

A006472


a(n) = n!*(n1)!/2^(n1).
(Formerly M3052)


29



1, 1, 3, 18, 180, 2700, 56700, 1587600, 57153600, 2571912000, 141455160000, 9336040560000, 728211163680000, 66267215894880000, 6958057668962400000, 834966920275488000000, 113555501157466368000000, 17373991677092354304000000, 2970952576782792585984000000
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OFFSET

1,3


COMMENTS

Product of first (n1) positive triangular numbers. Might be called a triangular factorial number.  Amarnath Murthy, May 19 2002, corrected by Alex Ratushnyak, Dec 03 2013
Number of ways of transforming n distinguishable objects into n singletons via a sequence of n1 refinements. Example: a(3)=3 because we have XYZ>XYZ>XYZ, XYZ>YXZ>XYZ and XYZ>ZXY>XYZ.  Emeric Deutsch, Jan 23 2005
From David Callan, Aug 27 2009: (Start)
With offset 0, a(n) = number of unordered increasing full binary trees of 2n edges with leaf set {n,n+1,...,2n}, where full binary means each nonleaf vertex has two children, increasing means the vertices are labeled 0,1,2,...,2n and each child is greater than its parent, unordered might as well mean ordered and each pair of sibling vertices is increasing left to right. For example, a(2)=3 counts the trees with edge lists {01,02,13,14}, {01,03,12,14}, {01,04,12,13}.
PROOF. Given such a tree of size n, to produce a tree of size n+1, two new leaves must be added to the leaf n. Choose any two of the leaf set {n+1,...,2n,2n+1,2n+2} for the new leaves and use the rest to replace the old leaves n+1,...,2n, maintaining relative order. Thus each tree of size n yields (n+2)choose2 trees of the next size up. Since the ratio a(n+1)/a(n)=(n+2)choose2, the result follows by induction.
Without the condition on the leaves, these trees are counted by the reduced tangent numbers A002105. (End)
a(n) = Sum(M(t)N(t)), where summation is over all rooted trees t with n vertices, M(t) is the number of ways to take apart t by sequentially removing terminal edges (see A206494) and N(t) is the number of ways to build up t from the onevertex tree by adding successively edges to the existing vertices (the ConnesMoscovici weight; see A206496). See Remark on p. 3801 of the Hoffman reference. Example: a(3) = 3; indeed, there are two rooted trees with 3 vertices: t' = the path rab and t" = V; we have M(t')=N(t')=1 and M(t") =1, N(t")=2, leading to M(t')N(t') + M(t")N(t")=3.  Emeric Deutsch, Jul 20 2012


REFERENCES

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 148.
L. Lovasz, Combin. Problems and Exercises, NorthHolland, 1979, p. 165.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n = 1..50
Karl Dienger, Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung, JahresBericht LudwigWilhelmGymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]
Daniel Dockery, Polygorials, Special "Factorials" of Polygonal Numbers.
Filippo Disanto and Thomas Wiehe, Some combinatorial problems on binary rooted trees occurring in population genetics, arXiv preprint arXiv:1112.1295 [math.CO], 20112012.
P. Erdős, R. K. Guy and J. W. Moon, On refining partitions, J. London Math. Soc., 9 (1975), 565570.
L. Ferretti, F. Disanto and T, Wiehe, The Effect of Single Recombination Events on Coalescent Tree Height and Shape, PLoS ONE 8(4): e60123.
O. Frank and K. Svensson, On probability distributions of singlelinkage dendrograms, Journal of Statistical Computation and Simulation, 12 (1981), 121131. (Annotated scanned copy)
M. E. Hoffman, Combinatorics of rooted trees and Hopf algebras, Trans. Amer. Math. Soc., 355, 2003, 37953811.
C. L. Mallows, Note to N. J. A. Sloane circa 1979.
F. Murtagh, Counting dendrograms: a survey, Discrete Applied Mathematics, 7 (1984), 191199.
Index entries for sequences related to factorial numbers


FORMULA

a(n) = a(n1)*A000217(n1).
a(n) = A010790(n1)/2^(n1).
a(n) = Polygorial(n, 3) = A000142(n)/A000079(n)*A000142(n+1) = n!/2^n*product(i+2, i=0..n1) = n!/2^n*pochhammer(2, n) = n!^2/2^n*(n+1) = Polygorial(n, 4)/2^n*(n+1).  Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
a(n1) = (1)^(n+1)/n^2/det(M_n) where M_n is the matrix M_(i, j)=abs(1/i1/j).  Benoit Cloitre, Aug 21 2003
From Ilya Gutkovskiy, Dec 15 2016: (Start)
a(n) ~ 4*Pi*n^(2*n)/(2^n*exp(2*n)).
Sum_{n>=1} 1/a(n) = BesselI(1,2*sqrt(2))/sqrt(2) = 2.3948330992734... (End)


MAPLE

A006472 := n>n!*(n1)!/2^(n1);


MATHEMATICA

FoldList[Times, 1, Accumulate[Range[20]]] (* Harvey P. Dale, Jan 10 2013 *)


PROG

(PARI) a(n) = n*(n1)!^2/2^(n1) \\ Charles R Greathouse IV, May 18 2015


CROSSREFS

Cf. A084939, A084940, A084941, A084942, A084943, A084944.
Sequence in context: A111465 A247029 A108994 * A132853 A259666 A084879
Adjacent sequences: A006469 A006470 A006471 * A006473 A006474 A006475


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane


STATUS

approved



