cp's OEIS Frontend

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

Showing 1-10 of 16 results. Next

A179233 Irregular triangle T(n,k) = A049019(n,k)/A096162(n,k) read along rows, 1<=k <= A000041(n).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 8, 3, 18, 1, 1, 10, 20, 30, 45, 40, 1, 1, 12, 30, 10, 45, 360, 15, 80, 270, 75, 1, 1, 14, 42, 70, 63, 630, 210, 315, 140, 2520, 420, 175, 1050, 126, 1, 1, 16, 56, 112, 35, 84, 1008, 1680, 630, 840, 224, 5040, 1680
Offset: 1

Views

Author

Alford Arnold, Jul 08 2010

Keywords

Comments

Each row n of A049019, of A096162 and of the triangle here has A000041(n) entries.

Examples

			A049019(.,.) begins 1; 1; 2, 1; 6, 6, 1; 8, 6, 36, 24, ...
A096162(.,.) begins 1; 1; 2, 1; 1, 6, 1; 1, 2, 2, 24 ...
so
T(.,.) begins ..... 1; 1; 1, 1; 6, 1, 1; 8, 3, 18, 1 ...
		

Crossrefs

Formula

T(n,k) = A049019(n,k) / A096162(n,k) = A048996(n,k) * A036040(n,k).
Sum_{k=1..A000041(n)} T(n,k) = A120774(n).

Extensions

Extended, and bivariate indices restored - R. J. Mathar, Jul 13 2010

A182779 a(n) = A049019(n) * A118851(n). Irregular table read by rows.

Original entry on oeis.org

1, 1, 2, 2, 3, 12, 6, 4, 24, 24, 72, 24, 5, 40, 120, 180, 360, 480, 120, 6, 60, 240, 180, 360, 2160, 720, 1440, 4320, 3600, 720, 7, 84, 420, 840, 630, 5040, 3780, 7560, 3360, 30240, 20160, 12600, 50400, 30240, 5040
Offset: 0

Views

Author

Alford Arnold, Dec 01 2010

Keywords

Comments

The sequences have shape A000041 and their respective row sums are A000670, A006906 and A006153.

Examples

			For n = 3 the values are (3,12,6) = (1,6,6)*(3,2,1).
Table starts:
1;
1;
2, 2;
3, 12, 6;
4, (24, 24), 72, 24;
5, (40, 120), (180, 360), 480, 120;
6, (60, 240, 180), (360, 2160, 720), (1440, 4320), 3600, 720;
7, (84, 420, 840), (630, 5040, 3780, 7560), (3360, 30240, 20160), (12600, 50400), 30240, 5040;
		

Crossrefs

Cf. A006153 (related to function composition), A133314 (signed version of A049019).

Formula

a(n) = A049019(n) * A118851(n).

Extensions

a(0) = 1 prepended by Peter Luschny, May 31 2020

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

Views

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

A133314 Coefficients of list partition transform: reciprocal of an exponential generating function (e.g.f.).

Original entry on oeis.org

1, -1, -1, 2, -1, 6, -6, -1, 8, 6, -36, 24, -1, 10, 20, -60, -90, 240, -120, -1, 12, 30, -90, 20, -360, 480, -90, 1080, -1800, 720, -1, 14, 42, -126, 70, -630, 840, -420, -630, 5040, -4200, 2520, -12600, 15120, -5040, -1, 16, 56, -168, 112, -1008, 1344, 70
Offset: 0

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Author

Tom Copeland, Oct 18 2007, Oct 29 2007, Nov 16 2007

Keywords

Comments

The list partition transform of a sequence a(n) for which a(0)=1 is illustrated by:
b_0 = 1
b_1 = -a_1
b_2 = -a_2 + 2 a_1^2
b_3 = -a_3 + 6 a_2 a_1 - 6 a_1^3
b_4 = -a_4 + 8 a_3 a_1 + 6 a_2^2 - 36 a_2 a_1^2 + 24 a_1^4
... .
The unsigned coefficients are A049019 with a leading 1. The sign is dependent on the partition as evident from inspection (replace a_n's by -1).
Expressed umbrally, i.e., with the umbral operation (a.)^n := a_n,
exp(a.x) exp(b.x) = exp[(a.+b.)x] = 1; i.e., (a.+b.)^n = 1 for n=0 and 0 for all other values of n.
Expressed recursively,
b_0 = 1, b_n = -Sum_{j=1..n} binomial(n,j) a_j b_{n-j}; which is conditionally self-inverse, i.e., the roles of a_k and b_k may be reversed with a_0 = b_0 = 1.
Expressed in matrix form, b_n form the first column of B = matrix inverse of A .
A = Pascal matrix diagonally multiplied by a_n, i.e., A_{n,k} = binomial(n,k)* a_{n-k}.
Some examples of reciprocal pairs of sequences under these operations are:
1) A084358 and -A000262 with the first term set to 1.
2) (1,-1,0,0,...) and (0!,1!,2!,3!,...) with the unsigned associated matrices A128229 and A094587.
3) (1,-1,-1,-1,...) and A000670.
5) (1,-2,-2,0,0,0,...) and (0! c_1,1! c_2,2! c_3,3! c_4,...) where c_n = A000129(n) with the associated matrices A110327 and A110330.
6) (1,-2,2,0,0,0,...) and (1!,2!,3!,4!,...).
7) Sequences of rising and signed lowering factorials form reciprocal pairs where a_n = (-1)^n m!/(m-n)! and b_n = (m-1+n)!/(m-1)! for m=0,1,2,... .
Denote the action of the list partition transform on the sequence a. or an invertible matrix M by LPT(a.) = b. or LPT(M)= M^(-1).
If the matrix equation M = exp(T) also holds, then exp[a.*T]*exp[b.*T] = exp[(a.+b.)*T] = I, the identity matrix, because (a.+b.)^n = delta_n, the Kronecker delta with delta_n = 1 and delta_n = 0 otherwise, i.e., (0)^n = delta_n.
Therefore, [exp(a.*T)]^(-1) = exp[b.*T] = exp[LPT(a.)*T] = LPT[exp(a.*T)].
The fundamental Pascal (A007318), unsigned Lah (A105278) and associated Laguerre matrices can be generated by exponentiation of special infinitesimal matrices (see A132440, A132710 and A132681) such that finding LPT(a.) amounts to multiplying the k'th diagonal of the fundamental matrices by a_k for every diagonal followed by matrix inversion and then extraction of the b_n factors from the first column (simplest for the Pascal formulas above).
Conversely, the inverses of matrices formed by diagonally multiplying the three fundamental matrices by a_k are given by diagonally multiplying the fundamental matrices by b_k.
If LPT(M) is defined differently as application of the top formula to a_n = M^n, then b_n = (-M)^n and the formalism could even be applied to more general sequences of matrices M., providing the reciprocal of exp[t*M.].
The group of fundamental lower triangular matrices M = exp(T) such that LPT[exp(a.*T)] = exp[LPT(a.)*T] = [exp[a.*T]]^(-1) are obtained by infinitesimal generator matrices of the form T =
0;
t(0), 0;
0, t(1), 0;
0, 0, t(2), 0;
0, 0, 0, t(3), 0;
... .
T^m has trivially vanishing terms except along the m'th subdiagonal, which is a sequence of generalized factorials:
[ t(0)*t(1)...t(m-2)*t(m-1), t(1)*t(2)...t(m-1)*t(m), t(2)*t(3)...t(m)*t(m+1), ... ].
Therefore the principal submatrices of T (given by setting t(j) = 0 for j > n-1) are nilpotent with at least [Tsub_n]^(n+1) = 0.
The general group of matrices GM[a.] = exp[a.*T] can also be obtained through diagonal multiplication of M = exp(T) by the sequence a_n, as in the Pascal matrix example above and their inverses by diagonal multiplication by b. = LPT(a.).
Weighted-mappings interpretation for the top partition equation:
Given n pre-nodes (Pre) and k post-nodes (Post), each Pre is connected to only one Post and each Post has at least one Pre connected to it (surjections or onto functions/maps). Weight each Post by -a_m where m is the number of connections to the Post.
Weight each map by the product of the Post weights and multiply by the number of maps that share the same connectivity. Sum over the possible mappings for n Pre. The result is b_n.
E.g., b_3 = [ 3 Pre to 1 Post ] + [ 3 Pre to 2 Post ] + [ 3 Pre to 3 Post ]
= [1 map with 1 Post with 3 connections] + [ 6 maps with 1 Post with 2 connections and 1 Post with 1 connection] + [6 maps with 3 Post with 1 connection each]
= -a_3 + 6 * [-a_2*(-a_1)] + 6 * [-a_1*(-a_1)*(-a_1)].
See A263633 for the complementary formulation for the reciprocal of o.g.f.s rather than e.g.f.s and computations of these partition polynomials as Gram determinants. - Tom Copeland, Dec 04 2016
The coefficients of the partition polynomials enumerate the faces of the convex, bounded polytopes called permutohedra, and the absolute value of the sum of the 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 unity. 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 the fundamental matrix chosen to be the lower triangular Pascal matrix M, the matrix MA whose n-th diagonals are multiplied by a_n (i.e., MA_{i,j} = PM_{i,j} * a_{i-j}) gives a matrix representation of the e.g.f. associated to the Appell polynomial sequence defined by e^{a.t}e^{xt}= e^{(a.+x)t} = e^{A.(x)t} where umbrally (A.(x))^n = A_n(x) = (a. + x)^n = sum_{k=0..n} binomial(n,k) a_k x^{n-k} are the associated Appell polynomials. Left multiplication of the column vector (1,x,x^2,..) by MA gives the Appell polynomial sequence, and multiplication of the two e.g.f.s e^{a.t} and e^{b.t} corresponds to multiplication of their respective matrix representations MA and MB. Forming the reciprocal of an e.g.f. corresponds to taking the matrix inverse of its matrix representation as noted above. A263634 gives an associated modified Pascal matrix representation of the raising operator for the Appell sequence. - Tom Copeland, Nov 13 2019
The diagonal of MA consists of all ones. Let MAN be the truncated square submatrix of MA containing the coefficients of the first N Appell polynomials A_k=(a.+x)^k = Sum(j=0 to k) MAN(k,j) x^j. Then by the Cayley-Hamilton theorem (I-MAN)^N = 0; therefore, MAN^(-1) = Sum(k=1 to N) binomial(N,k) (-MAN)^{k-1} = MBN, the inverse of MAN, containing the coefficients of the first N rows of the Appell polynomials B_k(x) = (b. + x)^k = Sum(j=0 to k) MBN(k,j) x^j, which are the umbral compositional inverses of the Appell row polynomials A_k(x) of MAN; that is, A_k(B.(x)) = x^k = B_k(A.(x)), where, e.g., (A.(x))^k = A_k(x). - Tom Copeland, May 13 2020
The use of the term 'list partition transform' resulted from one of my first uses of these partition polynomials in relating A000262 to A084358 with their simple e.g.f.s. Other appropriate names would be the permutohedra polynomials since they are refined Euler characteristics of the permutohedra or the reciprocal polynomials since they give the multiplicative inverses of e.g.f.s with a constant of 1. - Tom Copeland, Oct 09 2022

Examples

			Table starts:
[0] [ 1]
[1] [-1]
[2] [-1,  2]
[3] [-1,  6, -6]
[4] [-1,  8,  6, -36,  24]
[5] [-1, 10, 20, -60, -90,  240, -120]
[6] [-1, 12, 30, -90,  20, -360,  480, -90, 1080, -1800, 720]
		

Crossrefs

Programs

  • Mathematica
    b[0] = 1; b[n_] := b[n] = -Sum[Binomial[n, j]*a[j]*b[n-j], {j, 1, n}];
    row[0] = {1}; row[n_] := Coefficient[b[n], #]& /@ (Times @@ (a /@ #)&) /@ IntegerPartitions[n];
    Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 23 2014 *)
  • Sage
    def A133314_row(n): return [(-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in Partitions(n)]
    for n in (0..10): print(A133314_row(n)) # Peter Luschny, Sep 18 2015

Formula

b_{n-1} = (1/n)(d/da(1))p_n[a_1, a_2, ..., a_n] where p_n are the row partition polynomials of the cumulant generator A127671. - Tom Copeland, Oct 13 2012
(E.g.f. of matrix B) = (e.g.f. of b)·exp(xt) = exp(b.t)·exp(xt) = exp(xt)/exp(a.t) = (e.g.f. of A^(-1)) and (e.g.f. of matrix A) = exp(a.t)·exp(xt) = exp(xt)/exp(b.t) = (e.g.f. of B^(-1)), where the umbral evaluation of exp(b.t) = Sum{n >= 0} (b.t)^n / n! = Sum_{n >= 0} b_n t^n / n! is understood in the denominator. These e.g.f.s define Appell sequences of polynomials. - Tom Copeland, Mar 22 2014
Sum of the n-th row is (-1)^n. - Peter Luschny, Sep 18 2015
The unsigned coefficients for the partitions a_2*a_1^n for n >= 0 are the Lah numbers A001286. - Tom Copeland, Aug 06 2016
G.f.: 1 / (1 + Sum_{n > 0} a_n x^n/n!) = 1 / exp(a.x). - Tom Copeland, Oct 18 2016
Let a_1 = 1 + x + B_1 = x + 1/2 and a_n = B_n = (B.)^n, where B_n are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted signed polynomials of A019538: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... , p_n(x) = n * b_{n-1}. - Tom Copeland, Oct 18 2016
With a_n = 1/(n+1), b_n = B_n, the Bernoulli numbers. - Tom Copeland, Nov 08 2016
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 A145271; [E]^(-1), A356145, the inverse set to [E]; [P], the permutohedra polynomials of this entry; [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

More terms from Jean-François Alcover, Apr 23 2014

A074909 Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11
Offset: 0

Views

Author

Wouter Meeussen, Oct 01 2002

Keywords

Comments

This sequence counts the "almost triangular" partitions of n. A partition is triangular if it is of the form 0+1+2+...+k. Examples: 3=0+1+2, 6=0+1+2+3. An "almost triangular" partition is a triangular partition with at most 1 added to each of the parts. Examples: 7 = 1+1+2+3 = 0+2+2+3 = 0+1+3+3 = 0+1+2+4. Thus a(7)=4. 8 = 1+2+2+3 = 1+1+3+3 = 1+1+2+4 = 0+2+3+3 = 0+2+2+4 = 0+1+3+4 so a(8)=6. - Moshe Shmuel Newman, Dec 19 2002
The "almost triangular" partitions are the ones cycled by the operation of "Bulgarian solitaire", as defined by Martin Gardner.
Start with A007318 - I (I = Identity matrix), then delete right border of zeros. - Gary W. Adamson, Jun 15 2007
Also the number of increasing acyclic functions from {1..n-k+1} to {1..n+2}. A function f is acyclic if for every subset B of the domain the image of B under f does not equal B. For example, T(3,1)=4 since there are exactly 4 increasing acyclic functions from {1,2,3} to {1,2,3,4,5}: f1={(1,2),(2,3),(3,4)}, f2={(1,2),(2,3),(3,5)}, f3={(1,2),(2,4),(3,5)} and f4={(1,3),(2,4),(4,5)}. - Dennis P. Walsh, Mar 14 2008
Second Bernoulli polynomials are (from A164555 instead of A027641) B2(n,x) = 1; 1/2, 1; 1/6, 1, 1; 0, 1/2, 3/2, 1; -1/30, 0, 1, 2, 1; 0, -1/6, 0, 5/3, 5/2, 1; ... . Then (B2(n,x)/A002260) = 1; 1/2, 1/2; 1/6, 1/2, 1/3; 0, 1/4, 1/2, 1/4; -1/30, 0, 1/3, 1/2, 1/5; 0, -1/12, 0, 5/12, 1/2, 1/6; ... . See (from Faulhaber 1631) Jacob Bernoulli Summae Potestatum (sum of powers) in A159688. Inverse polynomials are 1; -1, 2; 1, -3, 3; -1, 4, -6, 4; ... = A074909 with negative even diagonals. Reflected A053382/A053383 = reflected B(n,x) = RB(n,x) = 1; -1/2, 1; 1/6, -1, 1; 0, 1/2, -3/2, 1; ... . A074909 is inverse of RB(n,x)/A002260 = 1; -1/2, 1/2; 1/6, -1/2, 1/3; 0, 1/4, -1/2, 1/4; ... . - Paul Curtz, Jun 21 2010
A054143 is the fission of the polynomial sequence (p(n,x)) given by p(n,x) = x^n + x^(n-1) + ... + x + 1 by the polynomial sequence ((x+1)^n). See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Reversal of A135278. - Philippe Deléham, Feb 11 2012
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From A238363, the operator equation d/d(:xD:)f(xD)={exp[d/d(xD)]-1}f(xD) = f(xD+1)-f(xD) follows. Choosing f(x) = x^n and using :xD:^n/n! = binomial(xD,n) and (xD)^n = Bell(n,:xD:), the Bell polynomials of A008277, it follows that the lower triangular matrix [padded A074909]
A) = [St2]*[dP]*[St1] = A048993*A132440*[padded A008275]
B) = [St2]*[dP]*[St2]^(-1)
C) = [St1]^(-1)*[dP]*[St1],
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 whereas [padded A074909]=A007318-I with I=identity matrix. - Tom Copeland, Apr 25 2014
T(n,k) generated by m-gon expansions in the case of odd m with "vertex to side" version or even m with "vertex to vertes" version. Refer to triangle expansions in A061777 and A101946 (and their companions for m-gons) which are "vertex to vertex" and "vertex to side" versions respectively. The label values at each iteration can be arranged as a triangle. Any m-gon can also be arranged as the same triangle with conditions: (i) m is odd and expansion is "vertex to side" version or (ii) m is even and expansion is "vertex to vertex" version. m*Sum_{i=1..k} T(n,k) gives the total label value at the n-th iteration. See also A247976. Vertex to vertex: A061777, A247618, A247619, A247620. Vertex to side: A101946, A247903, A247904, A247905. - Kival Ngaokrajang Sep 28 2014
From Tom Copeland, Nov 12 2014: (Start)
With P(n,x) = [(x+1)^(n+1)-x^(n+1)], the row polynomials of this entry, Up(n,x) = P(n,x)/(n+1) form an Appell sequence of polynomials that are the umbral compositional inverses of the Bernoulli polynomials B(n,x), i.e., B[n,Up(.,x)] = x^n = Up[n,B(.,x)] under umbral substitution, e.g., B(.,x)^n = B(n,x).
The e.g.f. for the Bernoulli polynomials is [t/(e^t - 1)] e^(x*t), and for Up(n,x) it's exp[Up(.,x)t] = [(e^t - 1)/t] e^(x*t).
Another g.f. is G(t,x) = log[(1-x*t)/(1-(1+x)*t)] = log[1 + t /(1 + -(1+x)t)] = t/(1-t*Up(.,x)) = Up(0,x)*t + Up(1,x)*t^2 + Up(2,x)*t^3 + ... = t + (1+2x)/2 t^2 + (1+3x+3x^2)/3 t^3 + (1+4x+6x^2+4x^3)/4 t^4 + ... = -log(1-t*P(.,x)), expressed umbrally.
The inverse, Ginv(t,x), in t of the g.f. may be found in A008292 from Copeland's list of formulas (Sep 2014) with a=(1+x) and b=x. This relates these two sets of polynomials to algebraic geometry, e.g., elliptic curves, trigonometric expansions, Chebyshev polynomials, and the combinatorics of permutahedra and their duals.
Ginv(t,x) = [e^((1+x)t) - e^(xt)] / [(1+x) * e^((1+x)t) - x * e^(xt)] = [e^(t/2) - e^(-t/2)] / [(1+x)e^(t/2) - x*e^(-t/2)] = (e^t - 1) / [1 + (1+x) (e^t - 1)] = t - (1 + 2 x) t^2/2! + (1 + 6 x + 6 x^2) t^3/3! - (1 + 14 x + 36 x^2 + 24 x^3) t^4/4! + ... = -exp[-Perm(.,x)t], where Perm(n,x) are the reverse face polynomials, or reverse f-vectors, for the permutahedra, i.e., the face polynomials for the duals of the permutahedra. Cf. A090582, A019538, A049019, A133314, A135278.
With L(t,x) = t/(1+t*x) with inverse L(t,-x) in t, and Cinv(t) = e^t - 1 with inverse C(t) = log(1 + t). Then Ginv(t,x) = L[Cinv(t),(1+x)] and G(t,x) = C[L[t,-(1+x)]]. Note L is the special linear fractional (Mobius) transformation.
Connections among the combinatorics of the permutahedra, simplices (cf. A135278), and the associahedra can be made through the Lagrange inversion formula (LIF) of A133437 applied to G(t,x) (cf. A111785 and the Schroeder paths A126216 also), and similarly for the LIF A134685 applied to Ginv(t,x) involving the simplicial Whitehouse complex, phylogenetic trees, and other structures. (See also the LIFs A145271 and A133932). (End)
R = x - exp[-[B(n+1)/(n+1)]D] = x - exp[zeta(-n)D] is the raising operator for this normalized sequence UP(n,x) = P(n,x) / (n+1), that is, R UP(n,x) = UP(n+1,x), where D = d/dx, zeta(-n) is the value of the Riemann zeta function evaluated at -n, and B(n) is the n-th Bernoulli number, or constant B(n,0) of the Bernoulli polynomials. The raising operator for the Bernoulli polynomials is then x + exp[-[B(n+1)/(n+1)]D]. [Note added Nov 25 2014: exp[zeta(-n)D] is abbreviation of exp(a.D) with (a.)^n = a_n = zeta(-n)]. - Tom Copeland, Nov 17 2014
The diagonals T(n, n-m), for n >= m, give the m-th iterated partial sum of the positive integers; that is A000027(n+1), A000217(n), A000292(n-1), A000332(n+1), A000389(n+1), A000579(n+1), A000580(n+1), A000581(n+1), A000582(n+1), ... . - Wolfdieter Lang, May 21 2015
The transpose gives the numerical coefficients of the Maurer-Cartan form matrix for the general linear group GL(n,1) (cf. Olver, but note that the formula at the bottom of p. 6 has an error--the 12 should be a 15). - Tom Copeland, Nov 05 2015
The left invariant Maurer-Cartan form polynomial on p. 7 of the Olver paper for the group GL^n(1) is essentially a binomial convolution of the row polynomials of this entry with those of A133314, or equivalently the row polynomials generated by the product of the e.g.f. of this entry with that of A133314, with some reindexing. - Tom Copeland, Jul 03 2018
From Tom Copeland, Jul 10 2018: (Start)
The first column of the inverse matrix is the sequence of Bernoulli numbers, which follows from the umbral definition of the Bernoulli polynomials (B.(0) + x)^n = B_n(x) evaluated at x = 1 and the relation B_n(0) = B_n(1) for n > 1 and -B_1(0) = 1/2 = B_1(1), so the Bernoulli numbers can be calculated using Cramer's rule acting on this entry's matrix and, therefore, from the ratios of volumes of parallelepipeds determined by the columns of this entry's square submatrices. - Tom Copeland, Jul 10 2018
Umbrally composing the row polynomials with B_n(x), the Bernoulli polynomials, gives (B.(x)+1)^(n+1) - (B.(x))^(n+1) = d[x^(n+1)]/dx = (n+1)*x^n, so multiplying this entry as a lower triangular matrix (LTM) by the LTM of the coefficients of the Bernoulli polynomials gives the diagonal matrix of the natural numbers. Then the inverse matrix of this entry has the elements B_(n,k)/(k+1), where B_(n,k) is the coefficient of x^k for B_n(x), and the e.g.f. (1/x) (e^(xt)-1)/(e^t-1). (End)

Examples

			T(4,2) = 0+0+1+3+6 = 10 = binomial(5, 2).
Triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7   8   9 10 11
0:  1
1:  1  2
2:  1  3  3
3:  1  4  6   4
4:  1  5 10  10   5
5:  1  6 15  20  15   6
6:  1  7 21  35  35  21   7
7:  1  8 28  56  70  56  28   8
8:  1  9 36  84 126 126  84  36  9
9:  1 10 45 120 210 252 210 120 45   10
10: 1 11 55 165 330 462 462 330 165  55 11
11: 1 12 66 220 495 792 924 792 495 220 66 12
... Reformatted. - _Wolfdieter Lang_, Nov 04 2014
.
Can be seen as the square array A(n, k) = binomial(n + k + 1, n) read by descending antidiagonals. A(n, k) is the number of monotone nondecreasing functions f: {1,2,..,k} -> {1,2,..,n}. - _Peter Luschny_, Aug 25 2019
[0]  1,  1,   1,   1,    1,    1,     1,     1,     1, ... A000012
[1]  2,  3,   4,   5,    6,    7,     8,     9,    10, ... A000027
[2]  3,  6,  10,  15,   21,   28,    36,    45,    55, ... A000217
[3]  4, 10,  20,  35,   56,   84,   120,   165,   220, ... A000292
[4]  5, 15,  35,  70,  126,  210,   330,   495,   715, ... A000332
[5]  6, 21,  56, 126,  252,  462,   792,  1287,  2002, ... A000389
[6]  7, 28,  84, 210,  462,  924,  1716,  3003,  5005, ... A000579
[7]  8, 36, 120, 330,  792, 1716,  3432,  6435, 11440, ... A000580
[8]  9, 45, 165, 495, 1287, 3003,  6435, 12870, 24310, ... A000581
[9] 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, ... A000582
		

Crossrefs

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n+1,k)))); # Muniru A Asiru, Jul 10 2018
    
  • Haskell
    a074909 n k = a074909_tabl !! n !! k
    a074909_row n = a074909_tabl !! n
    a074909_tabl = iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [1])) [1]
    -- Reinhard Zumkeller, Feb 25 2012
    
  • Magma
    /* As triangle */ [[Binomial(n+1,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2018
    
  • Maple
    A074909 := proc(n,k)
        if k > n or k < 0 then
            0;
        else
            binomial(n+1,k) ;
        end if;
    end proc: # Zerinvary Lajos, Nov 09 2006
  • Mathematica
    Flatten[Join[{1}, Table[Sum[Binomial[k, m], {k, 0, n}], {n, 0, 12}, {m, 0, n}] ]] (* or *) Flatten[Join[{1}, Table[Binomial[n, m], {n, 12}, {m, n}]]]
  • PARI
    print1(1);for(n=1,10,for(k=1,n,print1(", "binomial(n,k)))) \\ Charles R Greathouse IV, Mar 26 2013
    
  • Python
    from math import comb, isqrt
    def A074909(n): return comb(r:=(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),n-comb(r,2)) # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = Sum_{i=0..n} C(i, n-k) = C(n+1, k).
Row n has g.f. (1+x)^(n+1)-x^(n+1).
E.g.f.: ((1+x)*e^t - x) e^(x*t). The row polynomials p_n(x) satisfy dp_n(x)/dx = (n+1)*p_(n-1)(x). - Tom Copeland, Jul 10 2018
T(n, k) = T(n-1, k-1) + T(n-1, k) for k: 0Reinhard Zumkeller, Apr 18 2005
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. for column k (with leading zeros): x^(k-1)*(1/(1-x)^(k+1)-1), k >= 0. - Wolfdieter Lang, Nov 04 2014
Up(n, x+y) = (Up(.,x)+ y)^n = Sum_{k=0..n} binomial(n,k) Up(k,x)*y^(n-k), where Up(n,x) = ((x+1)^(n+1)-x^(n+1)) / (n+1) = P(n,x)/(n+1) with P(n,x) the n-th row polynomial of this entry. dUp(n,x)/dx = n * Up(n-1,x) and dP(n,x)/dx = (n+1)*P(n-1,x). - Tom Copeland, Nov 14 2014
The o.g.f. GF(x,t) = x / ((1-t*x)*(1-(1+t)x)) = x + (1+2t)*x^2 + (1+3t+3t^2)*x^3 + ... has the inverse GFinv(x,t) = (1+(1+2t)x-sqrt(1+(1+2t)*2x+x^2))/(2t(1+t)x) in x about 0, which generates the row polynomials (mod row signs) of A033282. The reciprocal of the o.g.f., i.e., x/GF(x,t), gives the free cumulants (1, -(1+2t) , t(1+t) , 0, 0, ...) associated with the moments defined by GFinv, and, in fact, these free cumulants generate these moments through the noncrossing partitions of A134264. The associated e.g.f. and relations to Grassmannians are described in A248727, whose polynomials are the basis for an Appell sequence of polynomials that are umbral compositional inverses of the Appell sequence formed from this entry's polynomials (distinct from the one described in the comments above, without the normalizing reciprocal). - Tom Copeland, Jan 07 2015
T(n, k) = (1/k!) * Sum_{i=0..k} Stirling1(k,i)*(n+1)^i, for 0<=k<=n. - Ridouane Oudra, Oct 23 2022

Extensions

I added an initial 1 at the suggestion of Paul Barry, which makes the triangle a little nicer but may mean that some of the formulas will now need adjusting. - N. J. A. Sloane, Feb 11 2003
Formula section edited, checked and corrected by Wolfdieter Lang, Nov 04 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

A048996 Irregular triangle read by rows. Preferred multisets: numbers refining A007318 using format described in A036038.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 3, 1, 1, 2, 2, 3, 3, 4, 1, 1, 2, 2, 1, 3, 6, 1, 4, 6, 5, 1, 1, 2, 2, 2, 3, 6, 3, 3, 4, 12, 4, 5, 10, 6, 1, 1, 2, 2, 2, 1, 3, 6, 6, 3, 3, 4, 12, 6, 12, 1, 5, 20, 10, 6, 15, 7, 1, 1, 2, 2, 2, 2, 3, 6, 6, 3, 3, 6, 1, 4, 12, 12, 12, 12, 4, 5, 20, 10, 30, 5, 6, 30, 20, 7, 21, 8, 1
Offset: 0

Views

Author

Keywords

Comments

This array gives in row n>=1 the multinomial numbers (call them M_0 numbers) m!/product((a_j)!,j=1..n) with the exponents of the partitions of n with number of parts m:=sum(a_j,j=1..n), given in the Abramowitz-Stegun order. See p. 831 of the given reference. See also the arrays for the M_1, M_2 and M_3 multinomial numbers A036038, A036039 and A036040 (or A080575).
For a signed version see A111786.
These M_0 multinomial numbers give the number of compositions of n >= 1 with parts corresponding to the partitions of n (in A-St order). See an n = 5 example below. The triangle with the summed entries of like number of parts m is A007318(n-1, m-1) (Pascal). - Wolfdieter Lang, Jan 29 2021

Examples

			Table starts:
[1]
[1]
[1, 1]
[1, 2, 1]
[1, 2, 1, 3, 1]
[1, 2, 2, 3, 3, 4, 1]
[1, 2, 2, 1, 3, 6, 1, 4, 6,  5, 1]
[1, 2, 2, 2, 3, 6, 3, 3, 4, 12, 4, 5, 10, 6, 1]
.
T(5,6) = 4 because there are four multisets using the first four digits {0,1,2,3}: 32100, 32110, 32210 and 33210
T(5,6) = 4 because there are 4 compositions of 5 that can be formed from the partition 2+1+1+1. - _Geoffrey Critzer_, May 19 2013
These 4 compositions 2+1+1+1, 1+2+1+1, 1+1+2+1 and 1+1+1+2 of 5 correspond to the 4 set partitions of [5] :={1,2,3,4,5}, with 4 blocks of consecutive numbers, namely {1,2},{3},{4},{5} and {1},{2,3},{4},{5} and {1},{2},{3,4},{5} and {1},{2},{3},{4,5}. - _Wolfdieter Lang_, May 30 2018
		

Crossrefs

Cf. A000670, A007318, A036035, A036038, A019538, A115621, A309004, A000079 (row sums), A000041 (row lengths).

Programs

  • Maple
    nmax:=9: with(combinat): for n from 1 to nmax do P(n):=sort(partition(n)): for r from 1 to numbpart(n) do B(r):=P(n)[r] od: for m from 1 to numbpart(n) do s:=0: j:=0: while sA036040(n, m) := (add(q(t), t=1..n))!/(mul(q(t)!, t=1..n)); od: od: seq(seq(A036040(n, m), m=1..numbpart(n)), n=1..nmax); # Johannes W. Meijer, Jul 14 2016
  • PARI
    C(sig)={my(S=Set(sig)); (#sig)!/prod(k=1, #S, (#select(t->t==S[k], sig))!)}
    Row(n)={apply(C, [Vecrev(p) | p<-partitions(n)])}
    { for(n=0, 7, print(Row(n))) } \\ Andrew Howroyd, Oct 18 2020
  • SageMath
    from collections import Counter
    def ASPartitions(n, k):
        Q = [p.to_list() for p in Partitions(n, length=k)]
        for q in Q: q.reverse()
        return sorted(Q)
    def A048996_row(n):
        h = lambda p: product(map(factorial, Counter(p).values()))
        return [factorial(len(p))//h(p) for k in (0..n) for p in ASPartitions(n, k)]
    for n in (1..10): print(A048996_row(n)) # Peter Luschny, Nov 02 2019 [corrected on notice from Sean A. Irvine, Apr 30 2022]
    

Formula

T(n,k) = A036040(n,k) * Factorial(A036043(n,k)) / A036038(n,k) = A049019(n,k) / A036038(n,k).
If the n-th partition is P, a(n) is the multinomial coefficient of the signature of P. - Franklin T. Adams-Watters, May 30 2006
T(n,k) = A309004(A036035(n,k)). - Andrew Howroyd, Oct 19 2020

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 17 2001
a(0)=1 prepended by Andrew Howroyd, Oct 19 2020

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

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

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
Showing 1-10 of 16 results. Next