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

A001286 Lah numbers: a(n) = (n-1)*n!/2.

Original entry on oeis.org

1, 6, 36, 240, 1800, 15120, 141120, 1451520, 16329600, 199584000, 2634508800, 37362124800, 566658892800, 9153720576000, 156920924160000, 2845499424768000, 54420176498688000, 1094805903679488000, 23112569077678080000, 510909421717094400000
Offset: 2

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Author

Keywords

Comments

Number of surjections from {1,...,n} to {1,...,n-1}. - Benoit Cloitre, Dec 05 2003
First Eulerian transform of 0,1,2,3,4,... . - Ross La Haye, Mar 05 2005
With offset 0 : determinant of the n X n matrix m(i,j)=(i+j+1)!/i!/j!. - Benoit Cloitre, Apr 11 2005
These numbers arise when expressing n(n+1)(n+2)...(n+k)[n+(n+1)+(n+2)+...+(n+k)] as sums of squares: n(n+1)[n+(n+1)] = 6(1+4+9+16+ ... + n^2), n(n+1)(n+2)(n+(n+1)+(n+2)) = 36(1+(1+4)+(1+4+9)+...+(1+4+9+16+ ... + n^2)), n(n+1)(n+2)(n+3)(n+(n+1)+(n+2)+(n+3)) = 240(...), ... . - Alexander R. Povolotsky, Oct 16 2006
a(n) is the number of edges in the Hasse diagram for the weak Bruhat order on the symmetric group S_n. For permutations p,q in S_n, q covers p in the weak Bruhat order if p,q differ by an adjacent transposition and q has one more inversion than p. Thus 23514 covers 23154 due to the transposition that interchanges the third and fourth entries. Cf. A002538 for the strong Bruhat order. - David Callan, Nov 29 2007
a(n) is also the number of excedances in all permutations of {1,2,...,n} (an excedance of a permutation p is a value j such p(j)>j). Proof: j is exceeded (n-1)! times by each of the numbers j+1, j+2, ..., n; now, Sum_{j=1..n} (n-j)(n-1)! = n!(n-1)/2. Example: a(3)=6 because the number of excedances of the permutations 123, 132, 312, 213, 231, 321 are 0, 1, 1, 1, 2, 1, respectively. - Emeric Deutsch, Dec 15 2008
(-1)^(n+1)*a(n) is the determinant of the n X n matrix whose (i,j)-th element is 0 for i = j, is j-1 for j>i, and j for j < i. - Michel Lagneau, May 04 2010
Row sums of the triangle in A030298. - Reinhard Zumkeller, Mar 29 2012
a(n) is the total number of ascents (descents) over all n-permutations. a(n) = Sum_{k=1..n} A008292(n,k)*k. - Geoffrey Critzer, Jan 06 2013
For m>=4, a(m-2) is the number of Hamiltonian cycles in a simple graph with m vertices which is complete, except for one edge. Proof: think of distinct round-table seatings of m persons such that persons "1" and "2" may not be neighbors; the count is (m-3)(m-2)!/2. See also A001710. - Stanislav Sykora, Jun 17 2014
Popularity of left (right) children in treeshelves. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n. See A278677, A278678 or A278679 for more definitions and examples. See A008292 for the distribution of the left (right) children in treeshelves. - Sergey Kirgizov, Dec 24 2016

Examples

			G.f. = x^2 + 6*x^3 + 36*x^4 + 240*x^5 + 1800*x^6 + 15120*x^7 + 141120*x^8 + ...
a(10) = (1+2+3+4+5+6+7+8+9)*(1*2*3*4*5*6*7*8*9) = 16329600. - _Reinhard Zumkeller_, May 15 2010
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 90, ex. 4.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 156.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 44.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A002868 is an essentially identical sequence.
Column 2 of |A008297|.
Third column (m=2) of triangle |A111596(n, m)|: matrix product of |S1|.S2 Stirling number matrices.
Cf. also A000110, A000111.

Programs

Formula

a(n) = Sum_{i=0..n-1} (-1)^(n-i-1) * i^n * binomial(n-1,i). - Yong Kong (ykong(AT)curagen.com), Dec 26 2000 [corrected by Amiram Eldar, May 02 2022]
E.g.f.: x^2/[2(1-x)^2]. - Ralf Stephan, Apr 02 2004
a(n+1) = (-1)^(n+1)*det(M_n) where M_n is the n X n matrix M_(i,j)=max(i*(i+1)/2,j*(j+1)/2). - Benoit Cloitre, Apr 03 2004
Row sums of table A051683. - Alford Arnold, Sep 29 2006
5th binomial transform of A135218: (1, 1, 1, 25, 25, 745, 3145, ...). - Gary W. Adamson, Nov 23 2007
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j) then a(n)=(-1)^n*f(n,2,-2), (n>=2). - Milan Janjic, Mar 01 2009
a(n) = A000217(n-1)*A000142(n-1). - Reinhard Zumkeller, May 15 2010
a(n) = (n+1)!*Sum_{k=1..n-1} 1/(k^2+3*k+2). - Gary Detlefs, Sep 14 2011
Sum_{n>=2} 1/a(n) = 2*(2 - exp(1) - gamma + Ei(1)) = 1.19924064599..., where gamma = A001620 and Ei(1) = A091725. - Ilya Gutkovskiy, Nov 24 2016
a(n+1) = a(n)*n*(n+1)/(n-1). - Chai Wah Wu, Apr 11 2018
Sum_{n>=2} (-1)^n/a(n) = 2*(gamma - Ei(-1)) - 2/e, where e = A001113 and Ei(-1) = -A099285. - Amiram Eldar, May 02 2022

A278677 a(n) = Sum_{k=0..n} A011971(n, k)*(k + 1). The Aitken-Bell triangle considered as a linear transform applied to the positive numbers.

Original entry on oeis.org

1, 5, 23, 109, 544, 2876, 16113, 95495, 597155, 3929243, 27132324, 196122796, 1480531285, 11647194573, 95297546695, 809490850313, 7126717111964, 64930685865768, 611337506786061, 5940420217001199, 59502456129204083, 613689271227219015, 6510381400140132872
Offset: 0

Views

Author

Sergey Kirgizov, Nov 26 2016

Keywords

Comments

Original name: Popularity of left children in treeshelves avoiding pattern T231 (with offset 2).
Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Classical Françon's bijection maps bijectively treeshelves into permutations. Pattern T231 illustrated below corresponds to a treeshelf constructed from permutation 231. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n.
a(n) is also the sum of the last entries in all blocks of all set partitions of [n-1]. a(4) = 23 because the sum of the last entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 3+5+5+4+6 = 23. - Alois P. Heinz, Apr 24 2017
a(n-2) is the number of lines that rhyme (with at least one earlier line) across all rhyme schemes counted by A000110. - Martin Fuller, Apr 20 2025

Examples

			Treeshelves of size 3:
      1  1          1    1       1        1
     /    \        /      \     / \      / \
    2      2      /        \   2   \    /   2
   /        \    2          2       3  3
  3          3    \        /
                   3      3
Pattern T231:
     1
    /
   /
  2
   \
    3
Treeshelves of size 3 that avoid pattern T231:
      1  1      1       1        1
     /    \      \     / \      / \
    2      2      \   2   \    /   2
   /        \      2       3  3
  3          3    /
                 3
Popularity of left children here is 5.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, m) option remember; `if`(n=0, [1, 0],
         (p-> p+[0, p[1]*n])(b(n-1, m+1))+m*b(n-1, m))
        end:
    a:= n-> b(n+1, 0)[2]:
    seq(a(n), n=0..22);  # Alois P. Heinz, Dec 15 2023
    # Using the generating function:
    gf := ((exp(z + exp(z)-1)*(z-1)) + exp(exp(z)-1))/z^2: ser := series(gf, z, 25):
    seq((n+2)!*coeff(ser, z, n), n=0..22);  # Peter Luschny, Feb 01 2025
  • Mathematica
    a[n_] := (n+3) BellB[n+2] - BellB[n+3];
    Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Dec 01 2018 *)
  • Python
    from sympy import bell
    HOW_MANY = 30
    print([(n + 3) * bell(n+2) - bell(n + 3) for n in range(HOW_MANY)])

Formula

E.g.f.: ((exp(z + exp(z)-1)*(z-1)) + exp(exp(z)-1))/z^2.
a(n) = (n + 3)*b(n + 2) - b(n + 3) where b(n) is the n-th Bell number (see A000110).
Asymptotics: a(n) ~ n*b(n).
a(n) = Sum_{k=1..n+1} A285595(n+1,k)/k. - Alois P. Heinz, Apr 24 2017
a(n) = Sum_{k=0..n} Stirling2(n+2, k+1) * (n+1-k). - Ilya Gutkovskiy, Apr 06 2021
a(n) ~ n*Bell(n)*(1 - 1/LambertW(n)). - Vaclav Kotesovec, Jul 28 2021
a(n) = Sum_{k=n+1..(n+1)*(n+2)/2} k * A367955(n+1,k). - Alois P. Heinz, Dec 11 2023

Extensions

New name and offset 0 by Peter Luschny, Feb 01 2025

A278678 Popularity of left children in treeshelves avoiding pattern T321.

Original entry on oeis.org

1, 4, 19, 94, 519, 3144, 20903, 151418, 1188947, 10064924, 91426347, 887296422, 9164847535, 100398851344, 1162831155151, 14198949045106, 182317628906283, 2455925711626404, 34632584722468115, 510251350142181470, 7840215226100517191, 125427339735162102104
Offset: 2

Views

Author

Sergey Kirgizov, Nov 26 2016

Keywords

Comments

Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Classical Françon's bijection maps bijectively treeshelves into permutations. Pattern T321 illustrated below corresponds to a treeshelf constructed from permutation 321. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n.

Examples

			Treeshelves of size 3:
      1  1          1    1       1        1
     /    \        /      \     / \      / \
    2      2      /        \   2   \    /   2
   /        \    2          2       3  3
  3          3    \        /
                   3      3
Pattern T321:
      1
     /
    2
   /
  3
Treeshelves of size 3 that avoid pattern T321:
  1          1    1       1        1
   \        /      \     / \      / \
    2      /        \   2   \    /   2
     \    2          2       3  3
      3    \        /
            3      3
Popularity of left children is 4.
		

Crossrefs

Programs

  • Maple
    b:= proc(u, o) option remember; `if`(u+o=0, 1,
          add(b(o-1+j, u-j), j=1..u))
        end:
    a:= n-> (n+1)*b(n+1, 0)-b(n+2, 0):
    seq(a(n), n=2..25);  # Alois P. Heinz, Oct 27 2017
  • Mathematica
    b[u_, o_] := b[u, o] = If[u+o == 0, 1, Sum[b[o-1+j, u-j], {j, 1, u}]];
    a[n_] := (n+1)*b[n+1, 0] - b[n+2, 0];
    Table[a[n], {n, 2, 25}] (* Jean-François Alcover, Nov 06 2017, after Alois P. Heinz *)
  • Python
    # by Taylor expansion
    from sympy import *
    from sympy.abc import z
    h = (-sin(z) + 1 + (z-1)*cos(z))/ (1-sin(z))**2
    NUMBER_OF_COEFFS = 20
    coeffs = Poly(series(h,n = NUMBER_OF_COEFFS)).coeffs()
    coeffs.reverse()
    # and remove first coefficient 1 that corresponds to O(n**k)
    coeffs.pop(0)
    print([coeffs[n]*factorial(n+2) for n in range(len(coeffs))])

Formula

E.g.f.: (-sin(z) + 1 + (z-1)*cos(z))/ (1-sin(z))^2.
a(n) = (n+1)*e(n) - e(n+1), where e(n) is the n-th Euler number (see A000111).
Asymptotic: a(n) ~ 8*(Pi-2) / Pi^3 * n^2 * (2/Pi)^n.

A131178 Non-plane increasing unary binary (0-1-2) trees where the nodes of outdegree 1 come in 2 colors.

Original entry on oeis.org

1, 2, 5, 16, 64, 308, 1730, 11104, 80176, 643232, 5676560, 54650176, 569980384, 6401959328, 77042282000, 988949446144, 13488013248256, 194780492544512, 2969094574403840, 47640794742439936, 802644553810683904, 14166772337295285248, 261410917571703825920
Offset: 1

Views

Author

Wenjin Woan, Oct 31 2007

Keywords

Comments

A labeled tree of size n is a rooted tree on n nodes that are labeled by distinct integers from the set {1,...,n}. An increasing tree is a labeled tree such that the sequence of labels along any branch starting at the root is increasing. Thus the root of an increasing tree will be labeled 1. In unary binary trees (sometimes called 0-1-2 trees) the outdegree of a node is either 0, 1 or 2. Here we are counting non-plane (where the subtrees stemming from a node are not ordered between themselves) increasing unary binary trees where the nodes of outdegree 1 come in two colors. An example is given below. - Peter Bala, Sep 01 2011
The number of plane increasing 0-1-2 trees on n nodes, where the nodes of outdegree 1 come in two colors, is equal to n!. Other examples of sequences counting increasing trees include A000111, A000670, A008544, A008545, A029768 and A080635. - Peter Bala, Sep 01 2011
Number of plane increasing 0-1-2 trees, where the nodes of outdegree 1 come in 2 colors, avoiding pattern T213. See A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 16*x^4 + 64*x^5 + 308*x^6 + 1730*x^7 + 11104*x^8 + ...
a(3) = 5: Denoting the two types of node of outdegree 1 by the letters a or b, the 5 possible trees are
.
.  1a    1b    1a    1b      1
.  |     |     |     |      / \
.  2a    2b    2b    2a    2   3
.  |     |     |     |
.  3     3     3     3
- _Peter Bala_, Sep 01 2011
		

Crossrefs

Programs

  • Maple
    E:=  (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)):
    S:= map(simplify,series(E,x,101)):
    seq(coeff(S,x,j)*j!, j=1..100); # Robert Israel, Nov 23 2016
  • Mathematica
    max = 25; f[x_] := (2*(Exp[Sqrt[2]*x] - 1))/((2 + Sqrt[2]) - (2 - Sqrt[2])*Exp[Sqrt[2]*x]); Drop[ Simplify[ CoefficientList[ Series[f[x], {x, 0, max}], x]*Range[0, max]!], 1] (* Jean-François Alcover, Oct 05 2011 *)
  • PARI
    x='x+O('x^66); /* that many terms */
    default(realprecision,1000); /* working with floats here */
    egf=(2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x));
    round(Vec(serlaplace(egf))) /* show terms */
    /* Joerg Arndt, Sep 01 2011 */
    
  • PARI
    /* the following program should be preferred. */
    Vec( serlaplace( serreverse( intformal( 1/(1+2*x+1/2*x^2) + O(x^66) ) ) ) )
    \\ Joerg Arndt, Mar 01 2014
    
  • PARI
    {a(n) = if( n<1, 0, n! * polcoeff( 2 / (-2 + quadgen(8) * (-1 + 2 / (1 - exp(-quadgen(8) * x + x * O(x^n))))), n))};

Formula

E.g.f.: A(x) = (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)) = x+2*x^2/2!+5*x^3/3!+16*x^4/4!+64*x^5/5!+....
From Peter Bala, Sep 01 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A' = 1+2*A+1/2*A^2 with A(0) = 0. It follows that the inverse function A(x)^-1 may be expressed as an integral A(x)^-1 = int {t = 0..x} 1/(1+2*t+1/2*t^2).
Applying [Dominici, Theorem 4.1] to invert the integral gives the following method for calculating the terms of the sequence: let f(x) = 1+2*x+1/2*x^2. Let D be the operator f(x)*d/dx. Then a(n) = D^n(f(x)) evaluated at x = 0. Compare with A000111(n+1) = D^n(1+x+x^2/2!) evaluated at x = 0.
(End)
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ) and m=1; (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - 1/2*x^2*(k+1)*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2013
a(n) ~ n! * 2^((n+3)/2) / log(3+2*sqrt(2))^(n+1). - Vaclav Kotesovec, Oct 08 2013
G.f.: conjecture: T(0)/(1-2*x) -1, where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-2*x*(k+1))*(1-2*x*(k+2))/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2013
E.g.f.: x/(T(0)-x), where T(k) = 4*k + 1 + x^2/(8*k+6 + x^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2013

Extensions

Terms >= 80176 from Peter Bala, Sep 01 2011
Changed offset to 1 to agree with name and example. - Michael Somos, Nov 23 2016
Showing 1-5 of 5 results.