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%I #706 Sep 23 2022 15:40:06
%S 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,
%T 1191,120,1,1,247,4293,15619,15619,4293,247,1,1,502,14608,88234,
%U 156190,88234,14608,502,1,1,1013,47840,455192,1310354,1310354,455192,47840,1013,1
%N Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.
%C 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
%C 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.
%C 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
%C 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
%C Subtriangle for k>=1 and n>=1 of triangle A123125. - _Philippe Deléham_, Oct 22 2006
%C 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
%C [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
%C From _Tom Copeland_, Oct 07 2008: (Start)
%C 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).
%C 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.).
%C 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.
%C (End)
%C 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
%C 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
%C For a natural refinement of A008292 with connections to compositional inversion and iterated derivatives, see A145271. - _Tom Copeland_, Nov 06 2008
%C 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
%C From Walther Janous (walther.janous(AT)tirol.com), Nov 01 2009: (Start)
%C 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:
%C S(p,x) = Sum_{j>=0} (j+1)^p*x^j, that is
%C S(p,x) = e(p,x)/(1-x)^(p+1), whenever |x| < 1 and p is a positive integer.
%C (Note the inconsistent use of T(n,k) in the section listing the formula section. I adhere tacitly to the first one.) (End)
%C 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
%C 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
%C 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
%C 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
%C 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
%C 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
%C 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
%C 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
%C 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
%C For relations to values of the Riemann zeta function at integral arguments, see A131758 and the Dupont reference. - _Tom Copeland_, Mar 19 2018
%C Normalized volumes of the hypersimplices, attributed to Laplace. (Cf. the De Loera et al. reference, p. 327.) - _Tom Copeland_, Jun 25 2018
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%H Takao Komatsu and Yuan Zhang, <a href="https://arxiv.org/abs/2101.04298">Weighted Sylvester sums on the Frobenius set in more variables</a>, arXiv:2101.04298 [math.NT], 2021. Mentions this sequence.
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%H M-H. Li and N-C. Wong, <a href="http://www.math.nsysu.edu.tw/~wong/papers/soa-SEAM-formatted.pdf">Sums of angles of star polygons and the Eulerian Numbers</a>, Southeast Asian Bulletin of Mathematics 2004.
%H A. Losev and Y. Manin, <a href="http://arxiv.org/abs/math/0001003">New moduli spaces of pointed curves and pencils of flat connections</a>, arXiv:0001003 [math.AG], 2000 (p. 8)
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%H Shi-Mei Ma, <a href="http://arxiv.org/abs/1304.6654">On gamma-vectors and the derivatives of the tangent and secant functions</a>, arXiv:1304.6654 [math.CO], 2013.
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%H Shi-Mei Ma, Qi Fang, Toufik Mansour, and Yeong-Nan Yeh, <a href="https://arxiv.org/abs/2104.09374">Alternating Eulerian polynomials and left peak polynomials</a>, arXiv:2104.09374, 2021
%H Shi-Mei Ma, Jun Ma, and Yeong-Nan Yeh, <a href="https://arxiv.org/abs/1802.02861">On certain combinatorial expansions of descent polynomials and the change of grammars</a>, arXiv:1802.02861 [math.CO], 2018.
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%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>
%H <a href="/index/Cor#core">Index entries for "core" sequences</a>
%F T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.
%F T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).
%F Row sums = n! = A000142(n) unless n=0. - _Michael Somos_, Mar 17 2011
%F 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
%F 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
%F From John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
%F Four characterizations of Eulerian numbers T(i, n):
%F 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).
%F 2. T(i, n) = Sum_{j=0..i} (-1)^j*binomial(n+1,j)*(i-j+1)^n for n>=1, i>=0.
%F 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.
%F 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)
%F O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x. - _Vladeta Jovovic_, Sep 02 2002
%F 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.
%F Sum_{k=1..n} T(n, k)*2^k = A000629(n). - _Philippe Deléham_, Jun 05 2004
%F From _Tom Copeland_, Oct 10 2007: (Start)
%F 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.
%F 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).
%F 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)).
%F (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)
%F 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
%F 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
%F 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
%F From _Peter Bala_, Sep 29 2011: (Start)
%F 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.
%F 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.
%F 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.
%F (End)
%F 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
%F 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
%F E.g.f.: (1-y) / (1 - y*exp( (1-y)*x )). - _Geoffrey Critzer_, Nov 10 2012
%F From _Peter Bala_, Mar 12 2013: (Start)
%F 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).
%F (End)
%F 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
%F From _Tom Copeland_, Sep 18 2014: (Start)
%F 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! + ...
%F 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).
%F 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).
%F 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).
%F 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 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)
%F 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
%F 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
%F For a fixed k, T(n,k) ~ k^n, proved by induction. - _Ran Pan_, Oct 12 2015
%F 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
%e The triangle T(n, k) begins:
%e n\k 1 2 3 4 5 6 7 8 9 10 ...
%e 1: 1
%e 2: 1 1
%e 3: 1 4 1
%e 4: 1 11 11 1
%e 5: 1 26 66 26 1
%e 6: 1 57 302 302 57 1
%e 7: 1 120 1191 2416 1191 120 1
%e 8: 1 247 4293 15619 15619 4293 247 1
%e 9: 1 502 14608 88234 156190 88234 14608 502 1
%e 10: 1 1013 47840 455192 1310354 1310354 455192 47840 1013 1
%e ... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
%e -----------------------------------------------------------------
%e 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
%e 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
%e 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
%e .
%e . 1o (1+t) 1o t 1o t
%e . | / \ / \
%e . | / \ / \
%e . 2o (1+t) 2o 3o 3o 2o
%e . |
%e . |
%e . 3o
%e .
%e The total number of trees is (1+t)^2 + t + t = 1 + 4*t + t^2.
%p 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:
%t t[n_, k_] = Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];
%t Flatten[Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* _Jean-François Alcover_, May 31 2011, after _Michael Somos_ *)
%t Flatten[Table[CoefficientList[(1-x)^(k+1)*PolyLog[-k, x]/x, x], {k, 1, 10}]] (* _Vaclav Kotesovec_, Aug 27 2015 *)
%t Table[Tally[
%t Count[#, x_ /; x > 0] & /@ (Differences /@
%t Permutations[Range[n]])][[;; , 2]], {n, 10}] (* _Li Han_, Oct 11 2020 *)
%o (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 */
%o (PARI) {T(n, k) = sum( j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))}; /* _Michael Somos_, Jul 19 1999 */
%o {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))}
%o (Haskell)
%o import Data.List (genericLength)
%o a008292 n k = a008292_tabl !! (n-1) !! (k-1)
%o a008292_row n = a008292_tabl !! (n-1)
%o a008292_tabl = iterate f [1] where
%o f xs = zipWith (+)
%o (zipWith (*) ([0] ++ xs) (reverse ks)) (zipWith (*) (xs ++ [0]) ks)
%o where ks = [1 .. 1 + genericLength xs]
%o -- _Reinhard Zumkeller_, May 07 2013
%o (Python)
%o from sympy import binomial
%o def T(n, k): return sum([(-1)**j*(k - j)**n*binomial(n + 1, j) for j in range(k + 1)])
%o for n in range(1, 11): print([T(n, k) for k in range(1, n + 1)]) # _Indranil Ghosh_, Nov 08 2017
%o (R)
%o T <- function(n, k) {
%o S <- numeric()
%o for (j in 0:k) S <- c(S, (-1)^j*(k-j)^n*choose(n+1, j))
%o return(sum(S))
%o }
%o for (n in 1:10){
%o for (k in 1:n) print(T(n,k))
%o } # _Indranil Ghosh_, Nov 08 2017
%o (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
%o (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
%o (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
%Y Columns k=2..8 are A000295, A000460, A000498, A000505, A000514, A001243, A001244.
%Y Cf. A019538, A028246, A048993, A048994, A049019, A086885, A090582, A129185, A131758, A139605, A173018.
%K nonn,tabl,nice,eigen,core,look
%O 1,5
%A _N. J. A. Sloane_, Mar 15 1996
%E Thanks to _Michael Somos_ for additional comments.
%E Further comments from _Christian G. Bower_, May 12 2000
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