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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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A008292 Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 26, 66, 26, 1, 1, 57, 302, 302, 57, 1, 1, 120, 1191, 2416, 1191, 120, 1, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1
Offset: 1

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Author

N. J. A. Sloane, Mar 15 1996

Keywords

Comments

The indexing used here follows that given in the classic books by Riordan and Comtet. For two other versions see A173018 and A123125. - N. J. A. Sloane, Nov 21 2010
Coefficients of Eulerian polynomials. Number of permutations of n objects with k-1 rises. Number of increasing rooted trees with n+1 nodes and k leaves.
T(n,k) = number of permutations of [n] with k runs. T(n,k) = number of permutations of [n] requiring k readings (see the Knuth reference). T(n,k) = number of permutations of [n] having k distinct entries in its inversion table. - Emeric Deutsch, Jun 09 2004
T(n,k) = number of ways to write the Coxeter element s_{e1}s_{e1-e2}s_{e2-e3}s_{e3-e4}...s_{e_{n-1}-e_n} of the reflection group of type B_n, using s_{e_k} and as few reflections of the form s_{e_i+e_j}, where i = 1, 2, ..., n and j is not equal to i, as possible. - Pramook Khungurn (pramook(AT)mit.edu), Jul 07 2004
Subtriangle for k>=1 and n>=1 of triangle A123125. - Philippe Deléham, Oct 22 2006
T(n,k)/n! also represents the n-dimensional volume of the portion of the n-dimensional hypercube cut by the (n-1)-dimensional hyperplanes x_1 + x_2 + ... x_n = k, x_1 + x_2 + ... x_n = k-1; or, equivalently, it represents the probability that the sum of n independent random variables with uniform distribution between 0 and 1 is between k-1 and k. - Stefano Zunino, Oct 25 2006
[E(.,t)/(1-t)]^n = n!*Lag[n,-P(.,t)/(1-t)] and [-P(.,t)/(1-t)]^n = n!*Lag[n, E(.,t)/(1-t)] umbrally comprise a combinatorial Laguerre transform pair, where E(n,t) are the Eulerian polynomials and P(n,t) are the polynomials in A131758. - Tom Copeland, Sep 30 2007
From Tom Copeland, Oct 07 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1*x + (2+t)*x^2/2! + (6+6*t+t^2)*x^3/3! + ... gives row polynomials for A090582, the reverse f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra (Postnikov et al.).
G((t+1)*x, -1/(t+1)) = 1 + (1+t)*x + (1+3*t+2*t^2)*x^2/2! + ... gives row polynomials for A028246.
(End)
A subexceedant function f on [n] is a map f:[n] -> [n] such that 1 <= f(i) <= i for all i, 1 <= i <= n. T(n,k) equals the number of subexceedant functions f of [n] such that the image of f has cardinality k [Mantaci & Rakotondrajao]. Example T(3,2) = 4: if we identify a subexceedant function f with the word f(1)f(2)...f(n) then the subexceedant functions on [3] are 111, 112, 113, 121, 122 and 123 and four of these functions have an image set of cardinality 2. - Peter Bala, Oct 21 2008
Further to the comments of Tom Copeland above, the n-th row of this triangle is the h-vector of the simplicial complex dual to a permutohedron of type A_(n-1). The corresponding f-vectors are the rows of A019538. For example, 1 + 4*x + x^2 = y^2 + 6*y + 6 and 1 + 11*x + 11*x^2 + x^3 = y^3 + 14*y^2 + 36*y + 24, where x = y + 1, give [1,6,6] and [1,14,36,24] as the third and fourth rows of A019538. The Hilbert transform of this triangle (see A145905 for the definition) is A047969. See A060187 for the triangle of Eulerian numbers of type B (the h-vectors of the simplicial complexes dual to permutohedra of type B). See A066094 for the array of h-vectors of type D. For tables of restricted Eulerian numbers see A144696 - A144699. - Peter Bala, Oct 26 2008
For a natural refinement of A008292 with connections to compositional inversion and iterated derivatives, see A145271. - Tom Copeland, Nov 06 2008
The polynomials E(z,n) = numerator(Sum_{k>=1} (-1)^(n+1)*k^n*z^(k-1)) for n >=1 lead directly to the triangle of Eulerian numbers. - Johannes W. Meijer, May 24 2009
From Walther Janous (walther.janous(AT)tirol.com), Nov 01 2009: (Start)
The (Eulerian) polynomials e(n,x) = Sum_{k=0..n-1} T(n,k+1)*x^k turn out to be also the numerators of the closed-form expressions of the infinite sums:
S(p,x) = Sum_{j>=0} (j+1)^p*x^j, that is
S(p,x) = e(p,x)/(1-x)^(p+1), whenever |x| < 1 and p is a positive integer.
(Note the inconsistent use of T(n,k) in the section listing the formula section. I adhere tacitly to the first one.) (End)
If n is an odd prime, then all numbers of the (n-2)-th and (n-1)-th rows are in the progression k*n+1. - Vladimir Shevelev, Jul 01 2011
The Eulerian triangle is an element of the formula for the r-th successive summation of Sum_{k=1..n} k^j which appears to be Sum_{k=1..n} T(j,k-1) * binomial(j-k+n+r, j+r). - Gary Detlefs, Nov 11 2011
Li and Wong show that T(n,k) counts the combinatorially inequivalent star polygons with n+1 vertices and sum of angles (2*k-n-1)*Pi. An equivalent formulation is: define the total sign change S(p) of a permutation p in the symmetric group S_n to be equal to Sum_{i=1..n} sign(p(i)-p(i+1)), where we take p(n+1) = p(1). T(n,k) gives the number of permutations q in S_(n+1) with q(1) = 1 and S(q) = 2*k-n-1. For example, T(3,2) = 4 since in S_4 the permutations (1243), (1324), (1342) and (1423) have total sign change 0. - Peter Bala, Dec 27 2011
Xiong, Hall and Tsao refer to Riordan and mention that a traditional Eulerian number A(n,k) is the number of permutations of (1,2...n) with k weak exceedances. - Susanne Wienand, Aug 25 2014
Connections to algebraic geometry/topology and characteristic classes are discussed in the Buchstaber and Bunkova, the Copeland, the Hirzebruch, the Lenart and Zainoulline, the Losev and Manin, and the Sheppeard links; to the Grassmannian, in the Copeland, the Farber and Postnikov, the Sheppeard, and the Williams links; and to compositional inversion and differential operators, in the Copeland and the Parker links. - Tom Copeland, Oct 20 2015
The bivariate e.g.f. noted in the formulas is related to multiplying edges in certain graphs discussed in the Aluffi-Marcolli link. See p. 42. - Tom Copeland, Dec 18 2016
Distribution of left children in treeshelves is given by a shift of the Eulerian numbers. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. See A278677, A278678 or A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016
The row polynomial P(n, x) = Sum_{k=1..n} T(n, k)*x^k appears in the numerator of the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} P(n, x)/(1 - x)^(n+2) for n >= 0 (with 0^0=1). See also triangle A131689 with a Mar 31 2017 comment for a rewritten form. For the e.g.f see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
For relations to Ehrhart polynomials, volumes of polytopes, polylogarithms, the Todd operator, and other special functions, polynomials, and sequences, see A131758 and the references therein. - Tom Copeland, Jun 20 2017
For relations to values of the Riemann zeta function at integral arguments, see A131758 and the Dupont reference. - Tom Copeland, Mar 19 2018
Normalized volumes of the hypersimplices, attributed to Laplace. (Cf. the De Loera et al. reference, p. 327.) - Tom Copeland, Jun 25 2018

Examples

			The triangle T(n, k) begins:
n\k 1    2     3      4       5       6      7     8    9 10 ...
1:  1
2:  1    1
3:  1    4     1
4:  1   11    11      1
5:  1   26    66     26       1
6:  1   57   302    302      57       1
7:  1  120  1191   2416    1191     120      1
8:  1  247  4293  15619   15619    4293    247     1
9:  1  502 14608  88234  156190   88234  14608   502    1
10: 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
-----------------------------------------------------------------
E.g.f. = (y) * x^1 / 1! + (y + y^2) * x^2 / 2! + (y + 4*y^2 + y^3) * x^3 / 3! + ... - _Michael Somos_, Mar 17 2011
Let n=7. Then the following 2*7+1=15 consecutive terms are 1(mod 7): a(15+i), i=0..14. - _Vladimir Shevelev_, Jul 01 2011
Row 3: The plane increasing 0-1-2 trees on 3 vertices (with the number of colored vertices shown to the right of a vertex) are
.
.   1o (1+t)         1o t         1o t
.   |                / \          / \
.   |               /   \        /   \
.   2o (1+t)      2o     3o    3o    2o
.   |
.   |
.   3o
.
The total number of trees is (1+t)^2 + t + t = 1 + 4*t + t^2.
		

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 106.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 243.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 254; 2nd. ed., p. 268.[Worpitzky's identity (6.37)]
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1998, Vol. 3, p. 47 (exercise 5.1.4 Nr. 20) and p. 605 (solution).
  • Meng Li and Ron Goldman. "Limits of sums for binomial and Eulerian numbers and their associated distributions." Discrete Mathematics 343.7 (2020): 111870.
  • Anthony Mendes and Jeffrey Remmel, Generating functions from symmetric functions, Preliminary version of book, available from Jeffrey Remmel's home page http://math.ucsd.edu/~remmel/
  • K. Mittelstaedt, A stochastic approach to Eulerian numbers, Amer. Math. Mnthly, 127:7 (2020), 618-628.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
  • R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Figure M3416, Academic Press, 1995.
  • H. S. Wall, Analytic Theory of Continued Fractions, Chelsea, 1973, see p. 208.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 101.

Crossrefs

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->Sum([0..k],j->(-1)^j*(k-j)^n*Binomial(n+1,j))))); # Muniru A Asiru, Jun 29 2018
    
  • Haskell
    import Data.List (genericLength)
    a008292 n k = a008292_tabl !! (n-1) !! (k-1)
    a008292_row n = a008292_tabl !! (n-1)
    a008292_tabl = iterate f [1] where
       f xs = zipWith (+)
         (zipWith (*) ([0] ++ xs) (reverse ks)) (zipWith (*) (xs ++ [0]) ks)
         where ks = [1 .. 1 + genericLength xs]
    -- Reinhard Zumkeller, May 07 2013
    
  • Magma
    Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Eulerian(n,k): k in [0..n-1]]: n in [1..10]]; // G. C. Greubel, Apr 15 2019
  • Maple
    A008292 := proc(n,k) option remember; if k < 1 or k > n then 0; elif k = 1 or k = n then 1; else k*procname(n-1,k)+(n-k+1)*procname(n-1,k-1) ; end if; end proc:
  • Mathematica
    t[n_, k_] = Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];
    Flatten[Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, May 31 2011, after Michael Somos *)
    Flatten[Table[CoefficientList[(1-x)^(k+1)*PolyLog[-k, x]/x, x], {k, 1, 10}]] (* Vaclav Kotesovec, Aug 27 2015 *)
    Table[Tally[
       Count[#, x_ /; x > 0] & /@ (Differences /@
          Permutations[Range[n]])][[;; , 2]], {n, 10}] (* Li Han, Oct 11 2020 *)
  • PARI
    {T(n, k) = if( k<1 || k>n, 0, if( n==1, 1, k * T(n-1, k) + (n-k+1) * T(n-1, k-1)))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {T(n, k) = sum( j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {A(n,c)=c^(n+c-1)+sum(i=1,c-1,(-1)^i/i!*(c-i)^(n+c-1)*prod(j=1,i,n+c+1-j))}
    
  • Python
    from sympy import binomial
    def T(n, k): return sum([(-1)**j*(k - j)**n*binomial(n + 1, j) for j in range(k + 1)])
    for n in range(1, 11): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 08 2017
    
  • R
    T <- function(n, k) {
      S <- numeric()
      for (j in 0:k) S <- c(S, (-1)^j*(k-j)^n*choose(n+1, j))
      return(sum(S))
    }
    for (n in 1:10){
      for (k in 1:n) print(T(n,k))
    } # Indranil Ghosh, Nov 08 2017
    
  • Sage
    [[sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in (0..k)) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Feb 23 2019
    

Formula

T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.
T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).
Row sums = n! = A000142(n) unless n=0. - Michael Somos, Mar 17 2011
E.g.f. A(x, q) = Sum_{n>0} (Sum_{k=1..n} T(n, k) * q^k) * x^n / n! = q * ( e^(q*x) - e^x ) / ( q*e^x - e^(q*x) ) satisfies dA / dx = (A + 1) * (A + q). - Michael Somos, Mar 17 2011
For a column listing, n-th term: T(c, n) = c^(n+c-1) + Sum_{i=1..c-1} (-1)^i/i!*(c-i)^(n+c-1)*Product_{j=1..i} (n+c+1-j). - Randall L Rathbun, Jan 23 2002
From John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
Four characterizations of Eulerian numbers T(i, n):
1. T(0, n)=1 for n>=1, T(i, 1)=0 for i>=1, T(i, n) = (n-i)T(i-1, n-1) + (i+1)T(i, n-1).
2. T(i, n) = Sum_{j=0..i} (-1)^j*binomial(n+1,j)*(i-j+1)^n for n>=1, i>=0.
3. Let C_n be the unit cube in R^n with vertices (e_1, e_2, ..., e_n) where each e_i is 0 or 1 and all 2^n combinations are used. Then T(i, n)/n! is the volume of C_n between the hyperplanes x_1 + x_2 + ... + x_n = i and x_1 + x_2 + ... + x_n = i+1. Hence T(i, n)/n! is the probability that i <= X_1 + X_2 + ... + X_n < i+1 where the X_j are independent uniform [0, 1] distributions. - See Ehrenborg & Readdy reference.
4. Let f(i, n) = T(i, n)/n!. The f(i, n) are the unique coefficients so that (1/(r-1)^(n+1)) Sum_{i=0..n-1} f(i, n) r^{i+1} = Sum_{j>=0} (j^n)/(r^j) whenever n>=1 and abs(r)>1. (End)
O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x. - Vladeta Jovovic, Sep 02 2002
Triangle T(n, k), n>0 and k>0, read by rows; given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] (positive integers interspersed with 0's) where DELTA is Deléham's operator defined in A084938.
Sum_{k=1..n} T(n, k)*2^k = A000629(n). - Philippe Deléham, Jun 05 2004
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n} E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
From the relations between the h- and f-polynomials of permutohedra and reciprocals of e.g.f.s described in A049019: (t-1)((t-1)d/dx)^n 1/(t-exp(x)) evaluated at x=0 gives the n-th Eulerian row polynomial in t and the n-th row polynomial in (t-1) of A019538 and A090582. From the Comtet and Copeland references in A139605: ((t+exp(x)-1)d/dx)^(n+1) x gives pairs of the Eulerian polynomials in t as the coefficients of x^0 and x^1 in its Taylor series expansion in x. - Tom Copeland, Oct 05 2008
G.f: 1/(1-x/(1-x*y/1-2*x/(1-2*x*y/(1-3*x/(1-3*x*y/(1-... (continued fraction). - Paul Barry, Mar 24 2010
If n is odd prime, then the following consecutive 2*n+1 terms are 1 modulo n: a((n-1)*(n-2)/2+i), i=0..2*n. This chain of terms is maximal in the sense that neither the previous term nor the following one are 1 modulo n. - _Vladimir Shevelev, Jul 01 2011
From Peter Bala, Sep 29 2011: (Start)
For k = 0,1,2,... put G(k,x,t) := x -(1+2^k*t)*x^2/2 +(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 0 and for A008517 when k = 1.
The e.g.f. B(x,t) := compositional inverse with respect to x of G(0,x,t) = (exp(x)-exp(x*t))/(exp(x*t)-t*exp(x)) = x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... satisfies the autonomous differential equation dB/dx = (1+B)*(1+t*B) = 1 + (1+t)*B + t*B^2.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the Eulerian polynomials: A(n,t) counts plane increasing trees on n vertices where each vertex has outdegree <= 2, the vertices of outdegree 1 come in 1+t colors and the vertices of outdegree 2 come in t colors. An example is given below. Cf. A008517. Applying [Dominici, Theorem 4.1] gives the following method for calculating the Eulerian polynomials: Let f(x,t) = (1+x)*(1+t*x) and let D be the operator f(x,t)*d/dx. Then A(n+1,t) = D^n(f(x,t)) evaluated at x = 0.
(End)
With e.g.f. A(x,t) = G[x,(t-1)]-1 in Copeland's 2008 comment, the compositional inverse is Ainv(x,t) = log(t-(t-1)/(1+x))/(t-1). - Tom Copeland, Oct 11 2011
T(2*n+1,n+1) = (2*n+2)*T(2*n,n). (E.g., 66 = 6*11, 2416 = 8*302, ...) - Gary Detlefs, Nov 11 2011
E.g.f.: (1-y) / (1 - y*exp( (1-y)*x )). - Geoffrey Critzer, Nov 10 2012
From Peter Bala, Mar 12 2013: (Start)
Let {A(n,x)} n>=1 denote the sequence of Eulerian polynomials beginning [1, 1 + x, 1 + 4*x + x^2, ...]. Given two complex numbers a and b, the polynomial sequence defined by R(n,x) := (x+b)^n*A(n+1,(x+a)/(x+b)), n >= 0, satisfies the recurrence equation R(n+1,x) = d/dx((x+a)*(x+b)*R(n,x)). These polynomials give the row generating polynomials for several triangles in the database including A019538 (a = 0, b = 1), A156992 (a = 1, b = 1), A185421 (a = (1+i)/2, b = (1-i)/2), A185423 (a = exp(i*Pi/3), b = exp(-i*Pi/3)) and A185896 (a = i, b = -i).
(End)
E.g.f.: 1 + x/(T(0) - x*y), where T(k) = 1 + x*(y-1)/(1 + (k+1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 07 2013
From Tom Copeland, Sep 18 2014: (Start)
A) Bivariate e.g.f. A(x,a,b)= (e^(ax)-e^(bx))/(a*e^(bx)-b*e^(ax)) = x + (a+b)*x^2/2! + (a^2+4ab+b^2)*x^3/3! + (a^3+11a^2b+11ab^2+b^3)x^4/4! + ...
B) B(x,a,b)= log((1+ax)/(1+bx))/(a-b) = x - (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 - (a^3+a^2b+ab^2+b^3)x^4/4 + ... = log(1+u.*x), with (u.)^n = u_n = h_(n-1)(a,b) a complete homogeneous polynomial, is the compositional inverse of A(x,a,b) in x (see Drake, p. 56).
C) A(x) satisfies dA/dx = (1+a*A)(1+b*A) and can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova).
D) The bivariate Eulerian row polynomials are generated by the iterated derivative ((1+ax)(1+bx)d/dx)^n x evaluated at x=0 (see A145271).
E) A(x,a,b)= -(e^(-ax)-e^(-bx))/(a*e^(-ax)-b*e^(-bx)), A(x,-1,-1) = x/(1+x), and B(x,-1,-1) = x/(1-x).
F) FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = (x+y+(a+b)xy)/(1-ab*xy) is called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline. (End)
For x > 1, the n-th Eulerian polynomial A(n,x) = (x - 1)^n * log(x) * Integral_{u>=0} (ceiling(u))^n * x^(-u) du. - Peter Bala, Feb 06 2015
Sum_{j>=0} j^n/e^j, for n>=0, equals Sum_{k=1..n} T(n,k)e^k/(e-1)^(n+1), a rational function in the variable "e" which evaluates, approximately, to n! when e = A001113 = 2.71828... - Richard R. Forberg, Feb 15 2015
For a fixed k, T(n,k) ~ k^n, proved by induction. - Ran Pan, Oct 12 2015
From A145271, multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n (1+a*x)*(1+b*x) evaluated at x= 0, i.e., g_0 = 1, g_1 = (a+b), g_2 = 2ab, and g_n = 0 otherwise, to obtain the tridiagonal matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then the m-th bivariate row polynomial of this entry is P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^(m-1) (1, a+b, 2ab, 0, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. Also, P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^m (0, 1, 0, ...)^T. - Tom Copeland, Aug 02 2016
Cumulatively summing a row generates the n starting terms of the n-th differences of the n-th powers. Applying the finite difference method to x^n, these terms correspond to those before constant n! in the lowest difference row. E.g., T(4,k) is summed as 0+1=1, 1+11=12, 12+11=23, 23+1=4!. See A101101, A101104, A101100, A179457. - Andy Nicol, May 25 2024

Extensions

Thanks to Michael Somos for additional comments.
Further comments from Christian G. Bower, May 12 2000

A130534 Triangle T(n,k), 0 <= k <= n, read by rows, giving coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in increasing powers of x. T(n,k) is also the unsigned Stirling number |s(n+1, k+1)|, denoting the number of permutations on n+1 elements that contain exactly k+1 cycles.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 6, 1, 24, 50, 35, 10, 1, 120, 274, 225, 85, 15, 1, 720, 1764, 1624, 735, 175, 21, 1, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 09 2007

Keywords

Comments

This triangle is an unsigned version of the triangle of Stirling numbers of the first kind, A008275, which is the main entry for these numbers. - N. J. A. Sloane, Jan 25 2011
Or, triangle T(n,k), 0 <= k <= n, read by rows given by [1,1,2,2,3,3,4,4,5,5,6,6,...] DELTA [1,0,1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938.
Reversal of A094638.
Equals A132393*A007318, as infinite lower triangular matrices. - Philippe Deléham, Nov 13 2007
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the exponential integrals E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + n*(n+1)/x^2 - n*(n+1)*(n+2)/x^3 + ...), see Abramowitz and Stegun. This formula follows from the general formula for the asymptotic expansion, see A163932. We rewrite E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) and observe that the T(n,m) are the polynomials coefficients in the denominators. Looking at the a(n,m) formula of A028421, A163932 and A163934, and shifting the offset given above to 1, we can write T(n-1,m-1) = a(n,m) = (-1)^(n+m)*Stirling1(n,m), see the Maple program.
The asymptotic expansion leads for values of n from one to eleven to known sequences, see the cross-references. With these sequences one can form the triangles A008279 (right-hand columns) and A094587 (left-hand columns).
See A163936 for information about the o.g.f.s. of the right-hand columns of this triangle.
(End)
The number of elements greater than i to the left of i in a permutation gives the i-th element of the inversion vector. (Skiena-Pemmaraju 2003, p. 69.) T(n,k) is the number of n-permutations that have exactly k 0's in their inversion vector. See evidence in Mathematica code below. - Geoffrey Critzer, May 07 2010
T(n,k) counts the rooted trees with k+1 trunks in forests of "naturally grown" rooted trees with n+2 nodes. This corresponds to sums of coefficients of iterated derivatives representing vectors, Lie derivatives, or infinitesimal generators for flow fields and formal group laws. Cf. links in A139605. - Tom Copeland, Mar 23 2014
A refinement is A036039. - Tom Copeland, Mar 30 2014
From Tom Copeland, Apr 05 2014: (Start)
With initial n=1 and row polynomials of T as p(n,x)=x(x+1)...(x+n-1), the powers of x correspond to the number of trunks of the rooted trees of the "naturally-grown" forest referred to above. With each trunk allowed m colors, p(n,m) gives the number of such non-plane colored trees for the forest with each tree having n+1 vertices.
p(2,m) = m + m^2 = A002378(m) = 2*A000217(m) = 2*(first subdiag of |A238363|).
p(3,m) = 2m + 3m^2 + m^3 = A007531(m+2) = 3*A007290(m+2) = 3*(second subdiag A238363).
p(4,m) = 6m + 11m^2 + 6m^3 + m^4 = A052762(m+3) = 4*A033487(m) = 4*(third subdiag).
From the Joni et al. link, p(n,m) also represents the disposition of n distinguishable flags on m distinguishable flagpoles.
The chromatic polynomial for the complete graph K_n is the falling factorial, which encodes the colorings of the n vertices of K_n and gives a shifted version of p(n,m).
E.g.f. for the row polynomials: (1-y)^(-x).
(End)
A relation to derivatives of the determinant |V(n)| of the n X n Vandermonde matrix V(n) in the indeterminates c(1) thru c(n):
|V(n)| = Product_{1<=jTom Copeland, Apr 10 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A094587 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well defined). See the Example section. (End)
For the relation of this rising factorial to the moments of Viennot's Laguerre stories, see the Hetyei link, p. 4. - Tom Copeland, Oct 01 2015
Can also be seen as the Bell transform of n! without column 0 (and shifted enumeration). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle  T(n,k) begins:
n\k         0        1        2       3       4      5      6     7    8  9 10
n=0:        1
n=1:        1        1
n=2:        2        3        1
n=3:        6       11        6       1
n=4:       24       50       35      10       1
n=5:      120      274      225      85      15      1
n=6:      720     1764     1624     735     175     21      1
n=7:     5040    13068    13132    6769    1960    322     28     1
n=8:    40320   109584   118124   67284   22449   4536    546    36    1
n=9:   362880  1026576  1172700  723680  269325  63273   9450   870   45  1
n=10: 3628800 10628640 12753576 8409500 3416930 902055 157773 18150 1320 55  1
[Reformatted and extended by _Wolfdieter Lang_, Feb 05 2013]
T(3,2) = 6 because there are 6 permutations of {1,2,3,4} that have exactly 2 0's in their inversion vector: {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {2, 1, 3, 4},{2, 3, 1, 4}, {2, 3, 4, 1}. The respective inversion vectors are {0, 0, 1}, {0, 1, 0}, {0, 2, 0}, {1, 0, 0}, {2, 0, 0}, {3, 0, 0}. - _Geoffrey Critzer_, May 07 2010
T(3,1)=11 since there are exactly 11 permutations of {1,2,3,4} with exactly 2 cycles, namely, (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(143), (12)(34), (13)(24), and (14)(23). - _Dennis P. Walsh_, Jan 25 2011
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
  / 1          \/1        \/1        \      / 1           \
  | 1  1       ||0 1      ||0 1      |      | 1  1        |
  | 2  2  1    ||0 1 1    ||0 0 1    |... = | 2  3  1     |
  | 6  6  3 1  ||0 2 2 1  ||0 0 1 1  |      | 6 11  6  1  |
  |24 24 12 4 1||0 6 6 3 1||0 0 2 2 1|      |24 50 35 10 1|
  |...         ||...      ||...      |      |...          |
(End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 93-94.
  • Sriram Pemmaraju and Steven Skiena, Computational Discrete Mathematics, Cambridge University Press, 2003, pp. 69-71. [Geoffrey Critzer, May 07 2010]

Crossrefs

See A008275, which is the main entry for these numbers; A094638 (reversed rows).
From Johannes W. Meijer, Oct 07 2009: (Start)
Row sums equal A000142.
The asymptotic expansions lead to A000142 (n=1), A000142(n=2; minus a(0)), A001710 (n=3), A001715 (n=4), A001720 (n=5), A001725 (n=6), A001730 (n=7), A049388 (n=8), A049389 (n=9), A049398 (n=10), A051431 (n=11), A008279 and A094587.
Cf. A163931 (E(x,m,n)), A028421 (m=2), A163932 (m=3), A163934 (m=4), A163936.
(End)
Cf. A136662.

Programs

  • Haskell
    a130534 n k = a130534_tabl !! n !! k
    a130534_row n = a130534_tabl !! n
    a130534_tabl = map (map abs) a008275_tabl
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    with(combinat): A130534 := proc(n,m): (-1)^(n+m)*stirling1(n+1,m+1) end proc: seq(seq(A130534(n,m), m=0..n), n=0..10); # Johannes W. Meijer, Oct 07 2009, revised Sep 11 2012
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0, ..) as column 0 (and shifts the enumeration).
    BellMatrix(n -> n!, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    Table[Table[ Length[Select[Map[ToInversionVector, Permutations[m]], Count[ #, 0] == n &]], {n, 0, m - 1}], {m, 0, 8}] // Grid (* Geoffrey Critzer, May 07 2010 *)
    rows = 10;
    t = Range[0, rows]!;
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)

Formula

T(0,0) = 1, T(n,k) = 0 if k > n or if n < 0, T(n,k) = T(n-1,k-1) + n*T(n-1,k). T(n,0) = n! = A000142(n). T(2*n,n) = A129505(n+1). Sum_{k=0..n} T(n,k) = (n+1)! = A000142(n+1). Sum_{k=0..n} T(n,k)^2 = A047796(n+1). T(n,k) = |Stirling1(n+1,k+1)|, see A008275. (x+1)(x+2)...(x+n) = Sum_{k=0..n} T(n,k)*x^k. [Corrected by Arie Bos, Jul 11 2008]
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. - Philippe Deléham, Nov 13 2007
For k=1..n, let A={a_1,a_2,...,a_k} denote a size-k subset of {1,2,...,n}. Then T(n,n-k) = Sum(Product_{i=1..k} a_i) where the sum is over all subsets A. For example, T(4,1)=50 since 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50. - Dennis P. Walsh, Jan 25 2011
The preceding formula means T(n,k) = sigma_{n-k}(1,2,3,..,n) with the (n-k)-th elementary symmetric function sigma with the indeterminates chosen as 1,2,...,n. See the Oct 24 2011 comment in A094638 with sigma called there a. - Wolfdieter Lang, Feb 06 2013
From Gary W. Adamson, Jul 08 2011: (Start)
n-th row of the triangle = top row of M^n, where M is the production matrix:
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 4, 6, 4, 1;
... (End)
Exponential Riordan array [1/(1 - x), log(1/(1 - x))]. Recurrence: T(n+1,k+1) = Sum_{i=0..n-k} (n + 1)!/(n + 1 - i)!*T(n-i,k). - Peter Bala, Jul 21 2014

A145271 Coefficients for expansion of (g(x)d/dx)^n g(x); refined Eulerian numbers for calculating compositional inverse of h(x) = (d/dx)^(-1) 1/g(x); iterated derivatives as infinitesimal generators of flows.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 11, 4, 7, 1, 1, 26, 34, 32, 15, 11, 1, 1, 57, 180, 122, 34, 192, 76, 15, 26, 16, 1, 1, 120, 768, 423, 496, 1494, 426, 294, 267, 474, 156, 56, 42, 22, 1, 1, 247, 2904, 1389, 4288, 9204, 2127, 496, 5946, 2829, 5142, 1206, 855, 768, 1344, 1038, 288, 56, 98, 64, 29, 1
Offset: 0

Views

Author

Tom Copeland, Oct 06 2008

Keywords

Comments

For more detail, including connections to Legendre transformations, rooted trees, A139605, A139002 and A074060, see Mathemagical Forests p. 9.
For connections to the h-polynomials associated to the refined f-polynomials of permutohedra see my comments in A008292 and A049019.
From Tom Copeland, Oct 14 2011: (Start)
Given analytic functions F(x) and FI(x) such that F(FI(x))=FI(F(x))=x about 0, i.e., they are compositional inverses of each other, then, with g(x) = 1/dFI(x)/dx, a flow function W(s,x) can be defined with the following relations:
W(s,x) = exp(s g(x)d/dx)x = F(s+FI(x)) ,
W(s,0) = F(s) ,
W(0,x) = x ,
dW(0,x)/ds = g(x) = F'[FI(x)] , implying
dW(0,F(x))/ds = g(F(x)) = F'(x) , and
W(s,W(r,x)) = F(s+FI(F(r+FI(x)))) = F(s+r+FI(x)) = W(s+r,x) . (See MF link below.) (End)
dW(s,x)/ds - g(x)dW(s,x)/dx = 0, so (1,-g(x)) are the components of a vector orthogonal to the gradient of W and, therefore, tangent to the contour of W, at (s,x) . - Tom Copeland, Oct 26 2011
Though A139605 contains A145271, the op. of A145271 contains that of A139605 in the sense that exp(s g(x)d/dx) w(x) = w(F(s+FI(x))) = exp((exp(s g(x)d/dx)x)d/du)w(u) evaluated at u=0. This is reflected in the fact that the forest of rooted trees assoc. to (g(x)d/dx)^n, FOR_n, can be generated by removing the single trunk of the planted rooted trees of FOR_(n+1). - Tom Copeland, Nov 29 2011
Related to formal group laws for elliptic curves (see Hoffman). - Tom Copeland, Feb 24 2012
The functional equation W(s,x) = F(s+FI(x)), or a restriction of it, is sometimes called the Abel equation or Abel's functional equation (see Houzel and Wikipedia) and is related to Schröder's functional equation and Koenigs functions for compositional iterates (Alexander, Goryainov and Kudryavtseva). - Tom Copeland, Apr 04 2012
g(W(s,x)) = F'(s + FI(x)) = dW(s,x)/ds = g(x) dW(s,x)/dx, connecting the operators here to presentations of the Koenigs / Königs function and Loewner / Löwner evolution equations of the Contreras et al. papers. - Tom Copeland, Jun 03 2018
The autonomous differential equation above also appears with a change in variable of the form x = log(u) in the renormalization group equation, or Beta function. See Wikipedia, Zinn-Justin equations 2.10 and 3.11, and Krajewski and Martinetti equation 21. - Tom Copeland, Jul 23 2020
A variant of these partition polynomials appears on p. 83 of Petreolle et al. with the indeterminates e_n there related to those given in the examples below by e_n = n!*(n'). The coefficients are interpreted as enumerating certain types of trees. See also A190015. - Tom Copeland, Oct 03 2022

Examples

			From _Tom Copeland_, Sep 19 2014: (Start)
Let h(x) = log((1+a*x)/(1+b*x))/(a-b); then, g(x) = 1/(dh(x)/dx) = (1+ax)(1+bx), so (0')=1, (1')=a+b, (2')=2ab, evaluated at x=0, and higher order derivatives of g(x) vanish. Therefore, evaluated at x=0,
R^0 g(x) =  1
R^1 g(x) =  a+b
R^2 g(x) = (a+b)^2 + 2ab = a^2 + 4 ab + b^2
R^3 g(x) = (a+b)^3 + 4*(a+b)*2ab = a^3 + 11 a^2*b + 11 ab^2 + b^3
R^4 g(x) = (a+b)^4 + 11*(a+b)^2*2ab + 4*(2ab)^2
         =  a^4 + 26 a^3*b + 66 a^2*b^2 + 26 ab^3 + b^4,
etc., and these bivariate Eulerian polynomials (A008292) are the first few coefficients of h^(-1)(x) = (e^(ax) - e^(bx))/(a*e^(bx) - b*e^(ax)), the inverse of h(x). (End)
Triangle starts:
  1;
  1;
  1,   1;
  1,   4,    1;
  1,  11,    4,    7,    1;
  1,  26,   34,   32,   15,   11,    1;
  1,  57,  180,  122,   34,  192,   76,  15,   26,   16,    1;
  1, 120,  768,  423,  496, 1494,  426, 294,  267,  474,  156,   56,  42,  22,    1;
  1, 247, 2904, 1389, 4288, 9204, 2127, 496, 5946, 2829, 5142, 1206, 855, 768, 1344, 1038, 288, 56, 98, 64, 29, 1;
		

References

  • D. S. Alexander, A History of Complex Dynamics: From Schröder to Fatou to Julia, Friedrich Vieweg & Sohn, 1994.
  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015.

Crossrefs

Cf. (A133437, A086810, A181289) = (LIF, reduced LIF, associated g(x)), where LIF is a Lagrange inversion formula. Similarly for (A134264, A001263, A119900), (A134685, A134991, A019538), (A133932, A111999, A007318).
Second column is A000295, subdiagonal is A000124, row sums are A000142, row lengths are A000041. - Peter Luschny, Jul 21 2016

Programs

  • Maple
    with(LinearAlgebra): with(ListTools):
    A145271_row := proc(n) local b, M, V, U, G, R, T;
    if n < 2 then return 1 fi;
    b := (n,k) -> `if`(k=1 or k>n+1,0,binomial(n-1,k-2)*g[n-k+1]);
    M := n -> Matrix(n, b):
    V := n -> Vector[row]([1, seq(0,i=2..n)]):
    U := n -> VectorMatrixMultiply(V(n), M(n)^(n-1)):
    G := n -> Vector([seq(g[i], i=0..n-1)]);
    R := n -> VectorMatrixMultiply(U(n), G(n)):
    T := Reverse([op(sort(expand(R(n+1))))]);
    seq(subs({seq(g[i]=1, i=0..n)},T[j]),j=1..nops(T)) end:
    for n from 0 to 9 do A145271_row(n) od; # Peter Luschny, Jul 20 2016

Formula

Let R = g(x)d/dx; then
R^0 g(x) = 1 (0')^1
R^1 g(x) = 1 (0')^1 (1')^1
R^2 g(x) = 1 (0')^1 (1')^2 + 1 (0')^2 (2')^1
R^3 g(x) = 1 (0')^1 (1')^3 + 4 (0')^2 (1')^1 (2')^1 + 1 (0')^3 (3')^1
R^4 g(x) = 1 (0')^1 (1')^4 + 11 (0')^2 (1')^2 (2')^1 + 4 (0')^3 (2')^2 + 7 (0')^3 (1')^1 (3')^1 + 1 (0')^4 (4')^1
R^5 g(x) = 1 (0') (1')^5 + 26 (0')^2 (1')^3 (2') + (0')^3 [34 (1') (2')^2 + 32 (1')^2 (3')] + (0')^4 [ 15 (2') (3') + 11 (1') (4')] + (0')^5 (5')
R^6 g(x) = 1 (0') (1')^6 + 57 (0')^2 (1')^4 (2') + (0')^3 [180 (1')^2 (2')^2 + 122 (1')^3 (3')] + (0')^4 [ 34 (2')^3 + 192 (1') (2') (3') + 76 (1')^2 (4')] + (0')^5 [15 (3')^2 + 26 (2') (4') + 16 (1') (5')] + (0')^6 (6')
where (j')^k = ((d/dx)^j g(x))^k. And R^(n-1) g(x) evaluated at x=0 is the n-th Taylor series coefficient of the compositional inverse of h(x) = (d/dx)^(-1) 1/g(x), with the integral from 0 to x.
The partitions are in reverse order to those in Abramowitz and Stegun p. 831. Summing over coefficients with like powers of (0') gives A008292.
Confer A190015 for another way to compute numbers for the array for each partition. - Tom Copeland, Oct 17 2014
Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n g(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then R^n g(x) = (1, 0, 0, 0, ...) [VP * S]^n (g_0, g_1, g_2, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. - Tom Copeland, Feb 10 2016 (An evaluation removed by author on Jul 19 2016. Cf. A139605 and A134685.)
Also, R^n g(x) = (1, 0, 0, 0, ...) [VP * S]^(n+1) (0, 1, 0, ...)^T in agreement with A139605. - Tom Copeland, Jul 21 2016
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the cycle index polynomials of A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences". - Tom Copeland, Feb 06 2018
A formula for computing the polynomials of each row of this matrix is presented as T_{n,1} on p. 196 of the Ihara reference in A139605. - Tom Copeland, Mar 25 2020
Indeterminate substitutions as illustrated in A356145 lead to [E] = [L][P] = [P][E]^(-1)[P] = [P][RT] and [E]^(-1) = [P][L] = [P][E][P] = [RT][P], where [E] contains the refined Eulerian partition polynomials of this entry; [E]^(-1), A356145, the inverse set to [E]; [P], the permutahedra polynomials of A133314; [L], the classic Lagrange inversion polynomials of A134685; and [RT], the reciprocal tangent polynomials of A356144. Since [L]^2 = [P]^2 = [RT]^2 = [I], the substitutional identity, [L] = [E][P] = [P][E]^(-1) = [RT][P], [RT] = [E]^(-1)[P] = [P][L][P] = [P][E], and [P] = [L][E] = [E][RT] = [E]^(-1)[L] = [RT][E]^(-1). - Tom Copeland, Oct 05 2022

Extensions

Title amplified by Tom Copeland, Mar 17 2014
R^5 and R^6 formulas and terms a(19)-a(29) added by Tom Copeland, Jul 11 2016
More terms from Peter Luschny, Jul 20 2016

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

A133437 Irregular triangle of coefficients of a partition transform for direct Lagrange inversion of an o.g.f., complementary to A134685 for an e.g.f.; normalized by the factorials, these are signed, refined face polynomials of the associahedra.

Original entry on oeis.org

1, -2, 12, -6, -120, 120, -24, 1680, -2520, 360, 720, -120, -30240, 60480, -20160, -20160, 5040, 5040, -720, 665280, -1663200, 907200, 604800, -60480, -362880, -181440, 20160, 40320, 40320, -5040, -17297280, 51891840, -39916800, -19958400, 6652800, 19958400, 6652800, -1814400, -1814400, -3628800, -1814400, 362880, 362880, 362880, -40320
Offset: 1

Views

Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = u(t) - u(0) = Sum_{n>=1} u_n * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -2 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 12 (2')^2 - 6 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -120 (2')^3 + 120 (1')(2')(3') - 24 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 1680 (2')^4 - 2520 (1') (2')^2 (3') + 360 (1')^2 (3')^2 + 720 (1')^2 (2') (4') - 120 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -30240 (2')^5 + 60480 (1') (2')^3 (3') - 20160 (1')^2 (2') (3')^2 - 20160 (1')^2 (2')^2 (4') + 5040 (1')^3 (3')(4') + 5040 (1')^3 (2')(5') - 720 (1')^4 (6') ] * t^6 / 6!
P(7,t) = (1')^(-13) * [ 665280 (2')^6 - 1663200 (1')(2')^4(3') + (1')^2 [907200 (2')^2(3')^2 + 604800 (2')^3(4')] - (1')^3 [60480 (3')^3 + 362880 (2')(3')(4') + 181440 (2')^2(5')] + (1')^4 [20160 (4')^2 + 40320 (3')(5') + 40320 (2')(6')] - 5040 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -17297280 (2')^7 + 51891840 (1')(2')^5(3') - (1')^2 [39916800 (2')^3(3')^2 + 19958400 (2')^4(4')] + (1')^3 [6652800 (2')(3')^3 + 19958400 (2')^2(3')(4') + 6652800 (2')^3(5')] - (1')^4 [1814400 (3')^2(4') + 1814400 (2')(4')^2 + 3628800 (2')(3')(5') + 1814400 (2')^2(6')] + (1')^5 [362880 (4')(5') + 362880 (3')(6') + 362880 (2')(7')] - 40320 (1')^6(8')] * t^8 / 8!
...
See A134685 for more information.
A111785 is obtained from A133437 by dividing through the bracketed terms of the P(n,t) by n! and unsigned A111785 is a refinement of A033282 and A126216. - Tom Copeland, Sep 28 2008
For relation to the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see the Loday and McCammond links. E.g., P(5,t) = (1')^(-9) * [ 14 (2')^4 - 21 (1') (2')^2 (3') + 6 (1')^2 (2') (4')+ 3 (1')^2 (3')^2 - 1 (1')^3 (5') ] * t^5 is related to the 3-D associahedron with 14 vertices (0-D faces), 21 edges (1-D faces), 6 pentagons (2-D faces), 3 rectangles (2-D faces), 1 3-D polytope (3-D faces). Summing over faces of the same dimension gives A033282 or A126216. - Tom Copeland, Sep 29 2008
A relation between this Lagrange inversion for an o.g.f. and partition polynomials formed from the "refined Lah numbers" A130561 is presented in the link "Lagrange a la Lah" along with umbral binary tree representations.
With f(x,t) = x + t*Sum_{n>=2} u_n*x^n, the compositional inverse in x is related to the velocity profile of particles governed by the inviscid Burgers's, or Hopf, eqn. See A001764 and A086810. - Tom Copeland, Feb 15 2014
Newton was aware of this power series expansion for series reversion. See the Ferraro reference p. 75 eqn. 52. - Tom Copeland, Jan 22 2017
The coefficients of the partition polynomials divided by the associated factorial enumerate the faces of the convex, bounded polytopes called the associahedra, and the absolute value of the sum of the renormalized coefficients gives the Euler characteristic of unity for each polytope; i.e., the absolute value of the sum of each row of the array is either n! (unnormalized) or unity (normalized). In addition, the signs of the faces alternate with dimension, and the coefficients of faces with the same dimension for each polytope have the same sign. - Tom Copeland, Nov 13 2019
With u_1 = 1 and the other u_n replaced by suitably signed partition polynomials of A263633, the partition polynomials enumerating the refined noncrossing partitions of A134264 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Nov 16 2019
Relations between associahedra and oriented n-simplices are presented by Halvorson and by Street. - Tom Copeland, Dec 08 2019
Let f(x;t,n) = x - t * x^(n+1), giving u_1 = (1') = 1 and u_(n+1) = (n+1) = -t. Then inverting in x with t a parameter gives 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). Comparing this with the same result in A134264 gives relations between the faces of associahedra and noncrossing partitions (and other combinatorial constructs related to these inversion formulas and those listed in A145271). - Tom Copeland, Jan 27 2020
From Tom Copeland, Mar 24 2020: (Start)
There is a mapping between the faces of K_n, the associahedron of dimension (n-1), and polygon dissections. The dissecting noncrossing diagonals (i.e., nonintersecting in the interior) form subpolygons. Assign the indeterminate x_k to a subpolygon where k = (number of vertices of the subpolygon) - 1. Multiply the x_k together to form the monomials for the inversion formula.
For the 3-dimensional associahedron K_4, the fundamental polygon is the hexagon, which can be dissected into pentagons, associated to x_4; tetragons, to x_3; and triangles, to x_2; for example, there are six distinguished partitions of the hexagon into one triangle and one pentagon, sharing two vertices, associated to the monomial 6 * x_2 * x_4 since the unshared vertex of the triangle can be moved consecutively from one vertex of the hexagon to the next. This term corresponds to 720 (1')^2 (2') (4') / 5! in P(5,t) above, denumerating the six pentagonal facets of K_4. (End)

References

  • G. Ferraro, The Rise and Development of the Theory of Series up to the Early 1820s, Springer Science and Business Media, 2007.
  • H. Halvorson (editor), Deep Beauty: Understanding the Quantum World Through Innovation, Cambridge Univ. Press, 2011.
  • H. Turnbull (editor), The Correspondence of Isaac Newton Vol. II 1676-1687, Cambridge Univ. Press, 1960, p. 147.

Crossrefs

Cf. A145271, (A086810, A181289) = (reduced array, associated g(x)).

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k, {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [ (e(2))! * (e(3))! * ... * (e(n))! ].
From Tom Copeland, Sep 06 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u./(1-u.t)^2]
= 1/((u_1) + 2*(u_2)*t + 3*(u_3)*t^2 + 4*(u_4)*t^3 + ...),
where Ev denotes umbral evaluation.
Then for the partition polynomials of A133437,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (End)
From Tom Copeland, Oct 20 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)] = exp[x*h(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*h(d/dt) = t* 1/[(u_1) + 2*(u_2)*d/dt + 3*(u_3)*(d/dt)^2 + ...] and
L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2 + (u_3)*(d/dt)^3 + ....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x + u_3 * x^2 + ... + u_n * x^(n-1)]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / t^n, i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 6 u2 u4 + 3 u3^2, b3 = -21 u2^2 u3, and b4 = 14 u2^4.
The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. A086810) implies that, for n > 2, PB(n, 0 * b1, 1 * b2, ..., (K-1) * bK, ...) = [(n+1)/2] * Sum_{k = 2..n-1} PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 14 u2^4 - 2 * 21 u2^2 u3 + 1 * 6 u2 u4 + 1 * 3 u3^2 - 0 * u5 = 42 u2^4 - 42 u2^2 u3 + 6 u2 u4 + 3 u3^2 = 3 * [2 * PS(2,1,u2) * PS(4,1,u2,...,u4) + PS(3,1,u2,u3)^2] = 3 * [ 2 * (-u2) (-5 u2^3 + 5 u2 u3 - u4) + (2 u2^2 - u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
From Tom Copeland, Sep 22 2016: (Start)
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix A007318 by f_m = m!*u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) f_{n+1-k}, or equivalently multiply the diagonals of A132159 by u_m. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314.
Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ...)^T * t^n/n! in agreement with A139605. (End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the refined Lah polynomials of A130561 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018
The derivative of the partition polynomials of A350499 with respect to a distinguished indeterminate give polynomials proportional to those of this entry. The connection of this derivative relation to the inviscid Burgers-Hopf evolution equation is given in a reference for that entry. - Tom Copeland, Feb 19 2022

Extensions

Missing coefficient in P(6,t) replaced by Tom Copeland, Nov 06 2008
P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016
Title modified by Tom Copeland, Jan 13 2020
Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 07 2024

A134685 Irregular triangle read by rows: coefficients C(j,k) of a partition transform for direct Lagrange inversion.

Original entry on oeis.org

1, -1, 3, -1, -15, 10, -1, 105, -105, 10, 15, -1, -945, 1260, -280, -210, 35, 21, -1, 10395, -17325, 6300, 3150, -280, -1260, -378, 35, 56, 28, -1, -135135, 270270, -138600, -51975, 15400, 34650, 6930, -2100, -1575, -2520, -630, 126, 84, 36, -1
Offset: 1

Views

Author

Tom Copeland, Jan 26 2008, Sep 13 2008

Keywords

Comments

Let f(t) = u(t) - u(0) = Ev[exp(u.* t) - u(0)] = log{Ev[(exp(z.* t))/z_0]} = Ev[-log(1- a.* t)], where the operator Ev denotes umbral evaluation of the umbral variables u., z. or a., e.g., Ev[a.^n + a.^m] = a_n + a_m . The relation between z_n and u_n is given in reference in A127671 and u_n = (n-1)! * a_n .
If u_1 is not equal to 0, then the compositional inverse for these expressions is given by g(t) = Sum_{j>=1} P(j,t) where, with u_n denoted by (n') for brevity,
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -(2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 10 (1')(2')(3') - (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 105 (1') (2')^2 (3') + 10 (1')^2 (3')^2 + 15 (1')^2 (2') (4') - (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 1260 (1') (2')^3 (3') - 280 (1')^2 (2') (3')^2 - 210 (1')^2 (2')^2 (4') + 35 (1')^3 (3')(4') + 21 (1')^3 (2')(5') - (1')^4 (6') ] * t^6 / 6!
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 17325 (1') (2')^4 (3') + (1')^2 [ 6300 (2')^2 (3')^2 + 3150 (2')^3 (4')] - (1')^3 [280 (3')^3 + 1260 (2')(3')(4') + 378 (2')^2(5')] + (1')^4 [35 (4')^2 + 56 (3')(5') + 28 (2')(6')] - (1')^5 (7') ] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 270270 (1') (2')^5 (3') - (1')^2 [ 138600 (2')^3 (3')^2 + 51975 (2')^4 (4')] + (1')^3 [15400 (2')(3')^3 + 34650 (2')^2(3')(4') + 6930 (2')^3(5')] - (1')^4 [2100 (3')^2(4') + 1575 (2')(4')^2 + 2520 (2')(3')(5') + 630 (2')^2(6') ] + (1')^5 [126 (4')(5') + 84 (3')(6') + 36 (2')(7')] - (1')^6 (8') ] * t^8 / 8!
...
Substituting ((m-1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n-1 listed in Abramowitz and Stegun on page 831 (in the reversed order) and the number of partitions in P(n,t) is given by A000041(n-1).
Combinatorial interpretations are given in the link.
From Tom Copeland, Jul 10 2018: (Start)
Coefficients occurring in prolongation for the special Euclidean group SE(2) and special affine group SA(2) in the Olver presentation on moving frames (MFP) in slides 33 and 42. These are a result of applying an iterated derivative of the form h(x)d/dx = d/dy as in this entry (more generally as g(x) d/dx as discussed in A145271). See also p. 6 of Olver's paper on contact forms, but note that the 12 should be a 15 in the formula for the compositional inverse of S(t).
Change variables in the MFP to obtain connections to the partition polynomials Prt_n = n! * P(n,1) above. Let delta and beta in the formulas for the equi-affine curves in MFP be L and B, respectively, and D_y = (1/(L-B*u_x)) d/dx = (1/w_x) d/dx. Then v_(yy) = (1/B) [-w_(xx)/(w_x)^3] in MFP (there is an overall sign error in MFP for v_(yy) and higher derivatives w.r.t. y), and (d/dy)^n v = v_n = (1/B)* [(1/w_1)*(d/dx)]^(n-2) [-w_2/(w_1)^3] for n > 1, with w_n = (d/dx)^n w. Consequently, in the partition polynomials Prt_n for n > 1 here substitute (n') = -B*u_n = w_n for n > 1 and (1') = L-B*u_1 = w_1, where u_n = (d/dx)^n u, and then divide by B. For example, v_4 = (1/B)*Prt_4 = (1/B)*4!*P(4,1) = (1/B) (L-B*u_n)^(-7) [-15*(-B*u_2)^3 + 10 (L-B*u_1)(-B*u_2)(-B*u_3) - (L-B*u_1)^2 (-B*u_4)], agreeing with v_4 in MFP except for the overall sign.
For the SE(2) transformation formulas in MFP, let w_x = cos(phi) + sin(phi)*u_x, and then the same transformations apply as above with cos(phi) and sin(phi) substituted for L and -B, respectively. (End)

Examples

			Examples and checks:
1) Let u_1 = -1 and u_n = 1 for n>1,
then f(t) = exp(u.*t) - u(0) = exp(t)-2t-1
and g(t) = [e.g.f. of signed A000311];
therefore, the row sums of unsigned [C(j,k)] are A000311 =
(0,1,1,4,26,236,2752,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),4!*P(4,1),...).
2) Let u_1 = -1 and u_n = (n-1)! for n>1,
then f(t) = -log(1-t)-2t
and g(t) = [e.g.f. of signed (0,A032188)]
with (0,A032188) = (0,1,1,5,41,469,6889,...) = (0,-P(1,1),2!*P(2,1),-3!P(3,1),...).
3) Let u_1 = -1 and u_n = (-1)^n (n-2)! for n>1, then
f(t) = (1+t) log(1+t) - 2t
and g(t) = [e.g.f. of signed (0,A074059)]
with (0,A074059) = (0,1,1,2,7,34,213,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),...).
4) Let u_1 = 1, u_2 = -1 and u_n = 0 for n>2,
then f(t) = t(1-t/2)
and g(t) = [e.g.f. of (0,A001147)] = 1 - (1-2t)^(1/2)
with (0,A001147) = (0,1,1,3,15,105,945...) =(0,P(1,1),2!*P(2,1),3!*P(3,1),...).
5) Let u_1 = 1, u_2 = -2 and u_n = 0 for n>2,
then f(t)= t(1-t)
and g(t) = t * [o.g.f. of A000108] = [1 - (1-4t)^(1/2)] / 2
with (0,A000108) = (0,1,1,2,5,14,42,...) = (0,P(1,1),P(2,1),P(3,1),...).
.
From _Peter Luschny_, Feb 19 2021: (Start)
Triangle starts:
 [1]  1;
 [2] -1;
 [3]  3,     -1;
 [4] -15,     10,    -1;
 [5]  105,   -105,   [10, 15],  -1;
 [6] -945,    1260,  [-280, -210], [35, 21],  -1;
 [7]  10395, -17325, [6300, 3150], [-280, -1260, -378], [35, 56, 28], -1;
 [8] -135135, 270270, [-138600, -51975], [15400, 34650, 6930], [-2100, -1575, -2520, -630], [126, 84, 36], -1
The coefficients can be seen as a refinement of the Ward numbers: Let R(n, k) = Sum T(n, k), where the sum collects adjacent terms with equal sign, as indicated by the square brackets in the table, then R(n+1, k+1) = (-1)^(n-k)*W(n, k), where W(n, k) are the Ward numbers A181996, for n >= 0 and 0 <= k <= n-1.  (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 831.
  • D. S. Alexander, A History of Complex Dynamics: From Schröder to Fatou to Julia, Friedrich Vieweg & Sohn, 1994, p. 10.
  • J. Riordan, Combinatorial Identities, Robert E. Krieger Pub. Co., 1979, (unsigned partition polynomials in Table 5.2 on p. 181, but may have errors).

Crossrefs

Cf. A145271, (A134991, A019538) = (reduced array, associated g(x)).
Cf. A181996 (Ward numbers).

Programs

  • Mathematica
    rows[n_] := {{1}}~Join~Module[{h = 1/(1 + Sum[u[k] y^k/k!, {k, n-1}] + O[y]^n), g = y, r}, r = Reap[Do[g = h D[g, y]; Sow[Expand[Normal@g /. {y -> 0}]], {k, n}]][[2, 1, ;;]]; Table[Coefficient[r[[k]], Product[u[t], {t, p}]], {k, 2, n}, {p, Reverse@Sort[Sort /@ IntegerPartitions[k-1]]}]];
    rows[8] // Flatten (* Andrei Zabolotskii, Feb 19 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [2!^e(2)*e(2)!*3!^e(3)*e(3)! ... n!^e(n)*e(n)! ].
From Tom Copeland, Sep 05 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u.*exp(u.*t)]
= 1/(u_1+(u_2)*t+(u_3)*t^2/2!+(u_4)*t^3/3!+...),
an e.g.f. for the partition polynomials of A133314
(signed A049019) with an index shift.
Then for the partition polynomials of A134685,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (Cf. A000311 and A134991)(End)
From Tom Copeland, Oct 30 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*h(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*h(d/dt) = t * 1/[u_1+(u_2)*d/dt+(u_3)*(d/dt)^2/2!+...], and
L = f(d/dt)=(u_1)*d/dt+(u_2)*(d/dt)^2/2!+(u_3)*(d/dt)^3/3!+....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2! + u_3 * x^2/3! + ... + u_n * x^(n-1)/n!]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix by u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) u_{n+1-k}. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314. - Tom Copeland, Jul 15 2016
Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. - Tom Copeland, Aug 27 2016
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 15 u2 u4 + 10 u3^2, b3 = -105 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2,..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 105 u2^2 u3 + 1 * 15 u2 u4 + 1 * 10 u3^2 - 0 * u5 = 315 u2^4 - 210 u2^2 u3 + 15 u2 u4 + 10 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 10 u2 u3 - u4) + 6!/(3!*3!) * (3 u2^2 - u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the Bell polynomials of A036040 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016
Terms in rows 5-8 reordered by Andrei Zabolotskii, Feb 19 2024

A080635 Number of permutations on n letters without double falls and without initial falls.

Original entry on oeis.org

1, 1, 1, 3, 9, 39, 189, 1107, 7281, 54351, 448821, 4085883, 40533129, 435847959, 5045745069, 62594829027, 828229153761, 11644113200031, 173331882039141, 2723549731505163, 45047085512477049, 782326996336904679, 14233537708408467549, 270733989894887810547
Offset: 0

Views

Author

Emanuele Munarini, Feb 28 2003

Keywords

Comments

A permutation w has a double fall at k if w(k) > w(k+1) > w(k+2) and has an initial fall if w(1) > w(2).
exp(x*(1-y+y^2)*D_y)*f(y)|_{y=0} = f(1-E(-x)) for any function f with a Taylor series. D_y means differentiation with respect to y and E(x) is the e.g.f. given below. For a proof of exp(x*g(y)*D_y)*f(y) = f(F^{-1}(x+F(y))) with the compositional inverse F^{-1} of F(y)=int(1/g(y),y) with F(0)=0 see, e.g., the Datolli et al. reference.
Number of increasing ordered trees on vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <= 2. - David Callan, Mar 30 2007
Number of increasing colored 1-2 trees of order n with choice of two colors for the right branches of the vertices of outdegree 2. - Wenjin Woan, May 21 2011

Examples

			E.g.f. = 1 + x + (1/2)*x^2 + (1/2)*x^3 + (3/8)*x^4 + (13/40)*x^5 + (21/80)*x^6 + ...
G.f. = 1 + x + x^2 + 3*x^3 + 9*x^4 + 39*x^5 + 189*x^6 + 1107*x^7 + ...
For n = 3: 123, 132, 231. For n = 4: 1234, 1243, 1324, 1342, 1423, 2314, 2341, 2413, 3412.
a(4)=9. The 9 plane (ordered) increasing unary-binary trees are
...................................................................
..4................................................................
..|................................................................
..3..........4...4...............4...4...............3...3.........
..|........./.....\............./.....\............./.....\........
..2....2...3.......3...2...3...2.......2...3...4...2.......2...4...
..|.....\./.........\./.....\./.........\./.....\./.........\./....
..1......1...........1.......1...........1.......1...........1.....
...................................................................
..3...4...4...3....................................................
...\./.....\./.....................................................
....2.......2......................................................
....|.......|......................................................
....1.......1......................................................
...................................................................
		

Crossrefs

Programs

  • Maple
    a:= proc(n) if n < 2 then 1 else n! * sum((sqrt(3)/(2*Pi*(k+1/3)))^(n+1), k=-infinity..infinity) fi end: seq(a(n), n=0..30); # Richard Ehrenborg, Dec 09 2013
    a := proc(n) option remember; local k; if n < 3 then 1 else
    add(binomial(n-1, k)*a(k)*a(n-k-1), k = 0..n-2) fi end:
    seq(a(n), n = 0..23); # Peter Luschny, May 24 2024
  • Mathematica
    Table[n!, {n, 0, 40}]*CoefficientList[Series[ (1 + 1/Sqrt[3] Tan[Sqrt[3]/2 x])/(1 - 1/Sqrt[3] Tan[Sqrt[3]/2 x]), {x, 0, 40}], x]
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1/2 + Sqrt[3]/2 Tan[ Pi/6 + Sqrt[3] x/2], {x, 0, n}]]; (* Michael Somos, May 22 2011 *)
    Join[{1}, FullSimplify[Table[3^((n+1)/2) * n! * (Zeta[n+1, 1/3] - (-1)^n*Zeta[n+1, 2/3]) / (2*Pi)^(n+1), {n, 1, 20}]]] (* Vaclav Kotesovec, Aug 06 2021 *)
  • Maxima
    a(n):=if n=0 then 1 else sum((-3)^((n-k)/2)*((-1)^(n-k)+1)*sum(binomial(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^(j)*stirling2(n,j+k),j,0,n-k),k,1,n); /* Vladimir Kruchinin, Feb 13 2019 */
  • PARI
    {a(n) = my(A); if( n<1, n==0, A = O(x); for( k=1, n, A = intformal( 1 + A + A^2)); n! * polcoeff( A, n))}; /* Michael Somos, Oct 04 2003 */
    
  • PARI
    {a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(2*cosh(x +x*O(x^n)) - 1) ) )), n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 22 2016
    
  • Sage
    @CachedFunction
    def c(n,k) :
        if n==k: return 1
        if k<1 or k>n: return 0
        return ((n-k)//2+1)*c(n-1,k-1)+2*k*c(n-1,k+1)
    def A080635(n):
        return add(c(n,k) for k in (0..n))
    [A080635(n) for n in (0..23)] # Peter Luschny, Jun 10 2014
    

Formula

E.g.f.: (1 + 1/sqrt(3) * tan(sqrt(3)/2 * x)) / (1 - 1/sqrt(3) * tan( sqrt(3)/2 * x)).
Recurrence: a(n+1) = (Sum_{k=0..n} binomial(n, k) * a(k) * a(n-k)) - a(n) + 0^n.
E.g.f.: A(x) satisfies A' = 1 - A + A^2. - Michael Somos, Oct 04 2003
E.g.f.: E(x) = (3*cos((1/2)*3^(1/2)*x) + (3^(1/2))*sin((1/2)*3^(1/2)*x))/(3*cos((1/2)*3^(1/2)*x) - (3^(1/2))* sin((1/2)*3^(1/2)*x)). See the Michael Somos comment. - Wolfdieter Lang, Sep 12 2005
O.g.f.: A(x) = 1+x/(1-x-2*x^2/(1-2*x-2*3*x^2/(1-3*x-3*4*x^2/(1-... -n*x-n*(n+1)*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
From Peter Bala: (Start)
An alternative form of the e.g.f. for this sequence taken from [Bergeron et al.] is
(1)... (sqrt(3)/2)*tan((sqrt(3)/2)*x+Pi/6) [with constant term 1/2].
By comparing the egf for this sequence with the egf for the Eulerian numbers A008292 we can show that
(2)... a(n) = A(n,w)/(1+w)^(n-1) for n >= 1,
where w = exp(2*Pi*i/3) and {A(n,x),n>=1} = [1, 1+x, 1+4*x+x^2, 1+11*x+11*x^2+x^3,...] denotes the sequence of Eulerian polynomials. Equivalently,
(3)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k)*(-1/2 + sqrt(3)*i/6)^(k-1) for n >= 1, and
(4)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} (-1/2 + sqrt(3)*i/6)^(k-1)* Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*j^n for n >= 1.
This explicit formula for a(n) may be used to obtain various congruence results. For example,
(5a)... a(p) == 1 (mod p) for prime p = 6*n+1,
(5b)... a(p) == -1 (mod p) for prime p = 6*n+5.
For the corresponding results for the case of non-plane unary-binary trees see A000111. For type B results see A001586. For a related sequence of polynomials see A185415. See also A185416 for a recursive method to compute this sequence. For forests of plane increasing unary binary trees see A185422 and A185423. (End)
O.g.f.: A(x) = x - (1/2)*x^2 + (1/2)*x^3 - (3/8)*x^4 + (13/40)*x^5 - (21/80)*x^6 + (123/560)*x^7 - (809/4480)*x^8 + (671/4480)*x^9 - (5541/44800)*x^10 + .... - Vladimir Kruchinin, Jan 18 2011
Let f(x) = 1+x+x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 0. - Peter Bala, Oct 06 2011
From Sergei N. Gladkovskii, May 06 2013 - Dec 24 2013: (Start)
Continued fractions:
G.f.: 1 + 1/Q(0), where Q(k) = 1/(x*(k+1)) - 1 - 1/Q(k+1).
E.g.f.: 1 + 2*x/(W(0)-x), where W(k) = 4*k + 2 - 3*x^2/W(k+1).
G.f.: 1 + x/Q(0), m=1, where Q(k) = 1 - m*x*(2*k+1) - m*x^2*(2*k+1)*(2*k+2)/( 1 - m*x*(2*k+2) - m*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - x*(k+1) - x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 + T(0)*x/(1-x), where T(k) = 1 - x^2*(k+1)*(k+2)/( x^2*(k+1)*(k+2) - (1-x*(k+1))*(1-x*(k+2))/T(k+1) ).
G.f.: 1 + x/(G(0)-x), where G(k) = 1 + x*(k+1) - x*(k+1)/(1 - x*(k+2)/G(k+1) ). (End)
a(n) ~ 3^(3*(n+1)/2) * n^(n+1/2) / (exp(n)*(2*Pi)^(n+1/2)). - Vaclav Kotesovec, Oct 05 2013
a(n) = n! * Sum_{k=-oo..oo} (sqrt(3)/(2*Pi*(k+1/3)))^(n+1) for n >= 1. - Richard Ehrenborg, Dec 09 2013
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = (sqrt(3)/2)*tan((sqrt(3)/2)*x + Pi/6) satisfies the differential equation A"(x) = 2*A(x)*A'(x) with A(0) = 1/2 and A'(0) = 1, leading to the recurrence a(0) = 1/2, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the sequence [1/2, 1, 1, 3, 9, 39, 189, 1107, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 0 and a(1) = 1 produces A000182. Cf. A002105, A234797. (End)
E.g.f.: exp( Series_Reversion( Integral 1/(2*cosh(x) - 1) dx ) ). - Paul D. Hanna, Feb 22 2016
a(n) = Sum_{k=1..n} (-3)^((n-k)/2)*((-1)^(n-k)+1)*Sum_{j=0..n-k} C(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^j*Stirling2(n,j+k),n>0, a(0)=1. - Vladimir Kruchinin, Feb 13 2019
For n > 0, a(n) = 3^((n+1)/2) * n! * (zeta(n+1, 1/3) - (-1)^n*zeta(n+1, 2/3)) / (2*Pi)^(n+1). - Vaclav Kotesovec, Aug 06 2021

Extensions

Several typos corrected by Olivier Gérard, Mar 26 2011

A133932 Coefficients of a partition transform for Lagrange inversion of -log(1 - u(.)*t), complementary to A134685 for an e.g.f.

Original entry on oeis.org

1, -1, 3, -2, -15, 20, -6, 105, -210, 40, 90, -24, -945, 2520, -1120, -1260, 420, 504, -120, 10395, -34650, 25200, 18900, -2240, -15120, -9072, 1260, 2688, 3360, -720, -135135, 540540, -554400, -311850, 123200, 415800, 166320, -50400, -56700, -120960, -75600, 18144, 20160, 25920, -5040
Offset: 1

Views

Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = -log(1 - u(.)*t) = Sum_{n>=1} (u_n / n) * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -1 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - 2 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 20 (1')(2')(3') - 6 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 210 (1') (2')^2 (3') + 40 (1')^2 (3')^2 + 90 (1')^2 (2') (4') - 24 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 2520 (1') (2')^3 (3') - 1120 (1')^2 (2') (3')^2 - 1260 (1')^2 (2')^2 (4') + 420 (1')^3 (3')(4') + 504 (1')^3 (2')(5') - 120 (1')^4 (6') ] * t^6 / 6!
See A134685 for more information.
From Tom Copeland, Sep 28 2016: (Start)
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 34650 (1')(2')^4(3') + (1')^2 [25200 (2')^2(3')^2 + 18900 (2')^3(4')] - (1')^3 [2240 (3')^3 + 15120 (2')(3')(4') + 9072 (2')^2(5')] + (1')^4 [1260 (4')^2 + 2688 (3')(5') + 3360 (2')(6')] - 720 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 540540 (1')(2')^5(3') - (1')^2 [554400 (2')^3(3')^2 + 311850 (2')^4(4')] + (1')^3 [123200 (2')(3')^3 + 415800 (2')^2(3')(4') + 166320 (2')^3(5')] - (1')^4 [50400 (3')^2(4') + 56700 (2')(4')^2 + 120960 (2')(3')(5') + 75600 (2')^2(6')] + (1')^5 [18144 (4')(5') + 20160 (3')(6') + 25920 (2')(7')] - 5040 (1')^6(8')] * t^8 / 8! (End)

Examples

			From _Tom Copeland_, Sep 18 2014: (Start)
Let f(x) = log((1-ax)/(1-bx))/(b-a) = -log(1-u.*x) = x + (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 + (a^3+a^2b+ab^2+a^3)x^4/4 + ... . with (u.)^n = u_n = h_(n-1)(a,b) the complete homogeneous polynomials in two indeterminates.
Then the inverse g(x) = (e^(ax)-e^(bx))/(a*e^(ax)-b*e^(bx)) = x - (a+b)x^2/2! + (a^2+4ab+b^2)x^3/3! - (a^3+11a^2b+11ab^2+b^3)x^4/4! + ... , where the bivariate polynomials are the Eulerian polynomials of A008292.
The inversion formula gives, e.g., P(3,x) = 3(u_2)^2 - 2u_3 = 3(h_1)^2 - 2h_2 = 3(a+b)^2 - 2(a^2 + ab + b^2) = a^2 + 4ab + b^2. (End)
		

Crossrefs

Cf. A145271 (A111999, A007318) = (reduced array, associated g(x)).

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k/(k+1), {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [ 2^e(2) (e(2))! * 3^e(3) (e(3))! * ... n^e(n) * (e(n))! ].
From Tom Copeland, Sep 06 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u./(1-u.t)]
= 1/((u_1) + (u_2)*t + (u_3)*t^2 + (u_4)*t^3 + ...),
where Ev denotes umbral evaluation.
Then for the partition polynomials of A133932,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (End)
From Tom Copeland, Oct 20 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)] = exp[x*h(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*h(d/dt) = t* 1/[(u_1) + (u_2)*d/dt + (u_3)*(d/dt)^2 + ...], and
L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2/2 + (u_3)*(d/dt)^3/3 + ....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2 + u_3 * x^2/3 + ... + u_n * x^(n-1)/n]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
From Tom Copeland, Sep 19 2016: (Start)
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix A007318 by f_m = (m-1)! u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) f_{n+1-k}, or equivalently, multiply the diagonals of A094587 by u_m. Then P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314.
With u_1 = 1, the first column of UP^(-1) with u_1 = 1 (with initial indices [0,0]) is composed of the row polynomials n! * OP_n(-u_2,...,-u_(n+1)), where OP_n(x[1],...,x[n]) are the row polynomials of A263633 for n > 0 and OP_0 = 1, which are related to those of A133314 as reciprocal o.g.f.s are related to reciprocal e.g.f.s; e.g., UP^(-1)[0,0] = 1, Up^(-1)[1,0] = -u_2, UP^(-1)[2,0] = 2! * (-u_3 + u_2^2) = 2! * OP_2(-u_2,-u_3).
Also, P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. (End)
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -24 u5, b2 = 90 u2 u4 + 40 u3^2, b3 = -210 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers's equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2, ..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1, u_1=1, u_2, ..., u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 210 u2^2 u3 + 1 * 90 u2 u4 + 1 * 40 u3^2 - 0 * -24 u5 = 315 u2^4 - 420 u2^2 u3 + 90 u2 u4 + 40 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 20 u2 u3 - 6 u4) + 6!/(3!*3!) * (3 u2^2 - 2 u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the refined Stirling polynomials of the first kind A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 08 2024

A248120 Triangle read by rows: Lagrange (compositional) inversion of a function in terms of the coefficients of the Taylor series expansion of its reciprocal, scaled version of A248927, n >= 1, k = 1..A000041(n-1).

Original entry on oeis.org

1, 2, 6, 3, 24, 36, 4, 120, 360, 60, 80, 5, 720, 3600, 1800, 1200, 300, 150, 6, 5040, 37800, 37800, 16800, 3150, 12600, 3150, 420, 630, 252, 7, 40320, 423360, 705600, 235200, 176400, 352800, 58800, 35280, 23520, 35280, 7056, 1960, 1176, 392, 8
Offset: 1

Views

Author

Tom Copeland, Oct 28 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 and a scaled version of A248927. A refinement of A055302, which enumerates the number of labeled rooted trees with n nodes and k leaves, with row sums A000169.
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) = [ 1 (0') ] * t
+ [ 2 (0') (1') ] * t^2/2!
+ [ 6 (0') (1')^2 + 3 (0')^2 (2') ] * t^3/3!
+ [24 (0') (1')^3 + 36 (0')^2 (1') (2') + 4 (0')^3 (3')] * t^4/4!
+ [120 (0') (1')^4 + 360 (0')^2 (1')^2 (2') + (0')^3 [60 (2')^2
+ 80 (1') (3')] + 5 (0')^4 (4')] * t^5/5!
+ [720 (0')(1')^5 + 3600 (0')^2 (1')^3(2') + (0')^3 [1800 (1')(2')^2 + 1200 ( 1')^2(3')] + (0')^4 [300 (2')(3') + 150 (1')(4')] + 6 (0')^5 (5')] * t^6/6! + ... .
Operating with [1/(n*(n-1))] d/d(1') = [1/(n*(n-1))] 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.
Each n-th partition polynomial here is n times the (n-1)-th partition polynomial of A248927.
From Tom Copeland, Nov 24 2014: (Start)
The n-th row is a mapping of the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n)]^(n-1) 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 * h_1] = 3 * h_2 + 6 * h_1^2 = A248120(3) with h_0 = 1. (Example corrected Jul 14 2015.)
For another example and relations to A134264 and A036038, see A134264. 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.
The Abramowitz and Stegun reference in A036038 gives 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)
As presented above and in the Copeland link, this entry is related to exponentiation of e.g.f.s and, therefore, to discussions in the Scott and Sokal preprint (see eqn. 3.1 on p. 10 and eqn. 3.62 p. 24). - Tom Copeland, Jan 17 2017

Examples

			Triangle begins
     1;
     2;
     6,     3;
    24,    36,     4;
   120,   360,    60,    80,    5;
   720,  3600,  1800,  1200,  300,   150,    6;
  5040, 37800, 37800, 16800, 3150, 12600, 3150, 420, 630, 252, 7;
  ...
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 A248927, "scaled" versions of this Lagrange inversion.
Cf. A036038.

Programs

  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); (n+1)*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)! 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!/(3!*2!)]/(2!)^2 = 1800 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = 5! * [6!/4!]/(2!*3!) = 300.
If the initial factorial and final denominator of T(n,k) are removed and the expression divided by j and the partitions reversed in order, then 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 are A000169(n) = n^(n-1) = n* 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 (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.)
Following the notes connected to the Lagrange reversion theorem in A248927, a generator for the n-th partition polynomial P_n of this entry is (d/dx)^(n-1) (h (x))^n, and -log(1-t*P.) = (t*Q.) / (1 - t*Q.), umbrally, where (Q.)^n = Q_n is the n-th partition polynomial of A248927. - Tom Copeland, Nov 25 2016
With h_0 = 1, the n-th partition polynomial is obtained as the n-th element (with initial index 0) of the first column of M^{n+1}, where M is the matrix with M_{i,j}= binomial(i,j) h_{i-j}, i.e., the lower triangular Pascal matrix with its n-th diagonal multiplied by h_n. This follows from the Lagrange inversion theorem and the relation between powers of matrices such as M and powers of formal Taylor series discussed in A133314. This is equivalent to repeated binomial convolution of the coefficients of the Taylor series with itself. - Tom Copeland, Nov 13 2019
T(n,k) = n*A248927(n,k). - Andrew Howroyd, Feb 02 2022

Extensions

Terms a(31) and beyond from Andrew Howroyd, Feb 02 2022

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

Views

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
Showing 1-10 of 12 results. Next