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A008292 Triangle of Eulerian numbers T(n,k) (n>=1, 1 <= k <= n) read by rows. 215
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 (list; table; graph; refs; listen; history; text; internal format)



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.

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.


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) = numer(sum((-1)^(n+1)*k^n*z^(k-1), k=1..infinity)) 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(T(n,k+1)*x^k, k = 0 to n-1) turn out to be also the numerators of the closed-form expressions of the infinite sums:

S(p,x):= SUM((j+1)^p*x^j, j = 0 to infinity), 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^j,k=1..n) which appears to be sum(T(j,k-1) * binomial(j-k+n+r,j+r), k=1..n). - 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


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Index entries for sequences related to rooted trees


T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.

T(n, k) = Sum (-1)^j * (k-j)^n * C(n+1, j), j=0..k.

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)*prod(j=1, i, n+c+1-j)). - 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) = (n-i)T(i-1, n-1) + (i+1)T(i, n-1).

2. T(i, n) = sum_{j=0}^{i} (-1)^j (n+1 combin 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}^{infinity} (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) * C(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; Lag(n,x,m), the associated Laguerre polynomials of order m; and C(x,y) = x!/[ y!*(x-y)! ].

For x = 0, the equation gives sum(j=0,...,n) E(n,j) * C(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!*C(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = sum(j=0,...,n) E(n,j) * C(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)]}.

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 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/1-2x/(1-2x*y/(1-3x/(1-3x*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.


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.: (y -1)/(y - exp( (y-1)*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).


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+a^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)


Triangle begins:












E.g.f. = (q) * x^1 / 1! + (q + q^2) * x^2 / 2! + (q + 4*q^2 + q^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



....|............... /.\........../.\..........

....|.............. /...\......../...\.........






The total number of trees is (1+t)^2+t+t = 1+4*t+t^2.


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:


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]] (* Jean-François Alcover, May 31 2011, after Michael Somos *)


(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 */

{A008292(c, n)=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))}


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


See A173018 and A123125 for other ways to index the entries of this triangle.

Cf. A000142, A014630, A030196, A055325.

Cf. A006551, A025585, A180056, A180057, A169972, A027656 A084938, A049061, A008275, A008277, A087127.

Columns 2 through 8: A000295, A000460, A000498, A000505, A000514, A001243, A001244.

Cf. A062253, noting that A008292 gives the (unsigned) polynomial coefficients of (kn). Also note that taking alternating differences of rows with an odd number of terms (e.g., 1=1, -1+4-1=2, 1-26+66-26+1=16, -1+120-1191+2416-1191+120-1=272, etc.) gives A000182. - Henry Bottomley, Jun 15 2001

Cf. A019538 (f-vectors type A permutohedra), A047969 (Hilbert transform), A060187 (h-vectors type B permutohedra), A066094. - Peter Bala, Oct 26 2008

Cf. A242783, A242784.

Sequence in context: A168287 A221987 A174526 * A174036 A157221 A146967

Adjacent sequences:  A008289 A008290 A008291 * A008293 A008294 A008295




N. J. A. Sloane, Mar 15 1996


Thanks to Michael Somos for additional comments.

Further comments from Christian G. Bower, May 12 2000



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Last modified December 18 23:18 EST 2014. Contains 252175 sequences.