

A008292


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


260



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
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OFFSET

1,5


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 k1 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_{e1e2}s_{e2e3}s_{e3e4}...s_{e_{n1}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 ndimensional volume of the portion of the ndimensional hypercube cut by the (n1)dimensional hyperplanes x_1 + x_2 + ... x_n = k, x_1 + x_2 + ... x_n = k1; or, equivalently, it represents the probability that the sum of n independent random variables with uniform distribution between 0 and 1 is between k1 and k.  Stefano Zunino, Oct 25 2006
[E(.,t)/(1t)]^n = n!*Lag[n,P(.,t)/(1t)] and [P(.,t)/(1t)]^n = n!*Lag[n, E(.,t)/(1t)] 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 + (1exp(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 fpolynomials for the permutohedra (see A019538).
G(x,t1) = 1 + 1*x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ...
gives row polynomials for A008292, the hpolynomials for permutohedra.
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 nth row of this triangle is the hvector of the simplicial complex dual to a permutohedron of type A_(n1). The corresponding fvectors 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 hvectors of the simplicial complexes dual to permutohedra of type B). See A066094 for the array of hvectors 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^(k1)) 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..n1} T(n,k+1)*x^k turn out to be also the numerators of the closedform expressions of the infinite sums:
S(p,x):= Sum_{j>=0}(j+1)^p*x^j, that is
S(p,x) = e(p,x)/(1x)^(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 (n2)th and (n1)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 rth successive summation of Sum_{k=1..n} k^j which appears to be Sum_{k=1..n} T(j,k1) * binomial(jk+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*kn1)*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*kn1. 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 PaoliMarcolli 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 (012) 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


REFERENCES

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D. Foata, Distributions eulériennes et mahoniennes sur le groupe des permutations, pp. 2749 of M. Aigner, editor, Higher Combinatorics, Reidel, Dordrecht, Holland, 1977.
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D. E. Knuth, The Art of Computer Programming. AddisonWesley, Reading, MA, 1998, Vol. 3, p. 47 (exercise 5.1.4 Nr. 20) and p. 605 (solution).
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/
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, AddisonWesley, 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.


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Index entries for sequences related to rooted trees
Index entries for "core" sequences


FORMULA

T(n, k) = k * T(n1, k) + (nk+1) * T(n1, k1), T(1, 1) = 1.
T(n, k) = Sum_{j=0..k} (1)^j * (kj)^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, nth term: T(c, n) = c^(n+c1) + Sum_{i=1..c1} (1)^i/i!*(ci)^(n+c1)*Product_{j=1..i} (n+c+1j).  Randall L. Rathbun (randallr(AT)abac.com), 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) = (ni)T(i1, n1) + (i+1)T(i, n1).
2. T(i, n) = Sum_{j=0..i} (1)^j*binomial(n+1,j)*(ij+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/(r1)^(n+1)) Sum_{i=0..n1} 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 nth row: (1x)^(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, jn) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,x, jn)) = 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, jn)) 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, jn)).
Clarification (Apr 19 2014): Here E(0,0) = S2(0,0) = 1, and for k>0, E(0,k) = E(k,0) = S2(0,k) = S2(k,0) = 0. (End)
From the relations between the h and fpolynomials of permutohedra and reciprocals of e.g.f.s described in A049019: (t1)((t1)d/dx)^n 1/(texp(x)) evaluated at x=0 gives the nth Eulerian row polynomial in t and the nth row polynomial in (t1) 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/(1x/(1x*y/12x/(12x*y/(13x/(13x*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((n1)*(n2)/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,(t1)]1 in Copeland's 2008 comment, the compositional inverse is Ainv(x,t)= log(t(t1)/(1+x))/(t1).  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.: (1y) / (1  y*exp( (1y)*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 = (1i)/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*(y1)/(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))/(ab) = 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_(n1)(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/(1x).
F) FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = (x+y+(a+b)xy)/(1ab*xy) is called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline. (End)
For x > 1, the nth 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/(e1)^(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 nth 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_(nk). Then the mth bivariate row polynomial of this entry is P(m,a,b) = (1, 0, 0, 0,..) [VP * S]^(m1) (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


EXAMPLE

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 012 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.


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(n1, k)+(nk+1)*procname(n1, k1) ; end if; end proc:


MATHEMATICA

len = 54; m = Ceiling[2 Sqrt[len]]; t[n_, k_] = Sum[(1)^j*(k  j)^n*Binomial[n + 1, j], {j, 0, k}]; Flatten[Table[t[n, k], {n, 1, m}, {k, 1, n}]][[1 ;; len]] (* JeanFrançois Alcover, May 31 2011, after Michael Somos *)
Flatten[Table[CoefficientList[(1x)^(k+1)*PolyLog[k, x]/x, x], {k, 1, 10}]] (* Vaclav Kotesovec, Aug 27 2015 *)


PROG

(PARI) {T(n, k) = if( k<1  k>n, 0, if( n==1, 1, k * T(n1, k) + (nk+1) * T(n1, k1)))}; /* Michael Somos, Jul 19 1999 */
(PARI) {T(n, k) = sum( j=0, k, (1)^j * (kj)^n * binomial( n+1, j))}; /* Michael Somos, Jul 19 1999 */
{A008292(c, n)=c^(n+c1)+sum(i=1, c1, (1)^i/i!*(ci)^(n+c1)*prod(j=1, i, n+c+1j))}
(Haskell)
import Data.List (genericLength)
a008292 n k = a008292_tabl !! (n1) !! (k1)
a008292_row n = a008292_tabl !! (n1)
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


CROSSREFS

Sequence in context: A221987 A285357 A174526 * A174036 A157221 A146967
Adjacent sequences: A008289 A008290 A008291 * A008293 A008294 A008295


KEYWORD

nonn,tabl,nice,eigen,core,changed


AUTHOR

N. J. A. Sloane, Mar 15 1996


EXTENSIONS

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


STATUS

approved



