cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A001147 Double factorial of odd numbers: a(n) = (2*n-1)!! = 1*3*5*...*(2*n-1).

Original entry on oeis.org

1, 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075, 13749310575, 316234143225, 7905853580625, 213458046676875, 6190283353629375, 191898783962510625, 6332659870762850625, 221643095476699771875, 8200794532637891559375, 319830986772877770815625
Offset: 0

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Comments

The solution to Schröder's third problem.
Number of fixed-point-free involutions in symmetric group S_{2n} (cf. A000085).
a(n-2) is the number of full Steiner topologies on n points with n-2 Steiner points. [corrected by Lyle Ramshaw, Jul 20 2022]
a(n) is also the number of perfect matchings in the complete graph K(2n). - Ola Veshta (olaveshta(AT)my-deja.com), Mar 25 2001
Number of ways to choose n disjoint pairs of items from 2*n items. - Ron Zeno (rzeno(AT)hotmail.com), Feb 06 2002
Number of ways to choose n-1 disjoint pairs of items from 2*n-1 items (one item remains unpaired). - Bartosz Zoltak, Oct 16 2012
For n >= 1 a(n) is the number of permutations in the symmetric group S_(2n) whose cycle decomposition is a product of n disjoint transpositions. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001
a(n) is the number of distinct products of n+1 variables with commutative, nonassociative multiplication. - Andrew Walters (awalters3(AT)yahoo.com), Jan 17 2004. For example, a(3)=15 because the product of the four variables w, x, y and z can be constructed in exactly 15 ways, assuming commutativity but not associativity: 1. w(x(yz)) 2. w(y(xz)) 3. w(z(xy)) 4. x(w(yz)) 5. x(y(wz)) 6. x(z(wy)) 7. y(w(xz)) 8. y(x(wz)) 9. y(z(wx)) 10. z(w(xy)) 11. z(x(wy)) 12. z(y(wx)) 13. (wx)(yz) 14. (wy)(xz) 15. (wz)(xy).
a(n) = E(X^(2n)), where X is a standard normal random variable (i.e., X is normal with mean = 0, variance = 1). So for instance a(3) = E(X^6) = 15, etc. See Abramowitz and Stegun or Hoel, Port and Stone. - Jerome Coleman, Apr 06 2004
Second Eulerian transform of 1,1,1,1,1,1,... The second Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} E(n,k)s(k), where E(n,k) is a second-order Eulerian number (A008517). - Ross La Haye, Feb 13 2005
Integral representation as n-th moment of a positive function on the positive axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/2)/sqrt(2*Pi*x) dx, n >= 0. - Karol A. Penson, Oct 10 2005
a(n) is the number of binary total partitions of n+1 (each non-singleton block must be partitioned into exactly two blocks) or, equivalently, the number of unordered full binary trees with n+1 labeled leaves (Stanley, ex 5.2.6). - Mitch Harris, Aug 01 2006
a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is i for iDavid Callan, Sep 25 2006
a(n) is the number of increasing ordered rooted trees on n+1 vertices where "increasing" means the vertices are labeled 0,1,2,...,n so that each path from the root has increasing labels. Increasing unordered rooted trees are counted by the factorial numbers A000142. - David Callan, Oct 26 2006
Number of perfect multi Skolem-type sequences of order n. - Emeric Deutsch, Nov 24 2006
a(n) = total weight of all Dyck n-paths (A000108) when each path is weighted with the product of the heights of the terminal points of its upsteps. For example with n=3, the 5 Dyck 3-paths UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD have weights 1*2*3=6, 1*2*2=4, 1*2*1=2, 1*1*2=2, 1*1*1=1 respectively and 6+4+2+2+1=15. Counting weights by height of last upstep yields A102625. - David Callan, Dec 29 2006
a(n) is the number of increasing ternary trees on n vertices. Increasing binary trees are counted by ordinary factorials (A000142) and increasing quaternary trees by triple factorials (A007559). - David Callan, Mar 30 2007
From Tom Copeland, Nov 13 2007, clarified in first and extended in second paragraph, Jun 12 2021: (Start)
a(n) has the e.g.f. (1-2x)^(-1/2) = 1 + x + 3*x^2/2! + ..., whose reciprocal is (1-2x)^(1/2) = 1 - x - x^2/2! - 3*x^3/3! - ... = b(0) - b(1)*x - b(2)*x^2/2! - ... with b(0) = 1 and b(n+1) = -a(n) otherwise. By the formalism of A133314, Sum_{k=0..n} binomial(n,k)*b(k)*a(n-k) = 0^n where 0^0 := 1. In this sense, the sequence a(n) is essentially self-inverse. See A132382 for an extension of this result. See A094638 for interpretations.
This sequence aerated has the e.g.f. e^(t^2/2) = 1 + t^2/2! + 3*t^4/4! + ... = c(0) + c(1)*t + c(2)*t^2/2! + ... and the reciprocal e^(-t^2/2); therefore, Sum_{k=0..n} cos(Pi k/2)*binomial(n,k)*c(k)*c(n-k) = 0^n; i.e., the aerated sequence is essentially self-inverse. Consequently, Sum_{k=0..n} (-1)^k*binomial(2n,2k)*a(k)*a(n-k) = 0^n. (End)
From Ross Drewe, Mar 16 2008: (Start)
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant but the pairs are not distinguishable, i.e., arrangements which are the same after permutations of the labels are equivalent.
If this sequence and A000680 are denoted by a(n) and b(n) respectively, then a(n) = b(n)/n! where n! = the number of ways of permuting the pair labels.
For example, there are 90 ways of arranging the elements of 3 pairs [1 1], [2 2], [3 3] when the pairs are distinguishable: A = { [112233], [112323], ..., [332211] }.
By applying the 6 relabeling permutations to A, we can partition A into 90/6 = 15 subsets: B = { {[112233], [113322], [221133], [223311], [331122], [332211]}, {[112323], [113232], [221313], [223131], [331212], [332121]}, ....}
Each subset or equivalence class in B represents a unique pattern of pair relationships. For example, subset B1 above represents {3 disjoint pairs} and subset B2 represents {1 disjoint pair + 2 interleaved pairs}, with the order being significant (contrast A132101). (End)
A139541(n) = a(n) * a(2*n). - Reinhard Zumkeller, Apr 25 2008
a(n+1) = Sum_{j=0..n} A074060(n,j) * 2^j. - Tom Copeland, Sep 01 2008
From Emeric Deutsch, Jun 05 2009: (Start)
a(n) is the number of adjacent transpositions in all fixed-point-free involutions of {1,2,...,2n}. Example: a(2)=3 because in 2143=(12)(34), 3412=(13)(24), and 4321=(14)(23) we have 2 + 0 + 1 adjacent transpositions.
a(n) = Sum_{k>=0} k*A079267(n,k).
(End)
Hankel transform is A137592. - Paul Barry, Sep 18 2009
(1, 3, 15, 105, ...) = INVERT transform of A000698 starting (1, 2, 10, 74, ...). - Gary W. Adamson, Oct 21 2009
a(n) = (-1)^(n+1)*H(2*n,0), where H(n,x) is the probabilists' Hermite polynomial. The generating function for the probabilists' Hermite polynomials is as follows: exp(x*t-t^2/2) = Sum_{i>=0} H(i,x)*t^i/i!. - Leonid Bedratyuk, Oct 31 2009
The Hankel transform of a(n+1) is A168467. - Paul Barry, Dec 04 2009
Partial products of odd numbers. - Juri-Stepan Gerasimov, Oct 17 2010
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
a(n) is the number of subsets of {1,...,n^2} that contain exactly k elements from {1,...,k^2} for k=1,...,n. For example, a(3)=15 since there are 15 subsets of {1,2,...,9} that satisfy the conditions, namely, {1,2,5}, {1,2,6}, {1,2,7}, {1,2,8}, {1,2,9}, {1,3,5}, {1,3,6}, {1,3,7}, {1,3,8}, {1,3,9}, {1,4,5}, {1,4,6}, {1,4,7}, {1,4,8}, and {1,4,9}. - Dennis P. Walsh, Dec 02 2011
a(n) is the leading coefficient of the Bessel polynomial y_n(x) (cf. A001498). - Leonid Bedratyuk, Jun 01 2012
For n>0: a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j)^2 for 1 <= i,j <= n. - Enrique Pérez Herrero, Jan 14 2013
a(n) is also the numerator of the mean value from 0 to Pi/2 of sin(x)^(2n). - Jean-François Alcover, Jun 13 2013
a(n) is the size of the Brauer monoid on 2n points (see A227545). - James Mitchell, Jul 28 2013
For n>1: a(n) is the numerator of M(n)/M(1) where the numbers M(i) have the property that M(n+1)/M(n) ~ n-1/2 (for example, large Kendell-Mann numbers, see A000140 or A181609, as n --> infinity). - Mikhail Gaichenkov, Jan 14 2014
a(n) = the number of upper-triangular matrix representations required for the symbolic representation of a first order central moment of the multivariate normal distribution of dimension 2(n-1), i.e., E[X_1*X_2...*X_(2n-2)|mu=0, Sigma]. See vignette for symmoments R package on CRAN and Phillips reference below. - Kem Phillips, Aug 10 2014
For n>1: a(n) is the number of Feynman diagrams of order 2n (number of internal vertices) for the vacuum polarization with one charged loop only, in quantum electrodynamics. - Robert Coquereaux, Sep 15 2014
Aerated with intervening zeros (1,0,1,0,3,...) = a(n) (cf. A123023), the e.g.f. is e^(t^2/2), so this is the base for the Appell sequence A099174 with e.g.f. e^(t^2/2) e^(x*t) = exp(P(.,x),t) = unsigned A066325(x,t), the probabilist's (or normalized) Hermite polynomials. P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for A066325(x,t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), where UP(n,t) are the polynomials of A066325 and, e.g., (P(.,t))^n = P(n,t). - Tom Copeland, Nov 15 2014
a(n) = the number of relaxed compacted binary trees of right height at most one of size n. A relaxed compacted binary tree of size n is a directed acyclic graph consisting of a binary tree with n internal nodes, one leaf, and n pointers. It is constructed from a binary tree of size n, where the first leaf in a post-order traversal is kept and all other leaves are replaced by pointers. These links may point to any node that has already been visited by the post-order traversal. The right height is the maximal number of right-edges (or right children) on all paths from the root to any leaf after deleting all pointers. The number of unbounded relaxed compacted binary trees of size n is A082161(n). See the Genitrini et al. link. - Michael Wallner, Jun 20 2017
Also the number of distinct adjacency matrices in the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017
From Christopher J. Smyth, Jan 26 2018: (Start)
a(n) = the number of essentially different ways of writing a probability distribution taking n+1 values as a sum of products of binary probability distributions. See comment of Mitch Harris above. This is because each such way corresponds to a full binary tree with n+1 leaves, with the leaves labeled by the values. (This comment is due to Niko Brummer.)
Also the number of binary trees with root labeled by an (n+1)-set S, its n+1 leaves by the singleton subsets of S, and other nodes labeled by subsets T of S so that the two daughter nodes of the node labeled by T are labeled by the two parts of a 2-partition of T. This also follows from Mitch Harris' comment above, since the leaf labels determine the labels of the other vertices of the tree.
(End)
a(n) is the n-th moment of the chi-squared distribution with one degree of freedom (equivalent to Coleman's Apr 06 2004 comment). - Bryan R. Gillespie, Mar 07 2021
Let b(n) = 0 for n odd and b(2k) = a(k); i.e., let the sequence b(n) be an aerated version of this entry. After expanding the differential operator (x + D)^n and normal ordering the resulting terms, the integer coefficient of the term x^k D^m is n! b(n-k-m) / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n. E.g., (x+D)^2 = x^2 + 2xD + D^2 + 1 with D = d/dx. The result generalizes to the raising (R) and lowering (L) operators of any Sheffer polynomial sequence by replacing x by R and D by L and follows from the disentangling relation e^{t(L+R)} = e^{t^2/2} e^{tR} e^{tL}. Consequently, these are also the coefficients of the reordered 2^n permutations of the binary symbols L and R under the condition LR = RL + 1. E.g., (L+R)^2 = LL + LR + RL + RR = LL + 2RL + RR + 1. (Cf. A344678.) - Tom Copeland, May 25 2021
From Tom Copeland, Jun 14 2021: (Start)
Lando and Zvonkin present several scenarios in which the double factorials occur in their role of enumerating perfect matchings (pairings) and as the nonzero moments of the Gaussian e^(x^2/2).
Speyer and Sturmfels (p. 6) state that the number of facets of the abstract simplicial complex known as the tropical Grassmannian G'''(2,n), the space of phylogenetic T_n trees (see A134991), or Whitehouse complex is a shifted double factorial.
These are also the unsigned coefficients of the x[2]^m terms in the partition polynomials of A134685 for compositional inversion of e.g.f.s, a refinement of A134991.
a(n)*2^n = A001813(n) and A001813(n)/(n+1)! = A000108(n), the Catalan numbers, the unsigned coefficients of the x[2]^m terms in the partition polynomials A133437 for compositional inversion of o.g.f.s, a refinement of A033282, A126216, and A086810. Then the double factorials inherit a multitude of analytic and combinatoric interpretations from those of the Catalan numbers, associahedra, and the noncrossing partitions of A134264 with the Catalan numbers as unsigned-row sums. (End)
Connections among the Catalan numbers A000108, the odd double factorials, values of the Riemann zeta function and its derivative for integer arguments, and series expansions of the reduced action for the simple harmonic oscillator and the arc length of the spiral of Archimedes are given in the MathOverflow post on the Riemann zeta function. - Tom Copeland, Oct 02 2021
b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b.)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022

Examples

			a(3) = 1*3*5 = 15.
From _Joerg Arndt_, Sep 10 2013: (Start)
There are a(3)=15 involutions of 6 elements without fixed points:
  #:    permutation           transpositions
  01:  [ 1 0 3 2 5 4 ]      (0, 1) (2, 3) (4, 5)
  02:  [ 1 0 4 5 2 3 ]      (0, 1) (2, 4) (3, 5)
  03:  [ 1 0 5 4 3 2 ]      (0, 1) (2, 5) (3, 4)
  04:  [ 2 3 0 1 5 4 ]      (0, 2) (1, 3) (4, 5)
  05:  [ 2 4 0 5 1 3 ]      (0, 2) (1, 4) (3, 5)
  06:  [ 2 5 0 4 3 1 ]      (0, 2) (1, 5) (3, 4)
  07:  [ 3 2 1 0 5 4 ]      (0, 3) (1, 2) (4, 5)
  08:  [ 3 4 5 0 1 2 ]      (0, 3) (1, 4) (2, 5)
  09:  [ 3 5 4 0 2 1 ]      (0, 3) (1, 5) (2, 4)
  10:  [ 4 2 1 5 0 3 ]      (0, 4) (1, 2) (3, 5)
  11:  [ 4 3 5 1 0 2 ]      (0, 4) (1, 3) (2, 5)
  12:  [ 4 5 3 2 0 1 ]      (0, 4) (1, 5) (2, 3)
  13:  [ 5 2 1 4 3 0 ]      (0, 5) (1, 2) (3, 4)
  14:  [ 5 3 4 1 2 0 ]      (0, 5) (1, 3) (2, 4)
  15:  [ 5 4 3 2 1 0 ]      (0, 5) (1, 4) (2, 3)
(End)
G.f. = 1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + 135135*x^7 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, (26.2.28).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 317.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 228, #19.
  • Hoel, Port and Stone, Introduction to Probability Theory, Section 7.3.
  • F. K. Hwang, D. S. Richards and P. Winter, The Steiner Tree Problem, North-Holland, 1992, see p. 14.
  • C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw-Hill, 1980, pages 466-467.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.6 and also p. 178.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer-Verlag, New York, 1999, p. 73.
  • G. Watson, The Theory of Bessel Functions, Cambridge Univ. Press, 1922.

Crossrefs

Cf. A086677; A055142 (for this sequence, |a(n+1)| + 1 is the number of distinct products which can be formed using commutative, nonassociative multiplication and a nonempty subset of n given variables).
Constant terms of polynomials in A098503. First row of array A099020.
Subsequence of A248652.
Cf. A082161 (relaxed compacted binary trees of unbounded right height).
Cf. A053871 (binomial transform).

Programs

  • GAP
    A001147 := function(n) local i, s, t; t := 1; i := 0; Print(t, ", "); for i in [1 .. n] do t := t*(2*i-1); Print(t, ", "); od; end; A001147(100); # Stefano Spezia, Nov 13 2018
    
  • Haskell
    a001147 n = product [1, 3 .. 2 * n - 1]
    a001147_list = 1 : zipWith (*) [1, 3 ..] a001147_list
    -- Reinhard Zumkeller, Feb 15 2015, Dec 03 2011
    
  • Magma
    A001147:=func< n | n eq 0 select 1 else &*[ k: k in [1..2*n-1 by 2] ] >; [ A001147(n): n in [0..20] ]; // Klaus Brockhaus, Jun 22 2011
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015
    
  • Maple
    f := n->(2*n)!/(n!*2^n);
    A001147 := proc(n) doublefactorial(2*n-1); end: # R. J. Mathar, Jul 04 2009
    A001147 := n -> 2^n*pochhammer(1/2, n); # Peter Luschny, Aug 09 2009
    G(x):=(1-2*x)^(-1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..19); # Zerinvary Lajos, Apr 03 2009; aligned with offset by Johannes W. Meijer, Aug 11 2009
    series(hypergeom([1,1/2],[],2*x),x=0,20); # Mark van Hoeij, Apr 07 2013
  • Mathematica
    Table[(2 n - 1)!!, {n, 0, 19}] (* Robert G. Wilson v, Oct 12 2005 *)
    a[ n_] := 2^n Gamma[n + 1/2] / Gamma[1/2]; (* Michael Somos, Sep 18 2014 *)
    Join[{1}, Range[1, 41, 2]!!] (* Harvey P. Dale, Jan 28 2017 *)
    a[ n_] := If[ n < 0, (-1)^n / a[-n], SeriesCoefficient[ Product[1 - (1 - x)^(2 k - 1), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 27 2017 *)
    (2 Range[0, 20] - 1)!! (* Eric W. Weisstein, Jul 22 2017 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(sum(binomial(n-1,i)*binomial(n-i-1,j)*a(i)*a(j)*a(n-i-j-1),j,0,n-i-1),i,0,n-1); /* Vladimir Kruchinin, May 06 2020 */
  • PARI
    {a(n) = if( n<0, (-1)^n / a(-n), (2*n)! / n! / 2^n)}; /* Michael Somos, Sep 18 2014 */
    
  • PARI
    x='x+O('x^33); Vec(serlaplace((1-2*x)^(-1/2))) \\ Joerg Arndt, Apr 24 2011
    
  • Python
    from sympy import factorial2
    def a(n): return factorial2(2 * n - 1)
    print([a(n) for n in range(101)])  # Indranil Ghosh, Jul 22 2017
    
  • Sage
    [rising_factorial(n+1,n)/2^n for n in (0..15)] # Peter Luschny, Jun 26 2012
    

Formula

E.g.f.: 1 / sqrt(1 - 2*x).
D-finite with recurrence: a(n) = a(n-1)*(2*n-1) = (2*n)!/(n!*2^n) = A010050(n)/A000165(n).
a(n) ~ sqrt(2) * 2^n * (n/e)^n.
Rational part of numerator of Gamma(n+1/2): a(n) * sqrt(Pi) / 2^n = Gamma(n+1/2). - Yuriy Brun, Ewa Dominowska (brun(AT)mit.edu), May 12 2001
With interpolated zeros, the sequence has e.g.f. exp(x^2/2). - Paul Barry, Jun 27 2003
The Ramanujan polynomial psi(n+1, n) has value a(n). - Ralf Stephan, Apr 16 2004
a(n) = Sum_{k=0..n} (-2)^(n-k)*A048994(n, k). - Philippe Deléham, Oct 29 2005
Log(1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + ...) = x + 5/2*x^2 + 37/3*x^3 + 353/4*x^4 + 4081/5*x^5 + 55205/6*x^6 + ..., where [1, 5, 37, 353, 4081, 55205, ...] = A004208. - Philippe Deléham, Jun 20 2006
1/3 + 2/15 + 3/105 + ... = 1/2. [Jolley eq. 216]
Sum_{j=1..n} j/a(j+1) = (1 - 1/a(n+1))/2. [Jolley eq. 216]
1/1 + 1/3 + 2/15 + 6/105 + 24/945 + ... = Pi/2. - Gary W. Adamson, Dec 21 2006
a(n) = (1/sqrt(2*Pi))*Integral_{x>=0} x^n*exp(-x/2)/sqrt(x). - Paul Barry, Jan 28 2008
a(n) = A006882(2n-1). - R. J. Mathar, Jul 04 2009
G.f.: 1/(1-x-2x^2/(1-5x-12x^2/(1-9x-30x^2/(1-13x-56x^2/(1- ... (continued fraction). - Paul Barry, Sep 18 2009
a(n) = (-1)^n*subs({log(e)=1,x=0},coeff(simplify(series(e^(x*t-t^2/2),t,2*n+1)),t^(2*n))*(2*n)!). - Leonid Bedratyuk, Oct 31 2009
a(n) = 2^n*gamma(n+1/2)/gamma(1/2). - Jaume Oliver Lafont, Nov 09 2009
G.f.: 1/(1-x/(1-2x/(1-3x/(1-4x/(1-5x/(1- ...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
The g.f. of a(n+1) is 1/(1-3x/(1-2x/(1-5x/(1-4x/(1-7x/(1-6x/(1-.... (continued fraction). - Paul Barry, Dec 04 2009
a(n) = Sum_{i=1..n} binomial(n,i)*a(i-1)*a(n-i). - Vladimir Shevelev, Sep 30 2010
E.g.f.: A(x) = 1 - sqrt(1-2*x) satisfies the differential equation A'(x) - A'(x)*A(x) - 1 = 0. - Vladimir Kruchinin, Jan 17 2011
a(n) = A123023(2*n). - Michael Somos, Jul 24 2011
a(n) = (1/2)*Sum_{i=1..n} binomial(n+1,i)*a(i-1)*a(n-i). See link above. - Dennis P. Walsh, Dec 02 2011
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*Stirling_1(n+k,k) [Kauers and Ko].
a(n) = A035342(n, 1), n >= 1 (first column of triangle).
a(n) = A001497(n, 0) = A001498(n, n), first column, resp. main diagonal, of Bessel triangle.
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n and sum of top row terms of M^(n-1), where M = a variant of the (1,2) Pascal triangle (Cf. A029635) as the following production matrix:
1, 2, 0, 0, 0, ...
1, 3, 2, 0, 0, ...
1, 4, 5, 2, 0, ...
1, 5, 9, 7, 2, ...
...
For example, a(3) = 15 is the left term in top row of M^3: (15, 46, 36, 8) and a(4) = 105 = (15 + 46 + 36 + 8).
(End)
G.f.: A(x) = 1 + x/(W(0) - x); W(k) = 1 + x + x*2*k - x*(2*k + 3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2011
a(n) = Sum_{i=1..n} binomial(n,i-1)*a(i-1)*a(n-i). - Dennis P. Walsh, Dec 02 2011
a(n) = A009445(n) / A014481(n). - Reinhard Zumkeller, Dec 03 2011
a(n) = (-1)^n*Sum_{k=0..n} 2^(n-k)*s(n+1,k+1), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = (2*n)4! = Gauss_factorial(2*n,4) = Product{j=1..2*n, gcd(j,4)=1} j. - Peter Luschny, Oct 01 2012
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k - 1)/(1 - x*(2*k + 2)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1 + x/Q(0), where Q(k) = 1 + (2*k - 1)*x - 2*x*(k + 1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) - 1 + 2*x*(2*k + 2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(2*k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k + 1)/(4*k + 2 - 2*x*(2*k + 1)*(4*k + 3)/(x*(4*k + 3) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
a(n) = (2*n - 3)*a(n-2) + (2*n - 2)*a(n-1), n > 1. - Ivan N. Ianakiev, Jul 08 2013
G.f.: G(0), where G(k) = 1 - x*(k+1)/(x*(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 04 2013
a(n) = 2*a(n-1) + (2n-3)^2*a(n-2), a(0) = a(1) = 1. - Philippe Deléham, Oct 27 2013
G.f. of reciprocals: Sum_{n>=0} x^n/a(n) = 1F1(1; 1/2; x/2), confluent hypergeometric Function. - R. J. Mathar, Jul 25 2014
0 = a(n)*(+2*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Sep 18 2014
a(n) = (-1)^n / a(-n) = 2*a(n-1) + a(n-1)^2 / a(n-2) for all n in Z. - Michael Somos, Sep 18 2014
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 3.
The sequence b(n) = A087547(n), beginning [1, 4, 52, 608, 12624, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} b(n)/a(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End)
E.g.f of the sequence whose n-th element (n = 1,2,...) equals a(n-1) is 1-sqrt(1-2*x). - Stanislav Sykora, Jan 06 2017
Sum_{n >= 1} a(n)/(2*n-1)! = exp(1/2). - Daniel Suteu, Feb 06 2017
a(n) = A028338(n, 0), n >= 0. - Wolfdieter Lang, May 27 2017
a(n) = (Product_{k=0..n-2} binomial(2*(n-k),2))/n!. - Stefano Spezia, Nov 13 2018
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} C(n-1,i)*C(n-i-1,j)*a(i)*a(j)*a(n-i-j-1), a(0)=1, - Vladimir Kruchinin, May 06 2020
From Amiram Eldar, Jun 29 2020: (Start)
Sum_{n>=1} 1/a(n) = sqrt(e*Pi/2)*erf(1/sqrt(2)), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(Pi/(2*e))*erfi(1/sqrt(2)), where erfi is the imaginary error function. (End)
G.f. of reciprocals: R(x) = Sum_{n>=0} x^n/a(n) satisfies (1 + x)*R(x) = 1 + 2*x*R'(x). - Werner Schulte, Nov 04 2024

Extensions

Removed erroneous comments: neither the number of n X n binary matrices A such that A^2 = 0 nor the number of simple directed graphs on n vertices with no directed path of length two are counted by this sequence (for n = 3, both are 13). - Dan Drake, Jun 02 2009

A036040 Irregular triangle of multinomial coefficients, read by rows (version 1).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 3, 6, 1, 1, 5, 10, 10, 15, 10, 1, 1, 6, 15, 10, 15, 60, 15, 20, 45, 15, 1, 1, 7, 21, 35, 21, 105, 70, 105, 35, 210, 105, 35, 105, 21, 1, 1, 8, 28, 56, 35, 28, 168, 280, 210, 280, 56, 420, 280, 840, 105, 70, 560, 420, 56, 210, 28, 1, 1, 9, 36, 84, 126, 36, 252
Offset: 1

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Author

Keywords

Comments

This is different from A080575 and A178867.
T(n,m) = count of set partitions of n with block lengths given by the m-th partition of n.
From Tilman Neumann, Oct 05 2008: (Start)
These are also the coefficients occurring in complete Bell polynomials, Faa di Bruno's formula (in its simplest form) and computation of moments from cumulants.
Though the Bell polynomials seem quite unwieldy, they can be computed easily as the determinant of an n-dimensional square matrix. (See, e.g., Coffey (2006) and program below.)
The complete Bell polynomial of the first n primes gives A007446. (End)
From Tom Copeland, Apr 29 2011: (Start)
A relation between partition polynomials formed from these "refined" Stirling numbers of the second kind and umbral operator trees and Lagrange inversion is presented in the link "Lagrange a la Lah".
For simple diagrams of the relation between connected graphs, cumulants, and A036040, see the references on statistical physics below. In some sense, these graphs are duals of the umbral bouquets presented in "Lagrange a la Lah". (End)
These M3 (Abramowitz-Stegun) partition polynomials are the complete Bell polynomials (see a comment above) with recurrence (see the Wikipedia link) B_0 = 1, B_n = Sum_{k=0..n-1} binomial(n-1,k) * B_{n-1-k}*x[k+1], n >= 1. - Wolfdieter Lang, Aug 31 2016
With the indeterminates (x_1, x_2, x_3,...) = (t, -c_2*t, -c_3*t, ...) with c_n > 0, umbrally B(n,a.) = B(n,t)|{t^n = a_n} = 0 and B(j,a.)B(k,a.) = B(j,t)B(k,t)|{t^n =a_n} = d_{j,k} >= 0 is the coefficient of x^j/j!*y^k/k! in the Taylor series expansion of the formal group law FGL(x,y) = f[f^{-1}(x)+f^{-1}(y)], where a_n are the inversion partition polynomials for calculating f(x) from the coefficients of the series expansion of f^{-1}(x) given in A134685. - Tom Copeland, Feb 09 2018
For applications to functionals in quantum field theory, see Figueroa et al., Brouder, Kreimer and Yeats, and Balduf. In the last two papers, the Bell polynomials with the indeterminates (x_1, x_2, x_3,...) = (c_1, 2!c_2, 3!c_3, ...) are equivalent to the partition polynomials of A130561 in the indeterminates c_n. - Tom Copeland, Dec 17 2019
From Tom Copeland, Oct 15 2020: (Start)
With a_n = n! * b_n = (n-1)! * c_n for n > 0, represent a function with f(0) = a_0 = b_0 = 1 as an
A) exponential generating function (e.g.f), or formal Taylor series: f(x) = e^{a.x} = 1 + Sum_{n > 0} a_n * x^n/n!
B) ordinary generating function (o.g.f.), or formal power series: f(x) = 1/(1-b.x) = 1 + Sum_{n > 0} b_n * x^n
C) logarithmic generating function (l.g.f): f(x) = 1 - log(1 - c.x) = 1 + Sum_{n > 0} c_n * x^n /n.
Expansions of log(f(x)) are given in
I) A127671 and A263634 for the e.g.f: log[ e^{a.*x} ] = e^{L.(a_1,a_2,...)x} = Sum_{n > 0} L_n(a_1,...,a_n) * x^n/n!, the logarithmic polynomials, cumulant expansion polynomials
II) A263916 for the o.g.f.: log[ 1/(1-b.x) ] = log[ 1 - F.(b_1,b_2,...)x ] = -Sum_{n > 0} F_n(b_1,...,b_n) * x^n/n, the Faber polynomials.
Expansions of exp(f(x)-1) are given in
III) A036040 for an e.g.f: exp[ e^{a.x} - 1 ] = e^{BELL.(a_1,...)x}, the Bell/Touchard/exponential partition polynomials, a.k.a. the Stirling partition polynomials of the second kind
IV) A130561 for an o.g.f.: exp[ b.x/(1-b.x) ] = e^{LAH.(b.,...)x}, the Lah partition polynomials
V) A036039 for an l.g.f.: exp[ -log(1-c.x) ] = e^{CIP.(c_1,...)x}, the cycle index polynomials of the symmetric groups S_n, a.k.a. the Stirling partition polynomials of the first kind.
Since exp and log are a compositional inverse pair, one can extract the indeterminates of the log set of partition polynomials from the exp set and vice versa. For a discussion of the relations among these polynomials and the combinatorics of connected and disconnected graphs/maps, see Novak and LaCroix on classical moments and cumulants and the two books on statistical mechanics referenced below. (End)
From Tom Copeland, Jun 12 2021: (Start)
These Bell polynomials and their relations to the Faa di Bruno Hopf bialgebra, correlation functions in quantum field theory, and the moment-cumulant duality are given on pp. 134 -144 of Zeidler.
An interpretation of the coefficients of the polynomials is given in expositions of the exponential formula, or principle, in Cameron et al., Duchamp, Duchamp et al., Labelle and Leroux, and Scott and Sokal along with some history. The simplest applications of this principle are given in A060540. (End)

Examples

			Triangle begins:
  1;
  1,  1;
  1,  3,  1;
  1,  4,  3,  6,  1;
  1,  5, 10, 10, 15, 10,  1;
  1,  6, 15, 10, 15, 60, 15, 20, 45, 15, 1;
  ...
The first partition of 3 (i.e., (3)) induces the set {{1, 2, 3}}, so T(3, 1) = 1; the second one (i.e., (2, 1)) the sets {{1, 2}, {3}}, {{1, 3}, {2}}, and {{2, 3}, {1}}, so T(3, 2) = 3; and the third one (i.e., (1, 1, 1)) the set {{1}, {2}, {3}}, so T(3, 1) = 1. - _Lorenzo Sauras Altuzarra_, Jun 20 2022
		

References

  • Abramowitz and Stegun, Handbook, p. 831, column labeled "M_3".
  • C. Itzykson and J. Drouffe, Statistical Field Theory Vol. 2, Cambridge Univ. Press, 1989, page 412.
  • S. Ma, Statistical Mechanics, World Scientific, 1985, page 205.
  • E. Zeidler, Quantum Field Theory II: Quantum Electrodynamics, Springer, 2009.

Crossrefs

See A080575 for another version.
Row sums are the Bell numbers A000110.
Cf. A000040, A007446, A178866 and A178867 (version 3).
Cf. A127671.
Cf. A060540 for the coefficients of the compositions e^{ x^m/m! }.

Programs

  • Maple
    with(combinat): nmax:=8: for n from 1 to nmax do P(n):=sort(partition(n)): for r from 1 to numbpart(n) do B(r):=P(n)[r] od: for m from 1 to numbpart(n) do s:=0: j:=0: while sA036040(n,m):= n!/(mul((t!)^q(t)*q(t)!,t=1..n)); od: od: seq(seq(A036040(n, m), m=1..numbpart(n)), n=1..nmax); # Johannes W. Meijer, Jun 21 2010, Jul 12 2016
  • Mathematica
    runs[li:{__Integer}] := ((Length/@ Split[ # ]))&[Sort@ li]; Table[temp=Map[Reverse, Sort@ (Sort/@ IntegerPartitions[w]), {1}]; Apply[Multinomial, temp, {1}]/Apply[Times, (runs/@ temp)!, {1}], {w, 6}]
  • MuPAD
    completeBellMatrix := proc(x,n) // x - vector x[1]...x[m], m>=n
    local i,j,M; begin
    M := matrix(n,n): // zero-initialized
    for i from 1 to n-1 do M[i,i+1] := -1: end_for:
    for i from 1 to n do for j from 1 to i do
        M[i,j] := binomial(i-1,j-1)*x[i-j+1]: end_for: end_for:
    return (M): end_proc:
    completeBellPoly := proc(x, n) begin
    return (linalg::det(completeBellMatrix (x,n))): end_proc:
    for i from 1 to 10 do print(i, completeBellPoly(x,i)): end_for:
    // Tilman Neumann, Oct 05 2008
    
  • PARI
    A036040_poly(n,V=vector(n,i,eval(Str('x,i))))={matdet(matrix(n,n,i,j,if(j<=i,binomial(i-1,j-1)*V[n-i+j],-(j==i+1))))} \\ Row n of the sequence is made of the coefficients of the monomials ordered by increasing total order (sum of powers) and then lexicographically. - M. F. Hasler, Nov 16 2013, updated Jul 12 2014
    
  • Sage
    from collections import Counter
    def ASPartitions(n, k):
        Q = [p.to_list() for p in Partitions(n, length=k)]
        for q in Q: q.reverse()
        return sorted(Q)
    def A036040_row(n):
        h = lambda p: product(map(factorial, Counter(p).values()))
        return [multinomial(p)//h(p) for k in (0..n) for p in ASPartitions(n, k)]
    for n in (1..10): print(A036040_row(n))
    # Peter Luschny, Dec 18 2016, corrected Apr 30 2022

Formula

E.g.f.: A(t) = exp(Sum_{k>=1} x[k]*(t^k)/k!).
T(n,m) is the coefficient of ((t^n)/n!)* x[1]^e(m,1)*x[2]^e(m,2)*...*x[n]^e(m,n) in A(t). Here the m-th partition of n, counted in Abramowitz-Stegun(A-St) order, is [1^e(m,1), 2^e(m,2), ..., n^e(m,n)] with e(m,j) >= 0 and if e(m, j)=0 then j^0 is not recorded.
a(n, m) = n!/Product_{j=1..n} j!^e(m,j)*e(m,j)!, with [1^e(m,1), 2^e(m,2), ..., n^e(m, n)] the m-th partition of n in the mentioned A-St order.
With the notation in the Lang reference, x(1) treated as a variable and D the derivative w.r.t. x(1), a raising operator for the polynomial S(n,x(1)) = P3_n(x[1], ..., x[n]) is R = Sum_{n>=0} x(n+1) D^n / n! ; i.e., R S(n, x(1)) = S(n+1, x(1)). The lowering operator is D; i.e., D S(n, x(1)) = n S(n-1, x(1)). The sequence of polynomials is an Appell sequence, so [S(.,x(1)) + y]^n = S(n, x(1) + y). For x(j) = (-1)^(j-1)* (j-1)! for j > 1, S(n, x(1)) = [x(1) - 1]^n + n [x(1) - 1]^(n-1). - Tom Copeland, Aug 01 2008
Raising and lowering operators are given for the partition polynomials formed from A036040 in the link in "Lagrange a la Lah Part I" on page 22. - Tom Copeland, Sep 18 2011
The n-th row is generated by the determinant of [Sum_{k=0..n-1} (x_(k+1)*(dP_n)^k/k!) - S_n], where dP_n is the n X n submatrix of A132440 and S_n is the n X n submatrix of A129185. The coefficients are flagged by the partitions of n represented by the monomials in the indeterminates x_k. Letting all x_n = t, generates the Bell / Touchard / exponential polynomials of A008277. - Tom Copeland, May 03 2014
The partition polynomials of A036039 are obtained by substituting (n-1)! x[n] for x[n] in the partition polynomials of this entry. - Tom Copeland, Nov 17 2015
-(n-1)! F(n, B(1, x[1]), B(2, x[1], x[2])/2!, ..., B(n, x[1], ..., x[n])/n!) = x[n] extracts the indeterminates of the complete Bell partition polynomials B(n, x[1], ..., x[n]) of this entry, where F(n, x[1], ..., x[n]) are the Faber polynomials of A263916. (Compare with A263634.) - Tom Copeland, Nov 29 2015; Sep 09 2016
T(n, m) = A127671(n, m)/A264753(n, m), n >= 1 and 1 <= m <= A000041(n). - Johannes W. Meijer, Jul 12 2016
From Tom Copeland, Sep 07 2016: (Start)
From the connections among the elementary Schur polynomials and the partition polynomials of A130561, A036039 and this array, the partition polynomials of this array satisfy (d/d(x_m)) P(n, x_1, ..., x_n) = binomial(n,m) * P(n-m, x_1, ..., x_(n-m)) with P(k, x_1, ..., x_n) = 0 for k < 0.
Just as in the discussion and example in A130561, the umbral compositional inverse sequence is given by the sequence P(n, x_1, -x_2, -x_3, ..., -x_n).
(End)
The partition polynomials with an index shift can be generated by (v(x) + d/dx)^n v(x). Cf. Guha, p. 12. - Tom Copeland, Jul 19 2018

Extensions

More terms from David W. Wilson
Additional comments from Wouter Meeussen, Mar 23 2003

A025035 Number of partitions of { 1, 2, ..., 3n } into sets of size 3.

Original entry on oeis.org

1, 1, 10, 280, 15400, 1401400, 190590400, 36212176000, 9161680528000, 2977546171600000, 1208883745669600000, 599606337852121600000, 356765771022012352000000, 250806337028474683456000000, 205661196363349240433920000000, 194555491759728381450488320000000
Offset: 0

Views

Author

Keywords

Comments

Row sums of A157703. - Johannes W. Meijer, Mar 07 2009
Number of bottom-row-increasing column-strict arrays of size 3 X n. - Ran Pan, Apr 10 2015
a(n) is the number of rooted semi-labeled or phylogenetic trees with n interior vertices and each interior vertex having out-degree 3. - Albert Alejandro Artiles Calix, Aug 12 2016

Examples

			G.f. = 1 + x + 10*x^2 + 280*x^3 + 15400*x^4 + 1401400*x^5 + ...
		

References

  • Erdos, Peter L., and L A. Szekely. "Applications of antilexicographic order. I. An enumerative theory of trees." Academic Press Inc. (1989): 488-96. Web. 4 July 2016.

Crossrefs

Column k=3 of A060540.

Programs

  • Magma
    [Factorial(3*n)/(Factorial(n)*6^n): n in [0..20]]; // Vincenzo Librandi, Apr 10 2015
  • Maple
    a := pochhammer(n+1, 2*n)/6^n: seq(a(n), n=0..15); # Peter Luschny, Nov 18 2019
  • Mathematica
    Select[Range[0, 39]! CoefficientList[Series[Exp[x^3/3!], {x, 0, 39}],  x], # > 0 &]  (* Geoffrey Critzer, Sep 24 2011 *)
    Table[(3 n)!/(n! (3!)^n), {n, 0, 15}] (* Michael De Vlieger, Aug 14 2016 *)
    a[ n_] := With[{m = 3 n}, If[ m < 0, 0, m! SeriesCoefficient[Exp[x^3/3!], {x, 0, m}]]]; (* Michael Somos, Nov 25 2016 *)
  • PARI
    {a(n) = if( n<0, 0, (3*n)! / n! / 6^n)}; /* Michael Somos, Mar 26 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, prod( i=0, n-1, binomial( 3*n - 3*i, 3)) / n!)}; /* Michael Somos, Feb 15 2011 */
    
  • Sage
    [rising_factorial(n+1,2*n)/6^n for n in (0..15)] # Peter Luschny, Jun 26 2012
    

Formula

a(n) = (3*n)!/(n!*(3!)^n). - Christian G. Bower, Sep 01 1998
Integral representation as n-th moment of a positive function on the positive axis: a(n) = Integral_{x>=0} x^n*sqrt(2/(3*x))*BesselK(1/3, 2*sqrt(2*x)/3)/Pi dx, for n>=0. - Karol A. Penson, Oct 05 2005
E.g.f.: exp(x^3/3!) (with interpolated zeros). - Paul Barry, May 26 2003
a(n) = Product_{i=0..n-1} binomial(3*n-3*i,3) / n! (equivalent to Christian Bower formula). - Olivier Gérard, Feb 14 2011
2*a(n) - (3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Dec 03 2012
a(n) ~ sqrt(3)*9^n*n^(2*n)/(2^n*exp(2*n)). - Ilya Gutkovskiy, Aug 12 2016
a(n) = Pochhammer(n + 1, 2*n)/6^n. - Peter Luschny, Nov 18 2019

A187783 De Bruijn's triangle, T(m,n) = (m*n)!/(n!^m) read by downward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 6, 1, 1, 1, 20, 90, 24, 1, 1, 1, 70, 1680, 2520, 120, 1, 1, 1, 252, 34650, 369600, 113400, 720, 1, 1, 1, 924, 756756, 63063000, 168168000, 7484400, 5040, 1
Offset: 0

Views

Author

Robert G. Wilson v, Jan 05 2013

Keywords

Comments

From Tilman Piesk, Oct 28 2014: (Start)
Number of permutations of a multiset that contains m different elements n times. These multisets have the signatures A249543(m,n-1) for m>=1 and n>=2.
In an m-dimensional Pascal tensor (the generalization of a symmetric Pascal matrix) P(x1,...,xn) = (x1+...+xn)!/(x1!*...*xn!), so the main diagonal of an m-dimensional Pascal tensor is D(n) = (m*n)!/(n!^m). These diagonals are the rows of this array (with m>0), which begins like this:
m\n:0 1 2 3 4 5
0: 1 1 1 1 1 1 ... A000012;
1: 1 1 1 1 1 1 ... A000012;
2: 1 2 6 20 70 252 ... A000984;
3: 1 6 90 1680 34650 756756 ... A006480;
4: 1 24 2520 369600 63063000 11732745024 ... A008977;
5: 1 120 113400 168168000 305540235000 623360743125120 ... A008978;
6: 1 720 7484400 137225088000 3246670537110000 88832646059788350720 ... A008979;
with columns: A000142 (n=1), A000680 (n=2), A014606 (n=3), A014608 (n=4), A014609 (n=5).
A089759 is the transpose of this matrix. A034841 is its diagonal. A141906 is its lower triangle. A120666 is the upper triangle of this matrix with indices starting from 1. A248827 are the diagonal sums (or the row sums of the triangle).
(End)

Examples

			T(3,5) = (3*5)!/(5!^3) = 756756 = A014609(3) = A006480(5) is the number of permutations of a multiset that contains 3 different elements 5 times, e.g., {1,1,1,1,1,2,2,2,2,2,3,3,3,3,3}.
		

Crossrefs

Cf. A089759 (transposed), A141906 (subtriangle), A120666 (subtriangle transposed), A060538 (1st row/column removed).
Main diagonal gives: A034841.
Row sums of the triangle: A248827.

Programs

  • Magma
    [Factorial(k*(n-k))/(Factorial(n-k))^k: k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 26 2022
    
  • Mathematica
    T[n_, k_]:= (k*n)!/(n!)^k; Table[T[n, k-n], {k, 9}, {n, 0, k-1}]//Flatten
  • SageMath
    def A187783(n,k): return gamma(k*(n-k)+1)/(factorial(n-k))^k
    flatten([[A187783(n,k) for k in range(n+1)] for n in range(11)]) # G. C. Greubel, Dec 26 2022

Formula

T(m,n) = (m*n)!/(n!)^m.
A060540(m,n) = T(m,n)/m! . - R. J. Mathar, Jun 21 2023

Extensions

Row m=0 prepended by Tilman Piesk, Oct 28 2014

A025036 Number of partitions of { 1, 2, ..., 4n } into sets of size 4.

Original entry on oeis.org

1, 1, 35, 5775, 2627625, 2546168625, 4509264634875, 13189599057009375, 59287247761257140625, 388035036597427985390625, 3546252199463894358484921875, 43764298393583920278062420859375, 709638098451963267308782154234765625, 14778213400262135041705388361938994140625
Offset: 0

Views

Author

Keywords

Comments

P-recursive. - Marni Mishna, Jul 11 2005

Examples

			a(1)=1: {1,2,3,4}.
One of the a(2)=35 partitions for n = 8: {1,2,3,4}{5,6,7,8}.
		

Crossrefs

Column k=4 of A060540.

Programs

  • Maple
    a := pochhammer(n + 1, 3*n) / 24^n:
    seq(a(n), n=0..13); # Peter Luschny, Nov 18 2019
  • Mathematica
    terms = 12; max = 4*(terms-1); DeleteCases[CoefficientList[Exp[x^4/4!] + O[x]^(max+1), x]*Range[0, max]!, 0] (* Jean-François Alcover, Jun 29 2018, after Paul Barry *)

Formula

a(n) = (4n)!/(n!(4!)^n). - Christian G. Bower, Sep 15 1998
E.g.f.: A(t) = Sum a(n)*t^(4n)/(4n!) = exp(t^4/4!); recurrence: 3*a(n) - (4*n-3)*(2*n-1)*(4*n-1)*a(n-1) = 0. - Marni Mishna, Jul 11 2005
Integral representation as n-th moment of a positive function on the positive axis in Maple notation: a(n)=int(x^n*(1/4*(2^(3/4)*hypergeom([], [5/4, 3/2], -3/32*x)*3^(3/4)*GAMMA(3/4)^2*x*Pi^(1/2)-2*hypergeom([], [3/4, 5/4], -3/32*x)*3^(1/2)*2^(1/2)*Pi*x^(3/4)*GAMMA(3/4)+hypergeom([], [1/2, 3/4], -3/32*x)*3^(1/4)*2^(3/4)*Pi^(3/2)*x^(1/2))/Pi^(3/2)/x^(5/4)/GAMMA(3/4)), x=0..infinity), n=0, 1..., with offset 1. -Karol A. Penson, Oct 06 2005
E.g.f.: exp(x^4/4!) (with interpolated zeros). - Paul Barry, May 26 2003
a(n) = Pochhammer(n+1, 3*n)/24^n. - Peter Luschny, Nov 18 2019
a(n) ~ 2^(5*n+1) * (n/e)^(3*n) / 3^n. - Amiram Eldar, Aug 28 2025

Extensions

Edited by N. J. A. Sloane, Aug 23 2008 at the suggestion of R. J. Mathar

A057599 a(n) = (n^2)!/(n!)^(n+1); number of ways of dividing n^2 labeled items into n unlabeled boxes of n items each.

Original entry on oeis.org

1, 1, 3, 280, 2627625, 5194672859376, 3708580189773818399040, 1461034854396267778567973305958400, 450538787986875167583433232345723106006796340625, 146413934927214422927834111686633731590253260933067148964500000000
Offset: 0

Views

Author

Henry Bottomley, Oct 06 2000

Keywords

Comments

Note that if n=p^k for p prime then a(n) is coprime to n (i.e., a(n) is not divisible by p).
a(n) is also the number of labelings for the simple graph K_n X K_n, the graph Cartesian product of the complete graph with itself. - Geoffrey Critzer, Oct 16 2016
a(n) is also the number of knockout tournament seedings with 2 rounds and n participants in each match. - Alexander Karpov, Dec 15 2017

Examples

			a(2)=3 since the possibilities are {{0,1},{2,3}}; {{0,2},{1,3}}; and {{0,3},{1,2}}.
		

Crossrefs

Main diagonal of A060540.

Programs

  • Maple
    a:= n-> (n^2)!/(n!)^(n+1):
    seq(a(n), n=0..10);  # Alois P. Heinz, Apr 29 2020
  • Mathematica
    Table[a[z_] := z^n/n!; (n^2)! Coefficient[Series[a[a[z]], {z, 0, n^2}],z^(n^2)], {n, 1, 10}] (* Geoffrey Critzer, Oct 16 2016 *)
  • PARI
    a(n) = (n^2)!/(n!)^(n+1); \\ Altug Alkan, Dec 17 2017

Formula

a(n) = A034841(n)/A000142(n).
a(n) ~ exp(n - 1/12) * n^((n-1)*(2*n-1)/2) / (2*Pi)^(n/2). - Vaclav Kotesovec, Nov 23 2018

A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+n*k+k, n*k+k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 5, 1, 1, 35, 28, 7, 1, 1, 126, 165, 55, 9, 1, 1, 462, 1001, 455, 91, 11, 1, 1, 1716, 6188, 3876, 969, 136, 13, 1, 1, 6435, 38760, 33649, 10626, 1771, 190, 15, 1, 1, 24310, 245157, 296010, 118755, 23751, 2925, 253, 17, 1, 1, 92378, 1562275
Offset: 0

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Comments

Main diagonal is A108288. Antidiagonal sums is A108289. Inverse binomial transforms of each row give triangle A108290. G.f. of row n multiplied by (1-x)^(n+1) equals g.f. of row n of triangle A108267 (rows sums of A108267 equal (n+1)^n).

Examples

			row 1: (2*n+1)/1!
row 2: (3*n+1)*(3*n+2)/2!
row 3: (4*n+1)*(4*n+2)*(4*n+3)/3!
row 4: (5*n+1)*(5*n+2)*(5*n+3)*(5*n+4)/4!
row 5: (6*n+1)*(6*n+2)*(6*n+3)*(6*n+4)*(6*n+5)/5!.
Table begins:
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
1,3,5,7,9,11,13,15,17,19,21,23,25,27,...
1,10,28,55,91,136,190,253,325,406,496,...
1,35,165,455,969,1771,2925,4495,6545,...
1,126,1001,3876,10626,23751,46376,82251,...
1,462,6188,33649,118755,324632,749398,...
1,1716,38760,296010,1344904,4496388,...
		

Crossrefs

Cf. A108290, A108267, A108288, A108289, A060544 (row 2), A015219 (row 3).
Rows include A000012, A001700, A025174. Columns include A000012, A005408, A060544. Main diagonal is A060545.

Programs

  • PARI
    T(n,k)=binomial(n+n*k+k,n*k+k)
    
  • PARI
    { i=0; write("b060543.txt", "0 1"); for (m=0, 20, for (k=0, m + 1, n=m - k + 1; write("b060543.txt", i++, " ", binomial(n + n*k + k, n*k + k))); ) } \\ Harry J. Smith, Jul 06 2009

Formula

a(n) = A060539(n, k)/n = A007318(nk, k)/n = A060540(n, k)/A060540(n-1, k).

Extensions

Entry revised by Paul D. Hanna, May 31 2005
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 17 2007

A082368 a(n) = (4*n-1)! / (n! * n! * n! * (n-1)! * 3!).

Original entry on oeis.org

1, 105, 15400, 2627625, 488864376, 96197645544, 19688264481600, 4148378852099625, 893864677761055000, 196056702961398759480, 43627992869961630486720, 9825387560922608865863400, 2235197406895366368301560000, 512889830640524227455318600000
Offset: 1

Views

Author

John A. Trono (jtrono(AT)smcvt.edu), May 10 2003

Keywords

Comments

Number of combinations that are possible when placing teams ranked #1 to #4*N in a single elimination tournament where there are four columns of N teams (as in the NCAA Men's Division-1 basketball tournament that is played in March) and each column is a separate regional tournament that produces one of the four semi-finalists. (The teams in the columns appear in sorted order and the relative positions of the four columns is irrelevant.)
Number of ways of dividing 4n labeled items into 4 unlabeled boxes with n items in each box. - Dan Parrish, Apr 09 2015

Examples

			8 ranked teams (n=2) in a four region, single elimination tournament generates 105 different possible tournament orderings, where the teams in each region are ordered from best to worst. (Teams would be matched up from top to bottom and continue towards the middle two for other matchups, when more than two teams are listed in each column.) 105 tournaments is too many to list here. As this formula applies to single elimination tournaments, this enumeration formula really only makes sense when n is even.
		

Crossrefs

Programs

  • Magma
    [Factorial(4*n-1) / (Factorial(n)*Factorial(n)* Factorial(n)*Factorial(n-1)*6): n in [1..15]]; // Vincenzo Librandi, Jun 16 2017
  • Maple
    [seq(binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24,n=1..17)]; # Zerinvary Lajos, Jun 25 2006
  • Mathematica
    Table[(4 n)! / (4! n!^4), {n, 30}] (* Vincenzo Librandi, Jun 16 2017 *)
  • PARI
    a(n)=(4*n)!/(4!*n!^4) \\ Charles R Greathouse IV, Apr 09 2015
    

Formula

a(n) = binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24. - Zerinvary Lajos, Jun 25 2006
a(n) = (4n)!/(4!*n!^4). - Dan Parrish, Apr 09 2015
From Robert Israel, Apr 09 2015: (Start)
a(n) = Gamma(2*n+1/2)*Gamma(n+1/2)*64^n/(24*Pi*(n!)^3).
a(n+1) = 8*(2*n+1)*(4*n+1)*(4*n+3)*a(n)/(n+1)^3.
G.f.: g(x) = x*hypergeom([1,5/4,3/2,7/4],[2,2,2],256*x) satisfies
x^4*(256*x-1)*g''''(x) + 5*x^3*(384*x-1)*g'''(x) + 4*x^2*(780*x-1)*g''(x) + 840*x^2*g'(x) = 0. (End)
From Karol A. Penson, Dec 31 2023: (Start)
a(n) = Integral_{x=0..256} x^n*W(x) dx, n>=0, where W(x) = x^(1/4)*hypergeometric3F2([1/4, 1/4, 1/4], [1/2, 3/4], x/256)/(96*Gamma(3/4)^4) - sqrt(x)*hypergeometric3F2([1/2, 1/2, 1/2], [3/4, 5/4], x/256)/(96*Pi^2) + Gamma(3/4)^4*x^(3/4)*hypergeometric3F2([3/4, 3/4, 3/4], [5/4, 3/2], x/256)/(768*Pi^4) is positive and unimodal on x = [0, 256]. It has a single maximum at approximately x = 31, and it goes to zero with W'(x) diverging, at both x = 0 and x = 256. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 256] is unique, as W(x) is the solution of the Hausdorff moment problem. (End)
a(n) = A008977(n)/24. - Vaclav Kotesovec, Feb 14 2024

Extensions

More terms from Zerinvary Lajos, Jun 25 2006

A060542 a(n) = (1/6)*multinomial(3*n;n,n,n).

Original entry on oeis.org

1, 15, 280, 5775, 126126, 2858856, 66512160, 1577585295, 37978905250, 925166131890, 22754499243840, 564121960420200, 14079683012144400, 353428777651788000, 8915829964229105280, 225890910734335847055, 5744976449471863238250, 146603287914300510042750
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Comments

Number of ways of dividing 3n labeled items into 3 unlabeled boxes with n items in each box.
From Antonio Campello (campello(AT)ime.unicamp.br), Nov 11 2009: (Start)
A060542(t) is the number of optimal [n,2,d] binary codes that correct at most t errors, i.e., having Hamming distance 2*t + 1 (achieved on length n = 3*t + 2). These codes are all isometric.
It is also the number of optimal [n,2,d] binary codes that detect 2*t + 1 errors, i.e., having Hamming distance 2t+2 (obtained by adding an overall parity check to the n = 3*t + 2 optimal codes). These codes are also all isometric.
For t = 0, we have the famous MDS, cyclic, simplex code {(000), (101), (110), (011)}. (End)
Also the number of distinct adjacency matrices of the complete tripartite graph K_{n,n,n}. - Eric W. Weisstein, Apr 21 2017

Crossrefs

Row 3 of A060540.

Programs

  • Maple
    a:= n-> combinat[multinomial](3*n,n$3)/3!:
    seq(a(n), n=1..18);  # Alois P. Heinz, Jul 29 2023
  • Mathematica
    Table[(3*n)!/(n!^3*6),{n,1,20}] (* Vaclav Kotesovec, Sep 23 2013 *)
    Table[Multinomial[n, n, n], {n, 20}]/6 (* Eric W. Weisstein, Apr 21 2017 *)
  • PARI
    { a=1/6; for (n=1, 100, write("b060542.txt", n, " ", a=a*3*(3*n - 1)*(3*n - 2)/n^2); ) } \\ Harry J. Smith, Jul 06 2009

Formula

a(n) = (3*n)!/((n!)^3*6) = a(n-1)*3*(3*n - 1)*(3*n - 2)/n^2 = A060540(3,n) = A006480(n)/6. - corrected by Vaclav Kotesovec, Sep 23 2013
a(n) ~ 3^(3*n-1/2)/(4*Pi*n). - Vaclav Kotesovec, Sep 23 2013
a(n) = 1/(8*n^3) * Sum_{k = 0..2*n} (-1)^(n+k) * k*(2*n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Oct 11 2024
a(n) = 1/(n^2) * Sum_{k = 0..n} (-1)^(n+k+1) * (n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Nov 03 2024

Extensions

Definition revised by N. J. A. Sloane, Feb 02 2009

A060538 Square array read by antidiagonals of number of ways of dividing n*k labeled items into n labeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 20, 90, 24, 1, 70, 1680, 2520, 120, 1, 252, 34650, 369600, 113400, 720, 1, 924, 756756, 63063000, 168168000, 7484400, 5040, 1, 3432, 17153136, 11732745024, 305540235000, 137225088000, 681080400, 40320, 1, 12870
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Examples

			       1        1        1        1
       2        6       20       70
       6       90     1680    34650
      24     2520   369600 63063000
		

Crossrefs

Subtable of A187783.
Rows include A000012, A000984, A006480, A008977, A008978 etc.
Columns include A000142, A000680, A014606, A014608, A014609 etc.
Main diagonal is A034841.

Programs

  • PARI
    T(n,k)=(n*k)!/k!^n;
    for(n=1, 6, for(k=1, 6, print1(T(n,k), ", ")); print) \\ Harry J. Smith, Jul 06 2009

Formula

T(n, k) = (nk)!/k!^n = T(n-1, k)*binomial(nk, k) = T(n-1, k)*A060539(n, k) = A060540(n, k)*A000142(k).
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