cp's OEIS Frontend

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

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

Original entry on oeis.org

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

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Author

N. J. A. Sloane, Mar 15 1996

Keywords

Comments

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

Examples

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

References

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

Crossrefs

Programs

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

Formula

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

Extensions

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

A147309 Riordan array [sec(x), log(sec(x) + tan(x))].

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 4, 0, 1, 5, 0, 10, 0, 1, 0, 40, 0, 20, 0, 1, 61, 0, 175, 0, 35, 0, 1, 0, 768, 0, 560, 0, 56, 0, 1, 1385, 0, 4996, 0, 1470, 0, 84, 0, 1, 0, 24320, 0, 22720, 0, 3360, 0, 120, 0, 1
Offset: 0

Views

Author

Paul Barry, Nov 05 2008

Keywords

Comments

Production array is [cosh(x),x] beheaded. Inverse is A147308. Row sums are A000111(n+1).
Unsigned version of A147308. - N. J. A. Sloane, Nov 07 2008
From Peter Bala, Jan 26 2011: (Start)
Define a polynomial sequence {Z(n,x)} n >= 0 by means of the recursion
(1)... Z(n+1,x) = 1/2*x*{Z(n,x-1)+Z(n,x+1)}
with starting condition Z(0,x) = 1. We call Z(n,x) the zigzag polynomial of degree n. This table lists the coefficients of these polynomials (for n >= 1) in ascending powers of x, row indices shifted by 1. The first few polynomials are
... Z(1,x) = x
... Z(2,x) = x^2
... Z(3,x) = x + x^3
... Z(4,x) = 4*x^2 + x^4
... Z(5,x) = 5*x + 10*x^3 + x^5.
The value Z(n,1) equals the zigzag number A000111(n). The polynomials Z(n,x) occur in formulas for the enumeration of permutations by alternating descents A145876 and in the enumeration of forests of non-plane unary binary labeled trees A147315.
{Z(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting Z(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
(2)... 1^[m]+...+(n-1)^[m] = (1/(m+1))*Sum_{k=0..m} (-1)^floor(k/2)*binomial(m+1,k)*B_k*n^[m+1-k],
where {B_k} k >= 0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For similarly defined polynomial sequences to Z(n,x) see A185415, A185417 and A185419. See also A185424.
(End)
[gd(x)^(-1)]^m = Sum_{n>=m} Tg(n,m)*(m!/n!)*x^n, where gd(x) is Gudermannian function, Tg(n+1,m+1)=T(n,m). - Vladimir Kruchinin, Dec 18 2011
The Bell transform of abs(E(n)), E(n) the Euler numbers. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

Examples

			Triangle begins
   1;
   0,  1;
   1,  0,   1;
   0,  4,   0,  1;
   5,  0,  10,  0,  1;
   0, 40,   0, 20,  0, 1;
  61,  0, 175,  0, 35, 0, 1;
		

Crossrefs

Programs

  • Maple
    Z := proc(n, x) option remember;
    description 'zigzag polynomials Z(n, x)'
    if n = 0 return 1 else return 1/2*x*(Z(n-1, x-1)+Z(n-1, x+1)) end proc:
    with(PolynomialTools):
    for n from 1 to 10 CoefficientList(Z(n, x), x); end do; # Peter Bala, Jan 26 2011
  • Mathematica
    t[n_, k_] := SeriesCoefficient[ 2^k*ArcTan[(E^x - 1)/(E^x + 1)]^k*n!/k!, {x, 0, n}]; Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten // Abs (* Jean-François Alcover, Jan 23 2015 *)
  • PARI
    T(n, k)=local(X); if(k<1 || k>n, 0, X=x+x*O(x^n); n!*polcoeff(polcoeff((tan(X)+1/cos(X))^y, n), k)) \\ Paul D. Hanna, Feb 06 2011
    
  • Sage
    R = PolynomialRing(QQ, 'x')
    @CachedFunction
    def zzp(n, x) :
        return 1 if n == 0 else x*(zzp(n-1, x-1)+zzp(n-1, x+1))/2
    def A147309_row(n) :
        x = R.gen()
        L = list(R(zzp(n, x)))
        del L[0]
        return L
    for n in (1..10) : print(A147309_row(n)) # Peter Luschny, Jul 22 2012
    
  • Sage
    # uses[bell_matrix from A264428]
    # Alternative: Adds a column 1,0,0,0, ... at the left side of the triangle.
    bell_matrix(lambda n: abs(euler_number(n)), 10) # Peter Luschny, Jan 18 2016

Formula

From Peter Bala, Jan 26 2011: (Start)
GENERATING FUNCTION
The e.g.f., upon including a constant term of '1', is given by:
(1) F(x,t) = (tan(t) + sec(t))^x = Sum_{n>=0} Z(n,x)*t^n/n! = 1 + x*t + x^2*t^2/2! + (x+x^3)*t^3/3! + ....
Other forms include
(2) F(x,t) = exp(x*arcsinh(tan(t))) = exp(2*x*arctanh(tan(t/2))).
(3) F(x,t) = exp(x*(t + t^3/3! + 5*t^5/5! + 61*t^7/7! + ...)),
where the coefficients [1,1,5,61,...] are the secant or zig numbers A000364.
ROW GENERATING POLYNOMIALS
One easily checks from (1) that
d/dt(F(x,t)) = 1/2*x*(F(x-1,t) + F(x+1,t))
and so the row generating polynomials Z(n,x) satisfy the recurrence relation
(4) Z(n+1,x) = 1/2*x*{Z(n,x-1) + Z(n,x+1)}.
The e.g.f. for the odd-indexed row polynomials is
(5) sinh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n+1,x)*t^(2n+1)/(2n+1)!.
The e.g.f. for the even-indexed row polynomials is
(6) cosh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n,x)*t^(2n)/(2n)!.
From sinh(2*x) = 2*sinh(x)*cosh(x) we obtain the identity
(7) Z(2n+1,2*x) = 2*Sum_{k=0..n} binomial(2n+1,2k)*Z(2k,x)*Z(2n-2k+1,x).
The zeros of Z(n,x) lie on the imaginary axis (use (4) and adapt the proof given in A185417 for the zeros of the polynomial S(n,x)).
BINOMIAL EXPANSION
The form of the e.g.f. shows that {Z(n,x)} n >= 0 is a sequence of polynomials of binomial type. In particular, we have the expansion
(8) Z(n,x+y) = Sum_{k=0..n} binomial(n,k)*Z(k,x)*Z(n-k,y).
The delta operator D* associated with this binomial type sequence is
(9) D* = D - D^3/3! + 5*D^5/5! - 61*D^7/7! + 1385*D^9/9! - ..., and satisfies
the relation
(10) tan(D*)+sec(D*) = exp(D).
The delta operator D* acts as a lowering operator on the zigzag polynomials:
(11) (D*)Z(n,x) = n*Z(n-1,x).
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(12) Z(p,x) = (-1)^((p-1)/2)*x (mod p).
More generally, for k = 1,2,3,...
(13) Z(p+k-1,x) = (-1)^((p-1)/2)*Z(k,x) (mod p).
RELATIONS WITH OTHER SEQUENCES
Row sums [1,1,2,5,16,61,...] are the zigzag numbers A000111(n) for n >= 1.
Column 1 (with 0's omitted) is the sequence of Euler numbers A000364.
A145876(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1,k-j)*Z(n,j).
A147315(n-1,k-1) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A185421(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A012123(n) = (-i)^n*Z(n,i) where i = sqrt(-1). A012259(n) = 2^n*Z(n,1/2).
(End)
T(n,m) = Sum(i=0..n-m, s(i)/(n-i)!*Sum(k=m..n-i, A147315(n-i,k)*Stirling1(k,m))), m>0, T(n,0) = s(n), s(n)=[1,0,1,0,5,0,61,0,1385,0,50521,...] (see A000364). - Vladimir Kruchinin, Mar 10 2011

A147315 L-matrix for Euler numbers A000111(n+1).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 11, 6, 1, 16, 45, 35, 10, 1, 61, 211, 210, 85, 15, 1, 272, 1113, 1351, 700, 175, 21, 1, 1385, 6551, 9366, 5901, 1890, 322, 28, 1, 7936, 42585, 70055, 51870, 20181, 4410, 546, 36, 1, 50521, 303271, 563970, 479345, 218925, 58107, 9240, 870, 45, 1
Offset: 0

Views

Author

Paul Barry, Nov 05 2008

Keywords

Comments

This is the inverse of the coefficient array for the orthogonal polynomials p(n,x) defined by: p(n,x)=if(n=-1,0,if(n=0,1,(x-n)p(n-1,x)-C(n,2)p(n-2,x))).
The Hankel array H for A000111(n+1) satisfies H=L*D*U with U the transpose of L.
Row sums are A000772(n+1) with e.g.f. dif(exp(-1)exp(sec(x)+tan(x)),x).
From Peter Bala, Jan 31 2011: (Start)
The following comments refer to the table with an offset of 1: i.e., both the row and column indexing starts at 1.
An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A000111(n) for n>=1 enumerates the number of increasing unordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2 (plane unary-binary trees in the notation of [Bergeron et al.])
The entry T(n,k) of the present table gives the number of forests of k increasing unordered trees on the vertex set {1,2,...,n} in which all outdegrees are <=2. See below for some examples.
For ordered forests of such trees see A185421. For forests of increasing ordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2, see A185422.
The Stirling number of the second kind Stirling2(n,k) is the number of partitions of the set [n] into k blocks. Arranging the elements in each block in ascending numerical order provides an alternative combinatorial interpretation for Stirling2(n,k) as counting forests of k increasing unary trees on n nodes. Thus we may view the present array, which counts increasing unary-binary trees, as generalized Stirling numbers of the second kind associated with A000111 or with the zigzag polynomials Z(n,x) of A147309 - see especially formulas (2) and (3) below.
See A145876 for generalized Eulerian numbers associated with A000111. (End)
The Bell transform of A000111(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

Examples

			Triangle begins
    1;
    1,    1;
    2,    3,    1;
    5,   11,    6,   1;
   16,   45,   35,  10,   1;
   61,  211,  210,  85,  15,  1;
  272, 1113, 1351, 700, 175, 21, 1;
  ...
The production array for L is the tridiagonal array
  1,  1;
  1,  2,  1;
  0,  3,  3,  1;
  0,  0,  6,  4,  1;
  0,  0,  0, 10,  5,  1;
  0,  0,  0,  0, 15,  6,  1;
  0,  0,  0,  0,  0, 21,  7,  1;
  0,  0,  0,  0,  0,  0, 28,  8,  1,;
  0,  0,  0,  0,  0,  0,  0, 36,  9,  1;
From _Peter Bala_, Jan 31 2011: (Start)
Examples of forests:
The diagrams below are drawn so that the leftmost child of a binary node has the maximum label.
T(4,1) = 5. The 5 forests consisting of a single non-plane increasing unary-binary tree on 4 nodes are
...4... ........ .......... ........... ...........
...|... ........ .......... ........... ...........
...3... .4...3.. .4........ ........4.. ........3..
...|... ..\./... ..\....... ......./... ......./...
...2... ...2.... ...3...2.. ..3...2.... ..4...2....
...|... ...|.... ....\./... ...\./..... ...\./.....
...1... ...1.... .....1.... ....1...... ....1......
T(4,2) = 11. The 11 forests consisting of two non-plane increasing unary-binary trees on 4 nodes are
......... ...3.....
.3...2... ...|.....
..\./.... ...2.....
...1...4. ...|.....
......... ...1...4.
.
......... ...4.....
.4...2... ...|.....
..\./.... ...2.....
...1...3. ...|.....
......... ...1...3.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...1...2. ...|.....
......... ...1...2.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...2...1. ...|.....
......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... .......... (End)
		

Crossrefs

Programs

  • Maple
    A147315 := proc(n,k) n!*exp(x*(sec(t)+tan(t)-1)) - 1: coeftayl(%,t=0,n) ; coeftayl(%,x=0,k) ; end proc:
    seq(seq(A147315(n,k),k=1..n),n=0..12) ; # R. J. Mathar, Mar 04 2011
    # second Maple program:
    b:= proc(u, o) option remember;
          `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
        end:
    g:= proc(n) option remember; expand(`if`(n=0, 1, add(
          g(n-j)*x*binomial(n-1, j-1)*b(j, 0), j=1..n)))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..n+1))(g(n+1)):
    seq(T(n), n=0..10);  # Alois P. Heinz, May 19 2021
  • Mathematica
    t[n_, k_] := t[n, k] = t[n-1, k-1] + (k+1)*t[n-1, k] + 1/2*(k+1)*(k+2)*t[n-1, k+1]; t[n_, k_] /; (n < 0 || k < 0 || k > n) = 0; t[0, 0] = t[1, 0] = 1; Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 21 2011, after PARI prog. *)
  • Maxima
    Co(n,k):=sum(binomial(k,j)*(if oddp(n-k+j) then 0 else if (n-k+j)/2A147315(n,m):=1/m!*sum((if oddp(n-k) then 0 else 2^(1-k)*sum((-1)^(floor((n+k)/2)-i)*binomial(k,i)*(2*i-k)^n,i,0,floor(k/2)))*(sum(Co(i,m)*binomial(k-i+m-1,m-1),i,1,k)),k,m,n); /* Vladimir Kruchinin, Feb 17 2011 */
    
  • Maxima
    T(n,m):=(sum(binomial(k+m,m)*((-1)^(n-k-m)+1)*sum(binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)*stirling2(n+1,j+k+m+1), j,0,n-k-m), k,0,n-m))/(m+1)!; /* Vladimir Kruchinin, May 17 2011 */
    
  • PARI
    {T(n,k)=if(k<0||k>n,0,if(n==0,1,T(n-1,k-1)+(k+1)*T(n-1,k)+(k+1)*(k+2)/2*T(n-1,k+1)))} /* offset=0 */
    
  • PARI
    {T(n,k)=local(X=x+x*O(x^(n+2)));(n+1)!*polcoeff(polcoeff(exp(y*((1+sin(X))/cos(X)-1))-1,n+1,x),k+1,y)} /* offset=0 */
    
  • PARI
    /* Generate from the production matrix P: */
    {T(n,k)=local(P=matrix(n,n,r,c,if(r==c-1,1,if(r==c,c,if(r==c+1,c*(c+1)/2)))));if(k<0||k>n,0,if(n==k,1,(P^n)[1,k+1]))}
    
  • Sage
    # uses[bell_matrix from A264428, A000111]
    # Adds a column 1,0,0,0, ... at the left side of the triangle.
    bell_matrix(lambda n: A000111(n+1), 10) # Peter Luschny, Jan 18 2016

Formula

From Peter Bala, Jan 31 2011: (Start)
The following formulas refer to the table with an offset of 1: i.e., both the row n and column k indexing start at 1.
GENERATING FUNCTION
E.g.f.:
(1)... exp(x*(sec(t)+tan(t)-1)) - 1 = Sum_{n>=1} R(n,x)*t^n/n!
= x*t + (x+x^2)*t^2/2! + (2*x+3*x^2+x^3)*t^3/3! + ....
TABLE ENTRIES
(2)... T(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j),
where Z(n,x) denotes the zigzag polynomials as described in A147309.
Compare (2) with the formula for the Stirling numbers of the second kind
(3)... Stirling2(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^n.
Recurrence relation
(4)... T(n+1,k) = T(n,k-1) + k*T(n,k) + (1/2)*k(k+1)*T(n,k+1).
ROW POLYNOMIALS
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x+x^2
R(3,x) = 2*x+3*x^2+x^3
They satisfy the recurrence
(5)... R(n+1,x) = x*{R(n,x)+R'(n,x) + (1/2)*R''(n,x)},
where ' indicates differentiation with respect to x. This should be compared with the recurrence satisfied by the Bell polynomials Bell(n,x)
(6)... Bell(n+1,x) = x*(Bell(n,x) + Bell'(n,x)). (End)
From Vladimir Kruchinin, Feb 17 2011: (Start)
Sum_{m=1..n} T(n,m) = A000772(n).
Sum_{m=1..2n-1} T(2n-1,m)* Stirling1(m,1) = A000364(n).
Let Co(n,k) = Sum_{j=1..k} binomial(k,j)*(if (n-k+j) is odd then 0 else if (n-k+j)/2
T(n,m) = m!* Sum_{k=m..n} (if n-k is odd then 0 else 2^(1-k)) * Sum_{i=0..floor(k/2)} (-1)^(floor((n+k)/2)-i) * binomial(k,i) * (2*i-k)^n)))) * Sum_{i=1..k} Co(i,m) * binomial(k-i+m-1,m-1), n>0.
(End)
T(n,m) = Sum_{k = 0..n-m} binomial(k+m,m)*((-1)^(n-k-m)+1)*Sum_{j=0..n-k-m} binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)* Stirling2(n+1,j+k+m+1)/(m+1)!. - Vladimir Kruchinin, May 17 2011
The row polynomials R(n,x) are given by D^n(exp(x*t)) evaluated at t = 0, where D is the operator (1+t+t^2/2!)*d/dt. Cf. A008277 and A094198. See also A185422. - Peter Bala, Nov 25 2011

Extensions

More terms from Michel Marcus, Mar 01 2014

A185422 Forests of k increasing plane unary-binary trees on n nodes. Generalized Stirling numbers of the second kind associated with A185415.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 9, 15, 6, 1, 39, 75, 45, 10, 1, 189, 459, 330, 105, 15, 1, 1107, 3087, 2709, 1050, 210, 21, 1, 7281, 23535, 23814, 11109, 2730, 378, 28, 1, 54351, 197235, 228285, 122850, 36099, 6174, 630, 36, 1
Offset: 1

Author

Peter Bala, Jan 28 2011

Keywords

Comments

An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing.
A080635 enumerates increasing plane (ordered) unary-binary trees with n nodes labeled from the set {1,2,...n}. The entry T(n,k) of the present table counts forests of k increasing plane unary-binary trees having n nodes in total. See below for an example.
The Stirling number of the second kind Stirling2(n,k) counts the partitions of the set [n] set into k blocks. Arranging the elements in each block in ascending numerical order provides an alternative combinatorial interpretation for Stirling2(n,k) as counting forests of k increasing unary trees on n nodes. Thus we may view the present array as generalized Stirling numbers of the second kind (associated with A080635 or with the polynomials P(n,x) of A185415 - see formulas (1) and (2) below).
For a table of ordered forests of increasing plane unary-binary trees see A185423. For the enumeration of forests and ordered forests in the non-plane case see A147315 and A185421.
The Bell transform of A080635(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

Examples

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....1
..2|....1......1
..3|....3......3......1
..4|....9.....15......6......1
..5|...39.....75.....45.....10......1
..6|..189....459....330....105.....15......1
..7|.1107...3087...2709...1050....210.....21......1
..
Examples of the recurrence:
T(5,1) = 39 = T(4,0)+1*T(4,1)+2*T(4,2) = 1*9+2*15;
T(6,3) = 330 = T(5,2)+3*T(5,3)+3*4*T(5,4) = 75+3*45+12*10.
Examples of forests:
T(4,1) = 9. The 9 plane increasing unary-binary trees on 4 nodes are shown in the example section of A080635.
T(4,2) = 15. The 15 forests consisting of two plane increasing unary-binary trees on 4 nodes consist of the 12 forests
......... ......... ...3.....
.2...3... .3...2... ...|.....
..\./.... ..\./.... ...2.....
...1...4. ...1...4. ...|.....
......... ......... ...1...4.
.
......... ......... ...4.....
.2...4... .4...2... ...|.....
..\./.... ..\./.... ...2.....
...1...3. ...1...3. ...|.....
......... ......... ...1...3.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...1...2. ...1...2. ...|.....
......... ......... ...1...2.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...2...1. ...2...1. ...|.....
......... ......... ...2...1.
.
and the three remaining forests
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... ..........
		

References

  • F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, in Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1922, pp. 24-48.

Crossrefs

Programs

  • Maple
    #A185422
    P := proc(n,x)
    description 'polynomial sequence P(n,x) A185415'
    if n = 0
    return 1
    else
    return x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
    end proc:
    with combinat:
    T:= (n,k) -> 1/k!*add ((-1)^(k-j)*binomial(k,j)*P(n,j),j = 0..k):
    for n from 1 to 10 do
    seq(T(n,k),k = 1..n);
    end do;
  • Mathematica
    t[n_, k_] := t[n, k] = t[n-1, k-1] + k*t[n-1, k] + k*(k+1)*t[n-1, k+1]; t[n_, n_] = 1; t[n_, k_] /; Not[1 <= k <= n] = 0; Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 22 2012, from given recurrence *)
  • PARI
    {T(n,k)=if(n<1||k<1||k>n,0,if(n==k,1,T(n-1,k-1)+k*T(n-1,k)+k*(k+1)*T(n-1,k+1)))}
    
  • PARI
    {T(n,k)=round(n!*polcoeff(polcoeff(exp(y*(-1/2+sqrt(3)/2*tan(sqrt(3)/2*x+Pi/6+x*O(x^n)))+y*O(y^k)),n,x),k,y))}
    
  • Sage
    # uses[bell_matrix from A264428]
    # Adds a column 1,0,0,0, ... at the left side of the triangle.
    bell_matrix(lambda n: A080635(n+1), 10) # Peter Luschny, Jan 18 2016

Formula

TABLE ENTRIES
(1)... T(n,k) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*P(n,j),
where P(n,x) are the polynomials described in A185415.
Compare (1) with the formula for the Stirling numbers of the second kind
(2)... Stirling2(n,k) = 1/k!*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*j^n.
RECURRENCE RELATION
(3)... T(n+1,k) = T(n,k-1) + k*T(n,k) + k*(k+1)*T(n,k+1).
GENERATING FUNCTION
Let E(t) = 1/2 + sqrt(3)/2*tan(sqrt(3)/2*t + Pi/6) be the e.g.f. for A080635.
The e.g.f. for the present triangle is
(4)... exp{x*(E(t)-1)} = Sum_{n>=0} R(n,x)*t^n/n!
= 1 + x*t + (x+x^2)*t^2/2! + (3*x+3*x^2+x^3)*t^3/3! + ....
ROW POLYNOMIALS
The row generating polynomials R(n,x) satisfy the recurrence
(5)... R(n+1,x) = x*{R(n,x)+R'(n,x)+R''(n,x)},
where the prime ' indicates differentiation with respect to x.
RELATIONS WITH OTHER SEQUENCES
Column 1 is A080635.
k!*T(n,k) counts ordered forests A185423(n,k).
The row polynomials R(n,x) are given by D^n(exp(x*t)) evaluated at t = 0, where D is the operator (1+t+t^2)*d/dt. Cf. A147315 and A008297. - Peter Bala, Nov 25 2011

A185423 Ordered forests of k increasing plane unary-binary trees with n nodes.

Original entry on oeis.org

1, 1, 2, 3, 6, 6, 9, 30, 36, 24, 39, 150, 270, 240, 120, 189, 918, 1980, 2520, 1800, 720, 1107, 6174, 16254, 25200, 25200, 15120, 5040, 7281, 47070, 142884, 266616, 327600, 272160, 141120, 40320, 54351, 394470, 1369710, 2948400, 4331880
Offset: 1

Author

Peter Bala, Jan 28 2011

Keywords

Comments

An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A080635 enumerates increasing plane (ordered) unary-binary trees with n nodes with labels drawn from the set {1,2,...,n}. The entry T(n,k) of the present table counts ordered forests of k increasing plane unary-binary trees with n nodes. See below for an example. See A185421 for the corresponding table in the non-plane case. See A185422 for the table for unordered forests of increasing plane unary-binary trees.

Examples

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....1
..2|....1......2
..3|....3......6......6
..4|....9.....30.....36.....24
..5|...39....150....270....240....120
..6|..189....918...1980...2520...1800....720
..7|.1107...6174..16254..25200..25200..15120...5040
..
Examples of recurrence relation:
T(5,3) = 270 = 3*(T(4,2) + T(4,3) + T(4,4)) = 3*(30+36+24);
T(6,1) = 1*(T(5,0) + T(5,1) + T(5,2)) = 39+150.
Examples of ordered forests:
T(4,2) = 30. The 15 unordered forests consisting of two plane increasing unary-binary trees on 4 nodes are shown below. We can order the trees in a forest in two ways giving rise to 30 ordered forests.
......... ......... ...3.....
.2...3... .3...2... ...|.....
..\./.... ..\./.... ...2.....
...1...4. ...1...4. ...|.....
......... ......... ...1...4.
.
......... ......... ...4.....
.2...4... .4...2... ...|.....
..\./.... ..\./.... ...2.....
...1...3. ...1...3. ...|.....
......... ......... ...1...3.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...1...2. ...1...2. ...|.....
......... ......... ...1...2.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...2...1. ...2...1. ...|.....
......... ......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... ..........
		

Crossrefs

Programs

  • Maple
    #A185423
    P := proc(n,x)
    description 'polynomial sequence P(n,x) A185415'
    if n = 0 return 1
    else
    return x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
    end proc:
    with combinat:
    T:=(n,k) -> add ((-1)^(k-j)*binomial(k,j)*P(n,j),j = 0..k):
    for n from 1 to 10 do
    seq(T(n,k),k = 1..n);
    end do;
  • PARI
    {T(n,k)=if(n<1||k<1||k>n,0,if(n==1,if(k==1,1,0),k*(T(n-1,k-1)+T(n-1,k)+T(n-1,k+1))))}

Formula

TABLE ENTRIES
(1)... T(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*P(n,j),
where P(n,x) are the polynomials described in A185415.
Compare (1) with the formula for the number of ordered set partitions
(2)... A019538(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*j^n.
RECURRENCE RELATION
(3)... T(n+1,k) = k*(T(n,k-1) + T(n,k) + T(n,k+1))
GENERATING FUNCTION
Let E(t) = 1/2 + (sqrt(3)/2)*tan(sqrt(3)/2*t + Pi/6) be the e.g.f. for A080635.
The e.g.f. for the present triangle is
(4)... 1/(1 + x*(1 - E(t))) = Sum {n >= 0} R(n,x)*t^n/n!
= 1 + x*t + (x + 2*x^2)*t^2/2! + (3*x + 6*x^2 + 6*x^3)*t^3/3! + ....
ROW POLYNOMIALS
The ordered Bell polynomials OB(n,x) are the row polynomials of A019538 given by
(5)... OB(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.
By comparing the e.g.f.s for A019538 and the present table we obtain the surprising identity
(6)... (-i*sqrt(3))^(n-1)*OB(n,x)/x = R(n,y)/y,
where i = sqrt(-1) and x = (sqrt(3)*i/3)*y + (-1/2+sqrt(3)*i/6).
RELATIONS WITH OTHER SEQUENCES
Column 1: A080635.
A185422(n,k) = T(n,k)/k!.
Setting y = 0 in (6) gives
(7)... A080635(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k) * (-1/2+sqrt(3)*i/6)^(k-1).
The row polynomials have their zeros on the vertical line Re(z) = -1/2 in the complex plane (this follows from the above connection with the ordered Bell polynomials, which have negative real zeros). - Peter Bala, Oct 23 2023
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