A050359 -id:A050359 - cp's OEIS frontend

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

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

A050352 Number of 4-level labeled linear rooted trees with n leaves.

Original entry on oeis.org

1, 1, 7, 73, 1015, 17641, 367927, 8952553, 248956855, 7788499561, 270732878647, 10351919533033, 431806658432695, 19512813265643881, 949587798053709367, 49512355251796513513, 2753726282896986372535, 162725978752448205162601

Offset: 0

Christian G. Bower, Oct 15 1999

nonn

G. C. Greubel, Table of n, a(n) for n = 0..370
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.
Index entries for sequences related to rooted trees

Cf. A000670, A050351 - A050359.

Equals 1/3 * A032033(n) for n>0.

[1] cat [(&+[3^(j-1)*Factorial(j)*StirlingSecond(n,j): j in [1..n]]) : n in [0..20]]; // G. C. Greubel, Jun 08 2020

seq(coeff(series( (3-2*exp(x))/(4-3*exp(x)), x, n+1)*n!, x, n), n = 0..20); # G. C. Greubel, Jun 08 2020

With[{nn=20}, CoefficientList[Series[(3-2Exp[x])/(4-3Exp[x]),{x,0,nn}], x]*Range[0,nn]!] (* Harvey P. Dale, Aug 16 2012 *)

a(n)=n!*if(n<0,0,polcoeff((3-2*exp(x))/(4-3*exp(x))+O(x^(n+1)),n))

{a(n)=if(n==0, 1, (1/12)*round(suminf(k=1, k^n * (3/4)^k *1.)))} \\ Paul D. Hanna, Nov 28 2014

[1]+[sum( 3^(j-1)*factorial(j)*stirling_number2(n,j) for j in (1..n)) for n in (1..20)] # G. C. Greubel, Jun 08 2020

E.g.f.: (3 - 2*exp(x))/(4 - 3*exp(x)).
a(n) is asymptotic to (1/12)*n!/log(4/3)^(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=4, so a(n)=(1/12)*sum(k>=0, (3/4)^k*k^n) (for n>0). - Benoit Cloitre, Jan 30 2003
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. 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)-4*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) = (1/12) * Sum_{k>=1} k^n * (3/4)^k for n>0. - Paul D. Hanna, Nov 28 2014
a(n) = Sum_{k=1..n} Stirling2(n, k) * k! * 3^(k-1). - Paul D. Hanna, Nov 28 2014, after Vladeta Jovovic in A050351
a(n) = 1 + 3 * Sum_{k=1..n-1} binomial(n,k) * a(k). - Ilya Gutkovskiy, Jun 08 2020

A050353 Number of 5-level labeled linear rooted trees with n leaves.

Original entry on oeis.org

1, 1, 9, 121, 2169, 48601, 1306809, 40994521, 1469709369, 59277466201, 2656472295609, 130952452264921, 7042235448544569, 410269802967187801, 25740278881968596409, 1730295054262416751321, 124066865052334175027769

Offset: 0

Christian G. Bower, Oct 15 1999

nonn

Michael De Vlieger, Table of n, a(n) for n = 0..360
Norihiro Nakashima, Shuhei Tsujie, Enumeration of Flats of the Extended Catalan and Shi Arrangements with Species, arXiv:1904.09748 [math.CO], 2019.
Index entries for sequences related to rooted trees

Cf. A000670, A050351 - A050359.

Equals 1/4 * A094417(n) for n>0.

R:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( (4-3*Exp(x))/(5-4*Exp(x)) ))); // G. C. Greubel, Jun 08 2020

seq(coeff(series( (4-3*exp(x))/(5-4*exp(x)), x, n+1)*n!, x, n), n = 0..20); # G. C. Greubel, Jun 08 2020

max = 16; f[x_] := (4-3*E^x) / (5-4*E^x); CoefficientList[ Series[ f[x], {x, 0, max}], x]*Range[0, max]! (* Jean-François Alcover, Nov 14 2011, after g.f. *)

a(n)=n!*if(n<0,0,polcoeff((4-3*exp(x))/(5-4*exp(x))+O(x^(n+1)),n))

{a(n)=if(n==0,1,(1/20)*round(suminf(k=1, k^n * (4/5)^k *1.)))} \\ Paul D. Hanna, Nov 28 2014

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

E.g.f.: (4 - 3*exp(x))/(5 - 4*exp(x)).
a(n) is asymptotic to (1/20)*n!/log(5/4)^(n+1). More generally if m>1, the number of m-level labeled linear rooted trees with n leaves is asymptotic to n!/log(m/(m-1))^(n+1)/(m^2-m). - 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=5, so a(n)=(1/20)*sum(k>=0, (4/5)^k*k^n) (for n>0). - Benoit Cloitre, Jan 30 2003
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 = 3/2. 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)-5*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) = (1/20) * Sum_{k>=1} k^n * (4/5)^k for n>0. - Paul D. Hanna, Nov 28 2014
a(n) = Sum_{k=1..n} Stirling2(n, k) * k! * 4^(k-1). - Paul D. Hanna, Nov 28 2014, after Vladeta Jovovic in A050351
a(n) = 1 + 4 * Sum_{k=1..n-1} binomial(n,k) * a(k). - Ilya Gutkovskiy, Jun 08 2020

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

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

A050358 Number of ordered factorizations of n with 3 levels of parentheses.

Original entry on oeis.org

1, 1, 1, 5, 1, 9, 1, 25, 5, 9, 1, 65, 1, 9, 9, 125, 1, 65, 1, 65, 9, 9, 1, 425, 5, 9, 25, 65, 1, 121, 1, 625, 9, 9, 9, 605, 1, 9, 9, 425, 1, 121, 1, 65, 65, 9, 1, 2625, 5, 65, 9, 65, 1, 425, 9, 425, 9, 9, 1, 1145, 1, 9, 65, 3125, 9, 121, 1, 65, 9, 121, 1, 4825, 1, 9, 65, 65, 9, 121

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).

The Dirichlet inverse is given by A050356, turning all but the first element of A050356 negative. - R. J. Mathar, Jul 15 2010

			6 = (((6))) = (((3*2))) = (((2*3))) = (((3)*(2))) = (((2)*(3))) = (((3))*((2))) = (((2))*((3))) = (((3)))*(((2))) = (((2)))*(((3))).

R. J. Mathar, Table of n, a(n) for n = 1..899

Cf. A002033, A050351-A050359. a(p^k)=5^(k-1). a(A002110)=A050353.

Dirichlet g.f.: (4-3*zeta(s))/(5-4*zeta(s)).
a(n) = A050359(A101296(n)). - R. J. Mathar, May 26 2017
Sum_{k=1..n} a(k) ~ -n^r / (16*r*Zeta'(r)), where r = 2.7884327053324956670606046076818023223650950899573090550836329583345... is the root of the equation Zeta(r) = 5/4. - Vaclav Kotesovec, Feb 02 2019

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

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A050352 Number of 4-level labeled linear rooted trees with n leaves.

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A050353 Number of 5-level labeled linear rooted trees with n leaves.

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

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

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A050358 Number of ordered factorizations of n with 3 levels of parentheses.

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