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).
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.
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
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
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
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
![plots of exceedances for permutations of [4]](https://oeis.org/wiki/File:Exceedances_4.png){kind=link}
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