

A008292


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


212



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)



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
Comment from Stefano Zunino, Oct 25 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.
[ 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
Comment 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 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) = numer(sum((1)^(n+1)*k^n*z^(k1), k=1..infinity)) for n >=1 lead directly to the triangle of Eulerian numbers.  Johannes W. Meijer, May 24 2009
Contribution 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 n1) turn out to be also the numerators of the closedform 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)/(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^j,k=1..n) which appears to be sum(T(j,k1) * binomial(jk+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*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


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. 196197.
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.
D. Foata, Distributions eule'riennes et mahoniennes sur le groupe des permutations, pp. 2749 of M. Aigner, editor, Higher Combinatorics, Reidel, Dordrecht, Holland, 1977.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. AddisonWesley, Reading, MA, 1990, p. 254; 2nd. ed., p. 268.
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/
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.


LINKS

T. D. Noe, Rows 1 to 100 of triangle, flattened.
J. Fernando Barbero G., Jesús Salas, Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. I. General Structure, arXiv:1307.2010, 2013
J. F. Barbero G., J. Salas and E. J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. II. Applications, arXiv preprint arXiv:1307.5624, 2013
Paul Barry, Eulerian polynomials as moments, via exponential Riordan arrays, Arxiv preprint 2011.
Paul Barry, Combinatorial polynomials as moments, Hankel transforms and exponential Riordan arrays, Arxiv preprint 2011.
D. Barsky, Analyse padique et suites classiques de nombres, Sem. Loth. Comb. B05b (1981) 121.
Hacene Belbachir, Mourad Rahmani and B. Sury, Alternating Sums of the Reciprocals of Binomial Coefficients, Journal of Integer Sequences, Vol. 15 (2012), #12.2.8.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.C. Raoult, Springer 1992, pp. 2448.
F. Cachazo, S. He, E. Y. Yuan, Scattering in Three Dimensions from Rational Maps, arXiv preprint arXiv:1306.2962, 2013
L. Carlitz et al., Permutations and sequences with repetitions by number of increases, J. Combin. Theory, 1 (1966), 350374, p. 351.
L. Carlitz, Eulerian numbers and operators, Collectanea Mathematica 24:2 (1973), pp. 175200.
Mircea I. Cirnu, Determinantal formulas for sum of generalized arithmeticgeometric series, Boletin de la Asociacion Matematica Venezolana, Vol. XVIII, No. 1 (2011), p. 13.
J. Desarmenien and D. Foata, The signed Eulerian Numbers
J. Desarmenien and D. Foata, The signed Eulerian numbers, Discrete Math. 99 (1992), no. 13, 4958.
E. Deutsch and B. E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, J. Num. Theory 117 (2006), 191215.
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA]
A. Dzhumadil'daev, D. Yeliussizov, Power sums of binomial coefficients, Journal of Integer Sequences, 16 (2013), Article 13.1.6
R. Ehrenborg, M. Readdy, E. Steingrímsson, Mixed Volumes and Slices of the Cube, J Comb. Theory, Series A 81, Issue 1, Jan. 1998, 121126.
FindStat  Combinatorial Statistic Finder, The number of exceedances (also excedences) of a permutation
D. Foata, M. Schutzenberger.  Theorie Geometrique des Polynomes Euleriens, Lecture Notes in Math., no.138, Springer Verlag 1970. [From Peter Bala, Oct 26 2008]
Dominique Foata and GuoNiu Han, Doubloons and new qtangent numbers, Quart. J. Math. 62 (2) (2011) 417432
Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
S. Garoufalidis and R. Kashaev, From state integrals to qseries, arXiv preprint arXiv:1304.2705, 2013
Ira Gessel, The Smith College diploma problem.
Jim Haglund and Mirko Visontai, Stable multivariate Eulerian polynomials and generalized Stirling permutations.
Herwig Hauser, Christoph Koutschan, Multivariate linear recurrences and power series division, Discrete Math. 312 (2012), no. 24, 35533560. MR2979485.
P. Hitczenko and S. Janson, Weighted random staircase tableaux, arXiv preprint arXiv:1212.5498, 2012.
Matthew Hubbard and Tom Roby, Pascal's Triangle From Top to Bottom
Svante Janson, EulerFrobenius numbers and rounding, arXiv preprint arXiv:1305.3512, 2013
Lucas Kang, Investigation of Rule 73 as Case Study of Class 4 LongDistance Cellular Automata, arXiv preprint arXiv:1310.3311, 2013
A. Kerber and K.J. Thuerlings, Eulerian numbers, Foulkes characters and Lefschetz characters of S_n, Séminaire Lotharingien, Vol. 8 (1984), 3136.
A. R. Kräuter, \"Uber die Permanente gewisser zirkulärer Matrizen...
D. H. Lehmer, Generalized Eulerian numbers, J. Combin. Theory Ser.A 32 (1982), no. 2, 195215. MR0654621 (83k:10026).
MH. Li and NC. Wong, Sums of angles of star polygons and the Eulerian Numbers, Southeast Asian Bulletin of Mathematics 2004.
ShiMei Ma, Some combinatorial sequences associated with contextfree grammars, arXiv:1208.3104v2 [math.CO].  From N. J. A. Sloane, Aug 21 2012
ShiMei Ma, On gammavectors and the derivatives of the tangent and secant functions, arXiv:1304.6654
ShiMei Ma, A family of twovariable derivative polynomials for tangent and secant, El J. Combinat. 20 (1) (2013) P11
S.M. Ma, T. Mansour, M. Schork. Normal ordering problem and the extensions of the Stirling grammar, arXiv preprint arXiv:1308.0169, 2013
ShiMei Ma, T. Mansour, D. Callan, Some combinatorial arrays related to the LotkaVolterra system, arXiv preprint arXiv:1404.0731, 2014
P. A. MacMahon, The divisors of numbers, Proc. London Math. Soc., (2) 19 (1920), 305340; Coll. Papers II, pp. 267302.
R. Mantaci and F. Rakotondrajao, A permutation representation that knows what "Eulerian" means, Discrete Mathematics and Theoretical Computer Science, 4 101108, (2001)
S. Parker, The Combinatorics of Functional Composition and Inversion, Dissertation, Brandeis Univ. (1993) [From Tom Copeland, Nov 10 2008]
A. Randrianarivony and J. Zeng, Une famille de polynomes qui interpole plusieurs suites..., Adv. Appl. Math. 17 (1996), 126.
J. Sack and H. Ulfarsson, Refined inversion statistics on permutations, Arxiv preprint arXiv:1106.1995, 2011.
R. Sprugnoli, Alternating Weighted Sums of Inverses of Binomial Coefficients, J. Integer Sequences, 15 (2012), #12.6.3.
S. Tanimoto, A study of Eulerian numbers by means of an operator on permutations, Europ. J. Combin., 24 (2003), 3343.
Eric Weisstein's World of Mathematics, Eulerian Number, Euler's Number Triangle, Eulerian Number
Susanne Wienand, plots of exceedances for permutations of [4]
L. K. Williams, Enumeration of totally positive Grassmann cells, arxiv:math.CO/0307271
Tingyao Xiong, Jonathan I. Hall, and HungPing Tsao, Combinatorial Interpretation of General Eulerian Numbers, Journal of Discrete Mathematics, (2014), Article ID 870596, 6 pages.
D. Yeliussizov, Permutation Statistics on Multisets, Ph.D. Dissertation, Computer Science, KazakhBritish Technical University, 2012
Index entries for sequences related to rooted trees


FORMULA

T(n, k) = k * T(n1, k) + (nk+1) * T(n1, k1), T(1, 1) = 1.
T(n, k) = Sum (1)^j * (kj)^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, nth term: T(c, n)=c^(n+c1)+sum(i=1, c1, (1)^i/i!*(ci)^(n+c1)*prod(j=1, i, n+c+1j)).  Randall L. Rathbun (randallr(AT)abac.com), Jan 23 2002
Four characterizations of Eulerian numbers T(i, n) from John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
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 (n+1 combin 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}^{infinity} (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) * 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!*(xy)! ].
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, 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.: (y 1)/(y  exp( (y1)*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


EXAMPLE

Triangle begins:
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
...
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 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 *)


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

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+41=2, 126+6626+1=16, 1+1201191+24161191+1201=272 etc.) gives A000182.  Henry Bottomley, Jun 15 2001
Cf. A019538 (fvectors type A permutohedra), A047969 (Hilbert transform), A060187 (hvectors 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


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



