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A050351 Number of 3-level labeled linear rooted trees with n leaves. 17
1, 1, 5, 37, 365, 4501, 66605, 1149877, 22687565, 503589781, 12420052205, 336947795317, 9972186170765, 319727684645461, 11039636939221805, 408406422098722357, 16116066766061589965, 675700891505466507541 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Lists of lists of sets.

REFERENCES

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.

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..390

S. Giraudo, Combinatorial operads from monoids, arXiv preprint arXiv:1306.6938 [math.CO], 2013.

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

FORMULA

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) satisifes 0 = 2 - A'(x) - 7*A(x) + 6*A(x)^2. - Michael Somos, Nov 28 2014

EXAMPLE

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

MAPLE

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

MATHEMATICA

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 *)

PROG

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

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

(Sage)

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

CROSSREFS

Cf. A000670, A050352-A050359.

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

Sequence in context: A234953 A025168 A084358 * A129137 A276232 A055869

Adjacent sequences:  A050348 A050349 A050350 * A050352 A050353 A050354

KEYWORD

nonn

AUTHOR

Christian G. Bower, Oct 15 1999

STATUS

approved

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Last modified February 17 20:05 EST 2018. Contains 299296 sequences. (Running on oeis4.)