):
seq(a(n), n = 0..16);  # Peter Luschny, May 21 2021

Table[n! SeriesCoefficient[1 - 1/(n - 1/(Exp[x] - 1)), {x, 0, n}], {n, 0, 16}]
Join[{1}, Table[Sum[StirlingS2[n, k] k! n^(k - 1), {k, n}], {n, 16}]]

{a(n) = if(n==0, 1, sum(k=0, n, k!*n^(k-1)*stirling(n, k, 2)))} \\ Seiichi Manyama, Jun 12 2020

a(0) = 1; a(n) = Sum_{k=1..n} Stirling2(n, k)*k!*n^(k-1).
a(n) = A257565(n, n).
From Vaclav Kotesovec, Oct 29 2018: (Start)
a(n) ~ exp(1/2) * n! * n^(n-1).
a(n) ~ sqrt(2*Pi) * n^(2*n - 1/2) / exp(n - 1/2). (End)
a(n) = F_{n}(n) / n for n >= 1, where F_{n}(x) is the Fubini polynomial. In other words: a(n) = A094420(n) / n for n >= 1. - Peter Luschny, May 21 2021

A050356 Number of ordered factorizations of n with 2 levels of parentheses.

Original entry on oeis.org

1, 1, 1, 4, 1, 7, 1, 16, 4, 7, 1, 40, 1, 7, 7, 64, 1, 40, 1, 40, 7, 7, 1, 208, 4, 7, 16, 40, 1, 73, 1, 256, 7, 7, 7, 292, 1, 7, 7, 208, 1, 73, 1, 40, 40, 7, 1, 1024, 4, 40, 7, 40, 1, 208, 7, 208, 7, 7, 1, 544, 1, 7, 40, 1024, 7, 73, 1, 40, 7, 73, 1, 1840, 1, 7, 40, 40, 7, 73, 1

Offset: 1

Christian G. Bower, Oct 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).

			For n=6 we have ((6)) = ((3*2)) = ((2*3)) = ((3)*(2)) = ((2)*(3)) = ((3))*((2)) = ((2))*((3)), thus a(6) = 7.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A002033, A050351, A050352, A050353, A050354, A050355, A050357, A050358, A050359.

A050356aux(n) = if(1==n,1/3, 3*sumdiv(n,d, if(dA050356aux(d), 0)));
A050356(n) = if(1==n,n,A050356aux(n)); \\ Antti Karttunen, May 19 2017, after the general recurrence given by Vladeta Jovovic May 25 2005 in A050354.

Dirichlet g.f.: (3-2*zeta(s))/(4-3*zeta(s)).
a(p^k) = 4^(k-1).
a(A002110(n)) = A050352(n).
Sum_{k=1..n} a(k) ~ -n^r / (9*r*Zeta'(r)), where r = 2.52138975790328306967497455387140053675965539610041801606891036... is the root of the equation Zeta(r) = 4/3. - Vaclav Kotesovec, Feb 02 2019

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A090353 G.f. satisfies A^4 = BINOMIAL(A^3).

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A050354 Number of ordered factorizations of n with one level of parentheses.

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A090355 G.f. satisfies A^4 = BINOMIAL(A)^3.

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A257565 Generalized Fubini numbers. Square array read by ascending antidiagonals, A(n,k) = 1 + k(Sum_{j=1..n-1} C(n,j)A(j,k)); n>=0 and k>=0.

A250914 E.g.f.: (18 - 17cosh(x)) / (25 - 24cosh(x)).