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.

Showing 1-7 of 7 results.

A000169 Number of labeled rooted trees with n nodes: n^(n-1).

Original entry on oeis.org

1, 2, 9, 64, 625, 7776, 117649, 2097152, 43046721, 1000000000, 25937424601, 743008370688, 23298085122481, 793714773254144, 29192926025390625, 1152921504606846976, 48661191875666868481, 2185911559738696531968, 104127350297911241532841, 5242880000000000000000000
Offset: 1

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Author

Keywords

Comments

Also the number of connected transitive subtree acyclic digraphs on n vertices. - Robert Castelo, Jan 06 2001
For any given integer k, a(n) is also the number of functions from {1,2,...,n} to {1,2,...,n} such that the sum of the function values is k mod n. - Sharon Sela (sharonsela(AT)hotmail.com), Feb 16 2002
The n-th term of a geometric progression with first term 1 and common ratio n: a(1) = 1 -> 1,1,1,1,... a(2) = 2 -> 1,2,... a(3) = 9 -> 1,3,9,... a(4) = 64 -> 1,4,16,64,... - Amarnath Murthy, Mar 25 2004
All rational solutions to the equation x^y = y^x, with x < y, are given by x = A000169(n+1)/A000312(n), y = A000312(n+1)/A007778(n), where n = 1, 2, 3, ... . - Nick Hobson, Nov 30 2006
a(n+1) is also the number of partial functions on n labeled objects. - Franklin T. Adams-Watters, Dec 25 2006
In other words, if A is a finite set of size n-1, then a(n) is the number of binary relations on A that are also functions. Note that a(n) = Sum_{k=0..n-1} binomial(n-1,k)*(n-1)^k = n^(n-1), where binomial(n-1,k) is the number of ways to select a domain D of size k from A and (n-1)^k is the number of functions from D to A. - Dennis P. Walsh, Apr 21 2011
This is the fourth member of a set of which the other members are the symmetric group, full transformation semigroup, and symmetric inverse semigroup. For the first three, see A000142, A000312, A002720. - Peter J. Cameron, Nov 03 2024.
More generally, consider the class of sequences of the form a(n) = (n*c(1)*...*c(i))^(n-1). This sequence has c(1)=1. A052746 has a(n) = (2*n)^(n-1), A052756 has a(n) = (3*n)^(n-1), A052764 has a(n) = (4*n)^(n-1), A052789 has a(n) = (5*n)^(n-1) for n>0. These sequences have a combinatorial structure like simple grammars. - Ctibor O. Zizka, Feb 23 2008
a(n) is equal to the logarithmic transform of the sequence b(n) = n^(n-2) starting at b(2). - Kevin Hu (10thsymphony(AT)gmail.com), Aug 23 2010
Also, number of labeled connected multigraphs of order n without cycles except one loop. See link below to have a picture showing the bijection between rooted trees and multigraphs of this kind. (Note that there are no labels in the picture, but the bijection remains true if we label the nodes.) - Washington Bomfim, Sep 04 2010
a(n) is also the number of functions f:{1,2,...,n} -> {1,2,...,n} such that f(1) = 1.
For a signed version of A000169 arising from the Vandermonde determinant of (1,1/2,...,1/n), see the Mathematica section. - Clark Kimberling, Jan 02 2012
Numerator of (1+1/(n-1))^(n-1) for n>1. - Jean-François Alcover, Jan 14 2013
Right edge of triangle A075513. - Michel Marcus, May 17 2013
a(n+1) is the number of n x n binary matrices with no more than a single one in each row. Partitioning the set of such matrices by the number k of rows with a one, we obtain a(n+1) = Sum_{k=0..n} binomial(n,k)*n^k = (n+1)^n. - Dennis P. Walsh, May 27 2014
Central terms of triangle A051129: a(n) = A051129(2*n-1,n). - Reinhard Zumkeller, Sep 14 2014
a(n) is the row sum of the n-th rows of A248120 and A055302, so it enumerates the monomials in the expansion of [x(1) + x(2) + ... + x(n)]^(n-1). - Tom Copeland, Jul 17 2015
For any given integer k, a(n) is the number of sums x_1 + ... + x_m = k (mod n) such that: x_1, ..., x_m are nonnegative integers less than n, the order of the summands does not matter, and each integer appears fewer than n times as a summand. - Carlo Sanna, Oct 04 2015
a(n) is the number of words of length n-1 over an alphabet of n letters. - Joerg Arndt, Oct 07 2015
a(n) is the number of parking functions whose largest element is n and length is n. For example, a(3) = 9 because there are nine such parking functions, namely (1,2,3), (1,3,2), (2,3,1), (2,1,3), (3,1,2), (3,2,1), (1,1,3), (1,3,1), (3,1,1). - Ran Pan, Nov 15 2015
Consider the following problem: n^2 cells are arranged in a square array. A step can be defined as going from one cell to the one directly above it, to the right of it or under it. A step above cannot be followed by a step below and vice versa. Once the last column of the square array is reached, you can only take steps down. a(n) is the number of possible paths (i.e., sequences of steps) from the cell on the bottom left to the cell on the bottom right. - Nicolas Nagel, Oct 13 2016
The rationals c(n) = a(n+1)/a(n), n >= 1, appear in the proof of G. Pólya's "elementary, but not too elementary, theorem": Sum_{n>=1} (Product_{k=1..n} a_k)^(1/n) < exp(1)*Sum_{n>=1} a_n, for n >= 1, with the sequence {a_k}{k>=1} of nonnegative terms, not all equal to 0. - _Wolfdieter Lang, Mar 16 2018
Coefficients of the generating series for the preLie operadic algebra. Cf. p. 417 of the Loday et al. paper. - Tom Copeland, Jul 08 2018
a(n)/2^(n-1) is the square of the determinant of the n X n matrix M_n with elements m(j,k) = cos(Pi*j*k/n). See Zhi-Wei Sun, Petrov link. - Hugo Pfoertner, Sep 19 2021
a(n) is the determinant of the n X n matrix P_n such that, when indexed [0, n), P(0, j) = 1, P(i <= j) = i, and P(i > j) = i-n. - C.S. Elder, Mar 11 2024

Examples

			For n=3, a(3)=9 because there are exactly 9 binary relations on A={1, 2} that are functions, namely: {}, {(1,1)}, {(1,2)}, {(2,1)}, {(2,2)}, {(1,1),(2,1)}, {(1,1),(2,2)}, {(1,2),(2,1)} and {(1,2),(2,2)}. - _Dennis P. Walsh_, Apr 21 2011
G.f. = x + 2*x^2 + 9*x^3 + 64*x^4 + 625*x^5 + 7776*x^6 + 117649*x^7 + ...
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 169.
  • Jonathan L. Gross and Jay Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 524.
  • Hannes Heikinheimo, Heikki Mannila and Jouni K. Seppnen, Finding Trees from Unordered 01 Data, in Knowledge Discovery in Databases: PKDD 2006, Lecture Notes in Computer Science, Volume 4213/2006, Springer-Verlag. - N. J. A. Sloane, Jul 09 2009
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 63.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 128.
  • 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).
  • Richard P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see page 25, Prop. 5.3.2, and p. 37, (5.52).

Crossrefs

Programs

  • Haskell
    a000169 n = n ^ (n - 1)  -- Reinhard Zumkeller, Sep 14 2014
    
  • Magma
    [n^(n-1): n in [1..20]]; // Vincenzo Librandi, Jul 17 2015
    
  • Maple
    A000169 := n -> n^(n-1);
    # second program:
    spec := [A, {A=Prod(Z, Set(A))}, labeled]; [seq(combstruct[count](spec, size=n), n=1..20)];
    # third program:
    A000169 := n -> add((-1)^(n+k-1)*pochhammer(n, k)*Stirling2(n-1, k), k = 0..n-1):
    seq(A000169(n), n = 1 .. 23);  # Mélika Tebni, May 07 2023
  • Mathematica
    Table[n^(n - 1), {n, 1, 20}] (* Stefan Steinerberger, Apr 01 2006 *)
    Range[0, 18]! CoefficientList[ Series[ -LambertW[-x], {x, 0, 18}], x] // Rest (* Robert G. Wilson v, updated by Jean-François Alcover, Oct 14 2019 *)
    (* Next, a signed version A000169 from the Vandermonde determinant of (1,1/2,...,1/n) *)
    f[j_] := 1/j; z = 12;
    v[n_] := Product[Product[f[k] - f[j], {j, 1, k - 1}], {k, 2, n}]
    Table[v[n], {n, 1, z}]
    1/%  (* A203421 *)
    Table[v[n]/v[n + 1], {n, 1, z - 1}]  (* A000169 signed *)
    (* Clark Kimberling, Jan 02 2012 *)
    a[n_]:=Det[Table[If[i==0,1,If[i<=j,i,i-n]],{i,0,n-1},{j,0,n-1}]]; Array[a,20] (* Stefano Spezia, Mar 12 2024 *)
  • MuPAD
    n^(n-1) $ n=1..20 /* Zerinvary Lajos, Apr 01 2007 */
    
  • PARI
    a(n) = n^(n-1)
    
  • Python
    def a(n): return n**(n-1)
    print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Sep 19 2021
    
  • Python
    from sympy import Matrix
    def P(n): return [[ (i-n if i > j else i) + (i == 0) for j in range(n) ] for i in range(n)]
    print(*(Matrix(P(n)).det() for n in range(1, 21)), sep=', ') # C.S. Elder, Mar 12 2024

Formula

The e.g.f. T(x) = Sum_{n>=1} n^(n-1)*x^n/n! satisfies T(x) = x*exp(T(x)), so T(x) is the functional inverse (series reversion) of x*exp(-x).
Also T(x) = -LambertW(-x) where W(x) is the principal branch of Lambert's function.
T(x) is sometimes called Euler's tree function.
a(n) = A000312(n-1)*A128434(n,1)/A128433(n,1). - Reinhard Zumkeller, Mar 03 2007
E.g.f.: LambertW(x)=x*G(0); G(k) = 1 - x*((2*k+2)^(2*k))/(((2*k+1)^(2*k)) - x*((2*k+1)^(2*k))*((2*k+3)^(2*k+1))/(x*((2*k+3)^(2*k+1)) - ((2*k+2)^(2*k+1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 30 2011
a(n) = Sum_{i=1..n} binomial(n-1,i-1)*i^(i-2)*(n-i)^(n-i). - Dmitry Kruchinin, Oct 28 2013
Limit_{n->oo} a(n)/A000312(n-1) = e. - Daniel Suteu, Jul 23 2016
From Amiram Eldar, Nov 20 2020: (Start)
Sum_{n>=1} 1/a(n) = A098686.
Sum_{n>=1} (-1)^(n+1)/a(n) = A262974. (End)
a(n) = Sum_{k=0..n-1} (-1)^(n+k-1)*Pochhammer(n, k)*Stirling2(n-1, k). - Mélika Tebni, May 07 2023
In terms of Eulerian numbers A340556(n,k) of the second order Sum_{m>=1} m^(m+n) z^m/m! = 1/(1-T(z))^(2n+1) * Sum_{k=0..n} A2(n,k) T(z)^k. - Marko Riedel, Jan 10 2024

A036038 Triangle of multinomial coefficients.

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 6, 12, 24, 1, 5, 10, 20, 30, 60, 120, 1, 6, 15, 20, 30, 60, 90, 120, 180, 360, 720, 1, 7, 21, 35, 42, 105, 140, 210, 210, 420, 630, 840, 1260, 2520, 5040, 1, 8, 28, 56, 70, 56, 168, 280, 420, 560, 336, 840, 1120, 1680, 2520, 1680, 3360, 5040, 6720
Offset: 1

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Author

Keywords

Comments

The number of terms in the n-th row is the number of partitions of n, A000041(n). - Amarnath Murthy, Sep 21 2002
For each n, the partitions are ordered according to A-St: first by length and then lexicographically (arranging the parts in nondecreasing order), which is different from the usual practice of ordering all partitions lexicographically. - T. D. Noe, Nov 03 2006
For this ordering of the partitions, for n >= 1, see the remarks and the C. F. Hindenburg link given in A036036. - Wolfdieter Lang, Jun 15 2012
The relation (n+1) * A134264(n+1) = A248120(n+1) / a(n) where the arithmetic is performed for matching partitions in each row n connects the combinatorial interpretations of this array to some topological and algebraic constructs of the two other entries. Also, these seem (cf. MOPS reference, Table 2) to be the coefficients of the Jack polynomial J(x;k,alpha=0). - Tom Copeland, Nov 24 2014
The conjecture on the Jack polynomials of zero order is true as evident from equation a) on p. 80 of the Stanley reference, suggested to me by Steve Kass. The conventions for denoting the more general Jack polynomials J(n,alpha) vary. Using Stanley's convention, these Jack polynomials are the umbral extensions of the multinomial expansion of (s_1*x_1 + s_2*x_2 + ... + s_(n+1)*x_(n+1))^n in which the subscripts of the (s_k)^j in the symmetric monomial expansions are finally ignored and the exponent dropped to give s_j(alpha) = j-th row polynomial of A094638 or |A008276| in ascending powers of alpha. (The MOPS table has some inconsistency between n = 3 and n = 4.) - Tom Copeland, Nov 26 2016

Examples

			1;
1, 2;
1, 3,  6;
1, 4,  6, 12, 24;
1, 5, 10, 20, 30, 60, 120;
1, 6, 15, 20, 30, 60,  90, 120, 180, 360, 720;
		

References

  • Abramowitz and Stegun, Handbook, p. 831, column labeled "M_1".

Crossrefs

Cf. A036036-A036040. Different from A078760. Row sums give A005651.
Cf. A183610 is a table of sums of powers of terms in rows.
Cf. A134264 and A248120.
Cf. A096162 for connections to A130561.

Programs

  • Maple
    nmax:=7: with(combinat): 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 sA036038(n, m) := n!/ (mul((t!)^q(t), t=1..n)); od: od: seq(seq(A036038(n, m), m=1..numbpart(n)), n=1..nmax); # Johannes W. Meijer, Jul 14 2016
  • Mathematica
    Flatten[Table[Apply[Multinomial, Reverse[Sort[IntegerPartitions[i],  Length[ #1]>Length[ #2]&]], {1}], {i,9}]] (* T. D. Noe, Nov 03 2006 *)
  • Sage
    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 A036038_row(n):
        return [multinomial(p) for k in (0..n) for p in ASPartitions(n, k)]
    for n in (1..10): print(A036038_row(n))
    # Peter Luschny, Dec 18 2016, corrected Apr 30 2022

Formula

The n-th row is the expansion of (x_1 + x_2 + ... + x_(n+1))^n in the basis of the monomial symmetric polynomials (m.s.p.). E.g., (x_1 + x_2 + x_3 + x_4)^3 = m[3](x_1,..,x_4) + 3*m[1,2](x_1,..,x_4) + 6*m[1,1,1](x_1,..,x_4) = (Sum_{i=1..4} x_i^3) + 3*(Sum_{i,j=1..4;i != j} x_i^2 x_j) + 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k). The number of indeterminates can be increased indefinitely, extending each m.s.p., yet the expansion coefficients remain the same. In each m.s.p., unique combinations of exponents and subscripts appear only once with a coefficient of unity. Umbral reduction by replacing x_k^j with x_j in the expansions gives the partition polynomials of A248120. - Tom Copeland, Nov 25 2016
From Tom Copeland, Nov 26 2016: (Start)
As an example of the umbral connection to the Jack polynomials: J(3,alpha) = (Sum_{i=1..4} x_i^3)*s_3(alpha) + 3*(Sum_{i,j=1..4;i!=j} x_i^2 x_j)*s_2(alpha)*s_1(alpha)+ 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k)*s_1(alpha)*s_1(alpha)*s_1(alpha) = (Sum_{i=1..4} x_i^3)*(1+alpha)*(1+2*alpha)+ 3*(sum_{i,j=1..4;i!=j} x_i^2 x_j)*(1+alpha) + 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k).
See the Copeland link for more relations between the multinomial coefficients and the Jack symmetric functions. (End)

Extensions

More terms from David W. Wilson and Wouter Meeussen

A134264 Coefficients T(j, k) of a partition transform for Lagrange compositional inversion of a function or generating series in terms of the coefficients of the power series for its reciprocal. Enumeration of noncrossing partitions and primitive parking functions. T(n,k) for n >= 1 and 1 <= k <= A000041(n-1), an irregular triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 6, 1, 1, 5, 5, 10, 10, 10, 1, 1, 6, 6, 3, 15, 30, 5, 20, 30, 15, 1, 1, 7, 7, 7, 21, 42, 21, 21, 35, 105, 35, 35, 70, 21, 1, 1, 8, 8, 8, 4, 28, 56, 56, 28, 28, 56, 168, 84, 168, 14, 70, 280, 140, 56, 140, 28, 1, 1, 9, 9, 9, 9, 36, 72
Offset: 1

Views

Author

Tom Copeland, Jan 14 2008

Keywords

Comments

Coefficients are listed in Abramowitz and Stegun order (A036036).
Given an invertible function f(t) analytic about t=0 (or a formal power series) with f(0)=0 and Df(0) not equal to 0, form h(t) = t / f(t) and let h_n denote the coefficient of t^n in h(t).
Lagrange inversion gives the compositional inverse about t=0 as g(t) = Sum_{j>=1} ( t^j * (1/j) * Sum_{permutations s with s(1) + s(2) + ... + s(j) = j - 1} h_s(1) * h_s(2) * ... * h_s(j) ) = t * T(1,1) * h_0 + Sum_{j>=2} ( t^j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) ), where H(j-1,k ; h_0,h_1,...) is the k-th partition for h_1 through h_(j-1) corresponding to n=j-1 on page 831 of Abramowitz and Stegun (ordered as in A&S) with (h_0)^(j-m)=(h_0)^(n+1-m) appended to each partition subsumed under n and m of A&S.
Denoting h_n by (n') for brevity, to 8th order in t,
g(t) = t * (0')
+ t^2 * [ (0') (1') ]
+ t^3 * [ (0')^2 (2') + (0') (1')^2 ]
+ t^4 * [ (0')^3 (3') + 3 (0')^2 (1') (2') + (0') (1')^3 ]
+ t^5 * [ (0')^4 (4') + 4 (0')^3 (1') (3') + 2 (0')^3 (2')^2 + 6 (0')^2 (1')^2 (2') + (0') (1')^4 ]
+ t^6 * [ (0')^5 (5') + 5 (0')^4 (1') (4') + 5 (0')^4 (2') (3') + 10 (0')^3 (1')^2 (3') + 10 (0')^3 (1') (2')^2 + 10 (0')^2 (1')^3 (2') + (0') (1')^5 ]
+ t^7 * [ (0')^6 (6') + 6 (0')^5 (1') (5') + 6 (0')^5 (2') (4') + 3 (0')^5 (3')^2 + 15 (0')^4 (1')^2 (4') + 30 (0')^4 (1') (2') (3') + 5 (0')^4 (2')^3 + 20 (0')^3 (1')^3 (3') + 30 (0')^3 (1')^2 (2')^2 + 15 (0')^2 (1')^4 (2') + (0') (1')^6]
+ t^8 * [ (0')^7 (7') + 7 (0')^6 (1') (6') + 7 (0')^6 (2') (5') + 7 (0')^6 (3') (4') + 21 (0')^5 (1')^2* (5') + 42 (0')^5 (1') (2') (4') + 21 (0')^5 (1') (3')^2 + 21 (0')^5 (2')^2 (3') + 35 (0')^4 (1')^3 (4') + 105 (0)^4 (1')^2 (2') (3') + 35 (0')^4 (1') (2')^3 + 35 (0')^3 (1')^4 (3') + 70 (0')^3 (1')^3 (2')^2 + 21 (0')^2 (1')^5 (2') + (0') (1')^7 ]
+ ..., where from the formula section, for example, T(8,1',2',...,7') = 7! / ((8 - (1'+ 2' + ... + 7'))! * 1'! * 2'! * ... * 7'!) are the coefficients of the integer partitions (1')^1' (2')^2' ... (7')^7' in the t^8 term.
A125181 is an extended, reordered version of the above sequence, omitting the leading 1, with alternate interpretations.
If the coefficients of partitions with the same exponent for h_0 are summed within rows, A001263 is obtained, omitting the leading 1.
From identification of the elements of the inversion with those on page 25 of the Ardila et al. link, the coefficients of the irregular table enumerate non-crossing partitions on [n]. - Tom Copeland, Oct 13 2014
From Tom Copeland, Oct 28-29 2014: (Start)
Operating with d/d(1') = d/d(h_1) on the n-th partition polynomial Prt(n;h_0,h_1,..,h_n) in square brackets above associated with t^(n+1) generates n * Prt(n-1;h_0,h_1,..,h_(n-1)); therefore, the polynomials are an Appell sequence of polynomials in the indeterminate h_1 when h_0=1 (a special type of Sheffer sequence).
Consequently, umbrally, [Prt(.;1,x,h_2,..) + y]^n = Prt(n;1,x+y,h_2,..); that is, Sum_{k=0..n} binomial(n,k) * Prt(k;1,x,h_2,..) * y^(n-k) = Prt(n;1,x+y,h_2,..).
Or, e^(x*z) * exp[Prt(.;1,0,h_2,..) * z] = exp[Prt(.;1,x,h_2,..) * z]. Then with x = h_1 = -(1/2) * d^2[f(t)]/dt^2 evaluated at t=0, the formal Laplace transform from z to 1/t of this expression generates g(t), the comp. inverse of f(t), when h_0 = 1 = df(t)/dt eval. at t=0.
I.e., t / (1 - t*(x + Prt(.;1,0,h_2,..))) = t / (1 - t*Prt(.;1,x,h_2,..)) = g(t), interpreted umbrally, when h_0 = 1.
(End)
Connections to and between arrays associated to the Catalan (A000108 and A007317), Riordan (A005043), Fibonacci (A000045), and Fine (A000957) numbers and to lattice paths, e.g., the Motzkin, Dyck, and Łukasiewicz, can be made explicit by considering the inverse in x of the o.g.f. of A104597(x,-t), i.e., f(x) = P(Cinv(x),t-1) = Cinv(x) / (1 + (t-1)*Cinv(x)) = x*(1-x) / (1 + (t-1)*x*(1-x)) = (x-x^2) / (1 + (t-1)*(x-x^2)), where Cinv(x) = x*(1-x) is the inverse of C(x) = (1 - sqrt(1-4*x)) / 2, a shifted o.g.f. for the Catalan numbers, and P(x,t) = x / (1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x / (1-t*x). Then h(x,t) = x / f(x,t) = x * (1+(t-1)Cinv(x)) / Cinv(x) = 1 + t*x + x^2 + x^3 + ..., i.e., h_1=t and all other coefficients are 1, so the inverse of f(x,t) in x, which is explicitly in closed form finv(x,t) = C(Pinv(x,t-1)), is given by A091867, whose coefficients are sums of the refined Narayana numbers above obtained by setting h_1=(1')=t in the partition polynomials and all other coefficients to one. The group generators C(x) and P(x,t) and their inverses allow associations to be easily made between these classic number arrays. - Tom Copeland, Nov 03 2014
From Tom Copeland, Nov 10 2014: (Start)
Inverting in x with t a parameter, let F(x;t,n) = x - t*x^(n+1). Then h(x) = x / F(x;t,n) = 1 / (1-t*x^n) = 1 + t*x^n + t^2*x^(2n) + t^3*x^(3n) + ..., so h_k vanishes unless k = m*n with m an integer in which case h_k = t^m.
Finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). [Added braces to disambiguate the formula. - N. J. A. Sloane, Oct 20 2015]
This relation reveals properties of the partitions and sums of the coefficients of the array. For n=1, h_k = t^k for all k, implying that the row sums are the Catalan numbers. For n = 2, h_k for k odd vanishes, implying that there are no blocks with only even-indexed h_k on the even-numbered rows and that only the blocks containing only even-sized bins contribute to the odd-row sums giving the Fuss-Catalan numbers for n=2. And so on, for n > 2.
These relations are reflected in any combinatorial structures enumerated by this array and the partitions, such as the noncrossing partitions depicted for a five-element set (a pentagon) in Wikipedia.
(End)
From Tom Copeland, Nov 12 2014: (Start)
An Appell sequence possesses an umbral inverse sequence (cf. A249548). The partition polynomials here, Prt(n;1,h_1,...), are an Appell sequence in the indeterminate h_1=u, so have an e.g.f. exp[Prt(.;1,u,h_2...)*t] = e^(u*t) * exp[Prt(.;1,0,h2,...)*t] with umbral inverses with an e.g.f e^(-u*t) / exp[Prt(.;1,0,h2,...)*t]. This makes contact with the formalism of A133314 (cf. also A049019 and A019538) and the signed, refined face partition polynomials of the permutahedra (or their duals), which determine the reciprocal of exp[Prt(.,0,u,h2...)*t] (cf. A249548) or exp[Prt(.;1,u,h2,...)*t], forming connections among the combinatorics of permutahedra and the noncrossing partitions, Dyck paths and trees (cf. A125181), and many other important structures isomorphic to the partitions of this entry, as well as to formal cumulants through A127671 and algebraic structures of Lie algebras. (Cf. relationship of permutahedra with the Eulerians A008292.)
(End)
From Tom Copeland, Nov 24 2014: (Start)
The n-th row multiplied by n gives the number of terms in the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n+1)]^n under the umbral mapping x(m)^j = h_j, for any m. E.g., [a + b + c]^2 = [a^2 + b^2 + c^2] + 2 * [a*b + a*c + b*c] is mapped to [3 * h_2] + 2 * [3 * h_1^2], and 3 * A134264(3) = 3 *(1,1)= (3,3) the number of summands in the two homogeneous polynomials in the square brackets. For n=3, [a + b + c + d]^3 = [a^3 + b^3 + ...] + 3 [a*b^2 + a*c^2 + ...] + 6 [a*b*c + a*c*d + ...] maps to [4 * h_3] + 3 [12 * h_1 * h_2] + 6 [4 * (h_1)^3], and the number of terms in the brackets is given by 4 * A134264(4) = 4 * (1,3,1) = (4,12,4).
The further reduced expression is 4 h_3 + 36 h_1 h_2 + 24 (h_1)^3 = A248120(4) with h_0 = 1. The general relation is n * A134264(n) = A248120(n) / A036038(n-1) where the arithmetic is performed on the coefficients of matching partitions in each row n.
Abramowitz and Stegun give combinatorial interpretations of A036038 and relations to other number arrays.
This can also be related to repeated umbral composition of Appell sequences and topology with the Bernoulli numbers playing a special role. See the Todd class link.
(End)
These partition polynomials are dubbed the Voiculescu polynomials on page 11 of the He and Jejjala link. - Tom Copeland, Jan 16 2015
See page 5 of the Josuat-Verges et al. reference for a refinement of these partition polynomials into a noncommutative version composed of nondecreasing parking functions. - Tom Copeland, Oct 05 2016
(Per Copeland's Oct 13 2014 comment.) The number of non-crossing set partitions whose block sizes are the parts of the n-th integer partition, where the ordering of integer partitions is first by total, then by length, then lexicographically by the reversed sequence of parts. - Gus Wiseman, Feb 15 2019
With h_0 = 1 and the other h_n replaced by suitably signed partition polynomials of A263633, the refined face partition polynomials for the associahedra of normalized A133437 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Sep 09 2019
Number of primitive parking functions associated to each partition of n. See Lemma 3.8 on p. 28 of Rattan. - Tom Copeland, Sep 10 2019
With h_n = n + 1, the d_k (A006013) of Table 2, p. 18, of Jong et al. are obtained, counting the n-point correlation functions in a quantum field theory. - Tom Copeland, Dec 25 2019
By inspection of the diagrams on Robert Dickau's website, one can see the relationship between the monomials of this entry and the connectivity of the line segments of the noncrossing partitions. - Tom Copeland, Dec 25 2019
Speicher has examples of the first four inversion partition polynomials on pp. 22 and 23 with his k_n equivalent to h_n = (n') here with h_0 = 1. Identifying z = t, C(z) = t/f(t) = h(t), and M(z) = f^(-1)(t)/t, then statement (3), on p. 43, of Theorem 3.26, C(z M(z)) = M(z), is equivalent to substituting f^(-1)(t) for t in t/f(t), and statement (4), M(z/C(z)) = C(z), to substituting f(t) for t in f^(-1)(t)/t. - Tom Copeland, Dec 08 2021
Given a Laurent series of the form f(z) = 1/z + h_1 + h_2 z + h_3 z^2 + ..., the compositional inverse is f^(-1)(z) = 1/z + Prt(1;1,h_1)/z^2 + Prt(2;1,h_1,h_2)/z^3 + ... = 1/z + h_1/z^2 + (h_1^2 + h_2)/z^3 + (h_1^3 + 3 h_1 h_2 + h_3)/z^4 + (h_1^4 + 6 h_1^2 h_2 + 4 h_1 h_3 + 2 h_2^2 + h_4)/z^5 + ... for which the polynomials in the numerators are the partition polynomials of this entry. For example, this formula applied to the q-expansion of Klein's j-invariant / function with coefficients A000521, related to monstrous moonshine, gives the compositional inverse with the coefficients A091406 (see He and Jejjala). - Tom Copeland, Dec 18 2021
The partition polynomials of A350499 'invert' the polynomials of this entry giving the indeterminates h_n. A multinomial formula for the coefficients of the partition polynomials of this entry, equivalent to the multinomial formula presented in the first four sentences of the formula section below, is presented in the MathOverflow question referenced in A350499. - Tom Copeland, Feb 19 2022

Examples

			1) With f(t) = t / (t-1), then h(t) = -(1-t), giving h_0 = -1, h_1 = 1 and h_n = 0 for n>1. Then g(t) = -t - t^2 - t^3 - ... = t / (t-1).
2) With f(t) = t*(1-t), then h(t) = 1 / (1-t), giving h_n = 1 for all n. The compositional inverse of this f(t) is g(t) = t*A(t) where A(t) is the o.g.f. for the Catalan numbers; therefore the sum over k of T(j,k), i.e., the row sum, is the Catalan number A000108(j-1).
3) With f(t) = (e^(-a*t)-1) / (-a), h(t) = Sum_{n>=0} Bernoulli(n) * (-a*t)^n / n! and g(t) = log(1-a*t) / (-a) = Sum_{n>=1} a^(n-1) * t^n / n. Therefore with h_n = Bernoulli(n) * (-a)^n / n!, Sum_{permutations s with s(1)+s(2)+...+s(j)=j-1} h_s(1) * h_s(2) * ... * h_s(j) = j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) = a^(j-1). Note, in turn, Sum_{a=1..m} a^(j-1) = (Bernoulli(j,m+1) - Bernoulli(j)) / j for the Bernoulli polynomials and numbers, for j>1.
4) With f(t,x) = t / (x-1+1/(1-t)), then h(t,x) = x-1+1/(1-t), giving (h_0)=x and (h_n)=1 for n>1. Then g(t,x) = (1-(1-x)*t-sqrt(1-2*(1+x)*t+((x-1)*t)^2)) / 2, a shifted o.g.f. in t for the Narayana polynomials in x of A001263.
5) With h(t)= o.g.f. of A075834, but with A075834(1)=2 rather than 1, which is the o.g.f. for the number of connected positroids on [n] (cf. Ardila et al., p. 25), g(t) is the o.g.f. for A000522, which is the o.g.f. for the number of positroids on [n]. (Added Oct 13 2014 by author.)
6) With f(t,x) = x / ((1-t*x)*(1-(1+t)*x)), an o.g.f. for A074909, the reverse face polynomials of the simplices, h(t,x) = (1-t*x) * (1-(1+t)*x) with h_0=1, h_1=-(1+2*t), and h_2=t*(1+t), giving as the inverse in x about 0 the o.g.f. (1+(1+2*t)*x-sqrt(1+(1+2*t)*2*x+x^2)) / (2*t*(1+t)*x) for signed A033282, the reverse face polynomials of the Stasheff polytopes, or associahedra. Cf. A248727. (Added Jan 21 2015 by author.)
7) With f(x,t) = x / ((1+x)*(1+t*x)), an o.g.f. for the polynomials (-1)^n * (1 + t + ... + t^n), h(t,x) = (1+x) * (1+t*x) with h_0=1, h_1=(1+t), and h_2=t, giving as the inverse in x about 0 the o.g.f. (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2)) / (2*x*t) for the Narayana polynomials A001263. Cf. A046802. (Added Jan 24 2015 by author.)
From _Gus Wiseman_, Feb 15 2019: (Start)
Triangle begins:
   1
   1
   1   1
   1   3   1
   1   4   2   6   1
   1   5   5  10  10  10   1
   1   6   6   3  15  30   5  20  30  15   1
   1   7   7   7  21  42  21  21  35 105  35  35  70  21   1
Row 5 counts the following non-crossing set partitions:
  {{1234}}  {{1}{234}}  {{12}{34}}  {{1}{2}{34}}  {{1}{2}{3}{4}}
            {{123}{4}}  {{14}{23}}  {{1}{23}{4}}
            {{124}{3}}              {{12}{3}{4}}
            {{134}{2}}              {{1}{24}{3}}
                                    {{13}{2}{4}}
                                    {{14}{2}{3}}
(End)
		

References

  • A. Nica and R. Speicher (editors), Lectures on the Combinatorics of Free Probability, London Mathematical Society Lecture Note Series: 335, Cambridge University Press, 2006 (see in particular, Eqn. 9.14 on p. 141, enumerating noncrossing partitions).

Crossrefs

(A001263,A119900) = (reduced array, associated g(x)). See A145271 for meaning and other examples of reduced and associated.
Other orderings are A125181 and A306438.
Cf. A119900 (e.g.f. for reduced W(x) with (h_0)=t and (h_n)=1 for n>0).
Cf. A248927 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A091867 and A125181, for relations to lattice paths and trees.
Cf. A249548 for use of Appell properties to generate the polynomials.
Cf. A133314, A049019, A019538, A127671, and A008292 for relations to permutahedra, Eulerians.
Cf. A006013.

Programs

  • Mathematica
    Table[Binomial[Total[y],Length[y]-1]*(Length[y]-1)!/Product[Count[y,i]!,{i,Max@@y}],{n,7},{y,Sort[Sort/@IntegerPartitions[n]]}] (* Gus Wiseman, Feb 15 2019 *)
  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!/((n-#v+1)!*prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!))}
    row(n)=[C(Vec(p)) | p<-partitions(n-1)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

Formula

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.
T(j,k) is [ P(j,m;a...) / j ] for the k-th partition of j-1 as described in the comments.
For example from g(t) above, T(5,4) = (5! / ((5-3)! * 2!)) / 5 = 6 for the 4th partition under n=5-1=4 with m=3 parts in A&S.
From Tom Copeland, Sep 30 2011: (Start)
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}
= [(h_0)-1+:1/(1-h.*x):]^2 / {(h_0)-:[h.x/(1-h.x)]^2:}
= [(h_0)+(h_1)x+(h_2)x^2+...]^2 / [(h_0)-(h_2)x^2-2(h_3)x^3-3(h_4)x^4-...], where :" ": denotes umbral evaluation of the expression within the colons and h. is an umbral coefficient.
Then for the partition polynomials of A134264,
Poly[n;h_0,...,h_(n-1)]=(1/n!)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t) = exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)), and g(t) gives A001263 with (h_0)=u and (h_n)=1 for n>0 and A000108 with u=1.
(End)
From Tom Copeland, Oct 20 2011: (Start)
With exp(x* PS(.,t)) = exp(t*g(x)) = exp(x*W(y)d/dy) exp(t*y) eval. at y=0, the raising (creation) and lowering (annihilation) operators defined by R PS(n,t) = PS(n+1,t) and L PS(n,t) = n*PS(n-1,t) are
R = t*W(d/dt) = t*((h_0) + (h_1)d/dt + (h_2)(d/dt)^2 + ...)^2 / ((h_0) - (h_2)(d/dt)^2 - 2(h_3)(d/dt)^3 - 3(h_4)(d/dt)^4 + ...), and
L = (d/dt)/h(d/dt) = (d/dt) 1/((h_0) + (h_1)*d/dt + (h_2)*(d/dt)^2 + ...)
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0 are the row polynomials of A134264. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
(End)
Using the formalism of A263634, the raising operator for the partition polynomials of this array with h_0 = 1 begins as R = h_1 + h_2 D + h_3 D^2/2! + (h_4 - h_2^2) D^3/3! + (h_5 - 5 h_2 h_3) D^4/4! + (h_6 + 5 h_2^3 - 7 h_3^2 - 9 h_2 h_4) D^5/5! + (h_7 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... with D = d/d(h_1). - Tom Copeland, Sep 09 2016
Let h(x) = x/f^{-1}(x) = 1/[1-(c_2*x+c_3*x^2+...)], with c_n all greater than zero. Then h_n are all greater than zero and h_0 = 1. Determine P_n(t) from exp[t*f^{-1}(x)] = exp[x*P.(t)] with f^{-1}(x) = x/h(x) expressed in terms of the h_n (cf. A133314 and A263633). Then P_n(b.) = 0 gives a recursion relation for the inversion polynomials of this entry a_n = b_n/n! in terms of the lower order inversion polynomials and P_j(b.)P_k(b.) = P_j(t)P_k(t)|{t^n = b_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)]. - Tom Copeland, Feb 09 2018
A raising operator for the partition polynomials with h_0 = 1 regarded as a Sheffer Appell sequence in h_1 is described in A249548. - Tom Copeland, Jul 03 2018

Extensions

Added explicit t^6, t^7, and t^8 polynomials and extended initial table to include the coefficients of t^8. - Tom Copeland, Sep 14 2016
Title modified by Tom Copeland, May 28 2018
More terms from Gus Wiseman, Feb 15 2019
Title modified by Tom Copeland, Sep 10 2019

A055302 Triangle of number of labeled rooted trees with n nodes and k leaves, n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 2, 0, 6, 3, 0, 24, 36, 4, 0, 120, 360, 140, 5, 0, 720, 3600, 3000, 450, 6, 0, 5040, 37800, 54600, 18900, 1302, 7, 0, 40320, 423360, 940800, 588000, 101136, 3528, 8, 0, 362880, 5080320, 16087680, 15876000, 5143824, 486864, 9144, 9, 0, 3628800
Offset: 1

Views

Author

Christian G. Bower, May 11 2000

Keywords

Comments

Beginning with the second row, dividing each row by n gives the mirror of row n-1 of A141618. Under the exponential transform, the mirror of A141618 is generated, relating the number of connected graphs here to the number of disconnected graphs associated with A141618 (cf. A127671 and A036040). - Tom Copeland, Oct 25 2014

Examples

			Triangle begins
     1,
     2,     0;
     6,     3,     0;
    24,    36,     4,     0;
   120,   360,   140,     5,    0;
   720,  3600,  3000,   450,    6, 0;
  5040, 37800, 54600, 18900, 1302, 7, 0;
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 313.

Crossrefs

Row sums give A000169. Columns 1 through 12: A000142, A055303-A055313. Cf. A055314.
Cf. A248120 for a natural refinement.

Programs

  • Maple
    T:= (n, k)-> (n!/k!)*Stirling2(n-1, n-k):
    seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Nov 13 2013
  • Mathematica
    Table[Table[n!/k! StirlingS2[n-1,n-k], {k,1,n}], {n,0,10}]//Grid  (* Geoffrey Critzer, Dec 01 2012 *)
  • PARI
    A055302(n,k)=n!/k!*stirling(n-1, n-k,2);
    for(n=1,10,for(k=1,n,print1(A055302(n,k),", "));print());
    \\ Joerg Arndt, Oct 27 2014

Formula

E.g.f. (relative to x) satisfies: A(x,y) = xy + x*exp(A(x,y)) - x. Divides by n and shifts up under exponential transform.
T(n,k) = (n!/k!)*Stirling2(n-1, n-k). - Vladeta Jovovic, Jan 28 2004
T(n,k) = A055314(n,k)*(n-k) + A055314(n,k+1)*(k+1). The first term is the number of such trees with root degree > 1 while the second term is the number of such trees with root degree = 1. This simplifies to the above formula by Vladeta Jovovic. - Geoffrey Critzer, Dec 01 2012
E.g.f.: G(x,t) = log[1 + t * N(x*t,1/t)], where N(x,t) is the e.g.f. of A141618. Also, G(x*t,1/t)= log[1 + N(x,t)/t] is the comp. inverse in x of x / [1 + t * (e^x - 1)]. - Tom Copeland, Oct 26 2014

A186366 Triangle read by rows: T(n,k) is the number of cycle-up-down permutations of {1,2,...,n} having k cycles (1<=k<=n).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 2, 7, 6, 1, 5, 20, 25, 10, 1, 16, 70, 105, 65, 15, 1, 61, 287, 490, 385, 140, 21, 1, 272, 1356, 2548, 2345, 1120, 266, 28, 1, 1385, 7248, 14698, 15204, 8715, 2772, 462, 36, 1, 7936, 43280, 93420, 105880, 69405, 26985, 6090, 750, 45, 1, 50521, 285571, 649715, 793210, 577225, 260337, 72765, 12210, 1155, 55, 1
Offset: 1

Views

Author

Emeric Deutsch, Feb 28 2011

Keywords

Comments

A permutation is said to be cycle-up-down if it is a product of up-down cycles. A cycle (b(1), b(2), ...) is said to be up-down if, when written with its smallest element in the first position, it satisfies b(1) < b(2) > b(3) < ... .
Sum of entries in row n is A000111(n+1) (the Euler or up-down numbers).
The row generating polynomial of row n is c_n(t) from the Johnson reference (p. 127).
T(n,1) = A000111(n-1).
Sum_{k=1..n} k*T(n,k) = A186367(n).
With a different offset, appears to be the same as the table of sub-permutations of a class of Andre permutations given by Disanto. - Peter Bala, Feb 11 2012. That is, this triangle appears to be identical to the triangle giving the number of binary increasing trees with n nodes and a "min-path" of length k. - N. J. A. Sloane, May 12 2012
The Bell transform of A000111. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016
An apparent signed version is presented on p. 6 of the Csikvari preprint, related to graph polynomials of the complete graphs K_n. - Tom Copeland, Jan 17 2017
The trivariate e.g.f. below may be expressed as H = [e^(-z) e^(xz) / (1-sin (z))]^t = [e^(z*(p.(x)-1))]^t = [e^(z*(p.(x-1)))]^t, where (p.(x))^n = p_n(x) are a sequence of Appell polynomials. For t=m, an integer, the formalism of A248120 related to the Hirzebruch criterion for convolutions applies and that of the Scott and Sokal preprint (see eqn. 3.1 on p. 10 and eqn. 3.62 on p. 24). - Tom Copeland, Jan 17 2017

Examples

			T(3,2)=3 because we have (1)(23), (12)(3), and (13)(2).
T(4,3)=6 because we have (1)(2)(34), (1)(23)(4), (1)(24)(3), (12)(3)(4), (13)(2)(4), and (14)(2)(3).
Triangle starts:
     1;
     1,    1;
     1,    3,     1;
     2,    7,     6,     1;
     5,   20,    25,    10,    1;
    16,   70,   105,    65,   15,    1;
    61,  287,   490,   385,  140,   21,   1;
   272, 1356,  2548,  2345, 1120,  266,  28,  1;
  1385, 7248, 14698, 15204, 8715, 2772, 462, 36, 1;
  ...
		

Crossrefs

Cf. A186367. Diagonals: A000111, A211602, A212258, A212259.
Cf. A248120.
T(2n,n) gives A344445.
T(n^2,n) gives A344532.

Programs

  • Maple
    G := 1/(1-sin(z))^t: Gser := simplify(series(G, z = 0, 16)): for n to 11 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n from 0 to 11 do seq(coeff(P[n], t, j), j = 1 .. n) end do; # yields sequence in triangular form
    # second Maple program:
    b:= proc(u, o) option remember; `if`(u+o=0, 1,
          add(b(o-1+j, u-j), j=1..u))
        end:
    g:= proc(n) option remember; expand(`if`(n=0, 1,
          add(g(n-j)*binomial(n-1, j-1)*x*b(j-1, 0), j=1..n)))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(g(n)):
    seq(T(n), n=1..12);  # Alois P. Heinz, May 21 2021
  • Mathematica
    rows = 12;
    a111[n_] := If[EvenQ[n], Abs[EulerE[n]], Abs[(2^(n+1)*(2^(n+1) - 1)* BernoulliB[n+1])/(n+1)]];
    BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
    B = BellMatrix[a111, rows];
    Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jul 23 2018, after Peter Luschny *)
  • Maxima
    T(n,m):=2*sum((stirling1(2*j-n,m)*(-1)^(m-n)*sum((2*i-2*j+n)^n*binomial(2*j-n,i)*(-1)^(j-i),i,0,(2*j-n)/2))/(2^(2*j-n)*(2*j-n)!),j,floor((n+m)/2),n); /* Vladimir Kruchinin, Mar 26 2013 */
    
  • Sage
    # uses[bell_matrix from A264428]
    # Adds a column 1,0,0,0, ... at the left side of the triangle.
    bell_matrix(lambda n: A000111(n), 10) # Peter Luschny, Jan 18 2016

Formula

E.g.f.: 1/(1-sin(z))^t.
The trivariate e.g.f. H(t,s,z) of the cycle-up-down permutations of {1,2,...,n} with respect to size (marked by z), number of cycles (marked by t), and number of fixed points (marked by x) is given by H(t,x,z)=exp((x-1)tz)/(1-sin z)^t.
T(n,m) = 2*sum(j=floor((n+m)/2)..n, (stirling1(2*j-n,m)*(-1)^(m-n)*sum(i=0..(2*j-n)/2, (2*i-2*j+n)^n*binomial(2*j-n,i)*(-1)^(j-i)))/(2^(2*j-n)*(2*j-n)!)). - Vladimir Kruchinin, Mar 26 2013

A248927 Triangle read by rows: T(n,k) are the coefficients of the Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its reciprocal, n >= 1, k = 1..A000041(n-1).

Original entry on oeis.org

1, 1, 2, 1, 6, 9, 1, 24, 72, 12, 16, 1, 120, 600, 300, 200, 50, 25, 1, 720, 5400, 5400, 2400, 450, 1800, 450, 60, 90, 36, 1, 5040, 52920, 88200, 29400, 22050, 44100, 7350, 4410, 2940, 4410, 882, 245, 147, 49, 1, 40320, 564480, 1411200, 376320, 705600, 940800
Offset: 1

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Author

Tom Copeland, Oct 16 2014

Keywords

Comments

Coefficients are listed in reverse graded colexicographic order (A228100). This is the reverse of Abramowitz and Stegun order (A036036).
Coefficients for Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its shifted reciprocal. Complementary to A134264 for formal power series. A refinement of A141618 with row sums A000272.
Given an invertible function f(t) analytic about t=0 with f(0)=0 and df(0)/dt not 0, form h(t) = t / f(t) and denote h_n = (n') as the coefficient of t^n/n! in h(t). Then the compositional inverse of f(t), g(t), as a formal Taylor series, or e.g.f., is given up to the first few orders by
g(t)/t = [ 1 (0') ]
+ [ 1 (0') (1') ] * t
+ [ 2 (0') (1')^2 + 1 (0')^2 (2') ] * t^2/2!
+ [ 6 (0') (1')^3 + 9 (0')^2 (1') (2') + 1 (0')^3 (3') ] * t^3/3!
+ [24 (0') (1')^4 + 72 (0')^2 (1')^2 (2') + (0')^3 [12 (2')^2
+ 16 (1') (3')] + (0')^4 (4')] * t^4/4!
+ [120 (0')(1')^5 + 600 (0')^2 (1')^3(2') + (0')^3 [300 (1')(2')^2 + 200 ( 1')^2(3')] + (0')^4 [50 (2')(3') + 25 (1')(4')] + (0')^5 (5')] * t^5/5! + [720 (0')(1')^6 + (0')^2 (1')^4(2')+(0')^3 [5400 (1')^2(2')^2 + 2400 (1')^3(3')] + (0')^4 [450 (2')^3+ 1800 (1')(2')(3') + 450( 1')^2(4')]+ (0')^5 [60 (3')^2 + 90 (2')(4') + 36 (1')(5')] + (0')^6 (6')] * t^6/6! + ...
..........
From Tom Copeland, Oct 28 2014: (Start)
Expressing g(t) as a Taylor series or formal e.g.f. in the indeterminates h_n generates a refinement of A055302, which enumerates the number of labeled root trees with n nodes and k leaves, with row sum A000169.
Operating with (1/n^2) d/d(1') = (1/n^2) d/d(h_1) on the n-th partition polynomial in square brackets above associated with t^n/n! generates the (n-1)-th partition polynomial.
Multiplying the n-th partition polynomial here by (n + 1) gives the (n + 1)-th partition polynomial of A248120. (End)
These are also the coefficients in the expansion of a series related to the Lagrange reversion theorem presented in Wikipedia of which the Lagrange inversion formula about the origin is a special case. Cf. Copeland link. - Tom Copeland, Nov 01 2016

Examples

			Triangle T(n,k) begins:
    1;
    1;
    2,    1;
    6,    9,    1;
   24,   72,   12,   16,   1;
  120,  600,  300,  200,  50,   25,   1;
  720, 5400, 5400, 2400, 450, 1800, 450, 60, 90, 36, 1;
  ...
For f(t) = e^t-1, h(t) = t/f(t) = t/(e^t-1), the e.g.f. for the Bernoulli numbers, and plugging the Bernoulli numbers into the Lagrange inversion formula gives g(t) = t - t^2/2 + t^3/3 + ... = log(1+t).
		

Crossrefs

Cf. A134264 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A036038.

Programs

  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!^2/((n-#v+1)!*prod(i=1, #S, my(x=S[i], c=#select(y->y==x, v)); x!^c*c!))}
    row(n)=[C(Vec(p)) | p<-Vecrev(partitions(n-1))]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 02 2022

Formula

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.
T(j,k) is [(j-1)!/j]* P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).
For example, from g(t) above, T(6,3) = [5!/6][6!/(3!*2!)]/(2!)^2 = 300 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = [5!/6][6!/4!]/(2!*3!) = 50.
If the initial factorial and final denominator are removed and the partitions reversed in order, A134264 is obtained, a refinement of the Narayana numbers.
For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums of A248927 are A000169(n)/n = n^(n-2) = A000272(n).
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/{d[x/[h_0+h_1*x+ ...]]/dx}. Then the partition polynomials above are given by (1/n)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t)= exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)). See A145271.
With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*W(y)d/dy] exp(t*y) eval. at y=0, the raising (creation) and lowering (annihilation) operators defined by R PS(n,t) = PS(n+1,t) and L PS(n,t)= n * PS(n-1,t) are R = t * W(d/dt) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).
Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
As noted in A248120 and A134264, this entry is given by the Hadamard product by partition of A134264 and A036038. For example, (1,4,2,6,1)*(1,4,6,12,24) = (1,16,12,72,24). - Tom Copeland, Nov 25 2016
T(n,k) = ((n-1)!)^2/((n-j)!*Product_{i>=1} s_i!*(i!)^s_i), where (1*s_1 + 2*s_2 + ... = n-1) is the k-th partition of n-1 and j = s_1 + s_2 ... is the number of parts. - Andrew Howroyd, Feb 02 2022

Extensions

Name edited and terms a(31) and beyond from Andrew Howroyd, Feb 02 2022

A231602 Triangular array read by rows: T(n,k) is the number of rooted labeled trees on n nodes that have exactly k nodes with outdegree = 1, n>=1, 0<=k<=n-1.

Original entry on oeis.org

1, 0, 2, 3, 0, 6, 4, 36, 0, 24, 65, 80, 360, 0, 120, 306, 1950, 1200, 3600, 0, 720, 4207, 12852, 40950, 16800, 37800, 0, 5040, 38424, 235592, 359856, 764400, 235200, 423360, 0, 40320, 573057, 2766528, 8481312, 8636544, 13759200, 3386880, 5080320, 0, 362880
Offset: 1

Views

Author

Geoffrey Critzer, Nov 11 2013

Keywords

Comments

T(n,k) is also the number of functions f:{1,2,...,n-1}->{1,2,...,n} that have exactly k elements whose preimage has cardinality = 1.
T(n,n-1) = n! = A000142(n).
Column k = 0 = A060356(n).
Row sums = n^(n-1) = A000169(n).
Refinement given by A248120. Sum coefficients of the partition polynomials with h_1 = (1') = t and all other h_n = (n') = 1 to obtain this entry. - Tom Copeland, Feb 01 2016

Examples

			1;
0, 2;
3, 0, 6;
4, 36, 0, 24;
65, 80, 360, 0, 120;
306, 1950, 1200, 3600, 0, 720;
4207, 12852, 40950, 16800, 37800, 0, 5040;
38424, 235592, 359856, 764400, 235200, 423360, 0, 40320;
....0..........0........
....|........./ \.......
....0........0...0......
.../ \.......|..........
..0   0......0..........
T(4,1) = 36.  Both of these graphs on 4 nodes have exactly 1 node that has outdegree = 1.  There are 12 + 24 = 36 labelings.
		

Crossrefs

Programs

  • Maple
    with(combinat): C:= binomial:
    b:= proc(t, i, u) option remember; `if`(t=0, 1,
          `if`(i<2, 0, b(t, i-1, u) +add(multinomial(t, t-i*j, i$j)
          *b(t-i*j, i-1, u-j)*u!/(u-j)!/j!, j=1..t/i)))
        end:
    T:= (n, k)-> C(n, k)*C(n-1, k)*k! *b(n-1-k$2, n-k):
    seq(seq(T(n, k), k=0..n-1), n=1..10);  # Alois P. Heinz, Nov 12 2013
  • Mathematica
    nn=8;Table[Table[Drop[Range[0,nn]!CoefficientList[Series[-ProductLog[x/(-1-x+x y)],{x,0,nn}],{x,y}],1][[r,c]],{c,1,r}],{r,1,nn}]//Grid

Formula

E.g.f. satisfies A(x,y) = y*x*A(x,y) + x*( exp(A(x,y)) - A(x,y) ).
Showing 1-7 of 7 results.