cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A130534 Triangle T(n,k), 0 <= k <= n, read by rows, giving coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in increasing powers of x. T(n,k) is also the unsigned Stirling number |s(n+1, k+1)|, denoting the number of permutations on n+1 elements that contain exactly k+1 cycles.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 6, 1, 24, 50, 35, 10, 1, 120, 274, 225, 85, 15, 1, 720, 1764, 1624, 735, 175, 21, 1, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

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Author

Philippe Deléham, Aug 09 2007

Keywords

Comments

This triangle is an unsigned version of the triangle of Stirling numbers of the first kind, A008275, which is the main entry for these numbers. - N. J. A. Sloane, Jan 25 2011
Or, triangle T(n,k), 0 <= k <= n, read by rows given by [1,1,2,2,3,3,4,4,5,5,6,6,...] DELTA [1,0,1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938.
Reversal of A094638.
Equals A132393*A007318, as infinite lower triangular matrices. - Philippe Deléham, Nov 13 2007
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the exponential integrals E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + n*(n+1)/x^2 - n*(n+1)*(n+2)/x^3 + ...), see Abramowitz and Stegun. This formula follows from the general formula for the asymptotic expansion, see A163932. We rewrite E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) and observe that the T(n,m) are the polynomials coefficients in the denominators. Looking at the a(n,m) formula of A028421, A163932 and A163934, and shifting the offset given above to 1, we can write T(n-1,m-1) = a(n,m) = (-1)^(n+m)*Stirling1(n,m), see the Maple program.
The asymptotic expansion leads for values of n from one to eleven to known sequences, see the cross-references. With these sequences one can form the triangles A008279 (right-hand columns) and A094587 (left-hand columns).
See A163936 for information about the o.g.f.s. of the right-hand columns of this triangle.
(End)
The number of elements greater than i to the left of i in a permutation gives the i-th element of the inversion vector. (Skiena-Pemmaraju 2003, p. 69.) T(n,k) is the number of n-permutations that have exactly k 0's in their inversion vector. See evidence in Mathematica code below. - Geoffrey Critzer, May 07 2010
T(n,k) counts the rooted trees with k+1 trunks in forests of "naturally grown" rooted trees with n+2 nodes. This corresponds to sums of coefficients of iterated derivatives representing vectors, Lie derivatives, or infinitesimal generators for flow fields and formal group laws. Cf. links in A139605. - Tom Copeland, Mar 23 2014
A refinement is A036039. - Tom Copeland, Mar 30 2014
From Tom Copeland, Apr 05 2014: (Start)
With initial n=1 and row polynomials of T as p(n,x)=x(x+1)...(x+n-1), the powers of x correspond to the number of trunks of the rooted trees of the "naturally-grown" forest referred to above. With each trunk allowed m colors, p(n,m) gives the number of such non-plane colored trees for the forest with each tree having n+1 vertices.
p(2,m) = m + m^2 = A002378(m) = 2*A000217(m) = 2*(first subdiag of |A238363|).
p(3,m) = 2m + 3m^2 + m^3 = A007531(m+2) = 3*A007290(m+2) = 3*(second subdiag A238363).
p(4,m) = 6m + 11m^2 + 6m^3 + m^4 = A052762(m+3) = 4*A033487(m) = 4*(third subdiag).
From the Joni et al. link, p(n,m) also represents the disposition of n distinguishable flags on m distinguishable flagpoles.
The chromatic polynomial for the complete graph K_n is the falling factorial, which encodes the colorings of the n vertices of K_n and gives a shifted version of p(n,m).
E.g.f. for the row polynomials: (1-y)^(-x).
(End)
A relation to derivatives of the determinant |V(n)| of the n X n Vandermonde matrix V(n) in the indeterminates c(1) thru c(n):
|V(n)| = Product_{1<=jTom Copeland, Apr 10 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A094587 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well defined). See the Example section. (End)
For the relation of this rising factorial to the moments of Viennot's Laguerre stories, see the Hetyei link, p. 4. - Tom Copeland, Oct 01 2015
Can also be seen as the Bell transform of n! without column 0 (and shifted enumeration). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle  T(n,k) begins:
n\k         0        1        2       3       4      5      6     7    8  9 10
n=0:        1
n=1:        1        1
n=2:        2        3        1
n=3:        6       11        6       1
n=4:       24       50       35      10       1
n=5:      120      274      225      85      15      1
n=6:      720     1764     1624     735     175     21      1
n=7:     5040    13068    13132    6769    1960    322     28     1
n=8:    40320   109584   118124   67284   22449   4536    546    36    1
n=9:   362880  1026576  1172700  723680  269325  63273   9450   870   45  1
n=10: 3628800 10628640 12753576 8409500 3416930 902055 157773 18150 1320 55  1
[Reformatted and extended by _Wolfdieter Lang_, Feb 05 2013]
T(3,2) = 6 because there are 6 permutations of {1,2,3,4} that have exactly 2 0's in their inversion vector: {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {2, 1, 3, 4},{2, 3, 1, 4}, {2, 3, 4, 1}. The respective inversion vectors are {0, 0, 1}, {0, 1, 0}, {0, 2, 0}, {1, 0, 0}, {2, 0, 0}, {3, 0, 0}. - _Geoffrey Critzer_, May 07 2010
T(3,1)=11 since there are exactly 11 permutations of {1,2,3,4} with exactly 2 cycles, namely, (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(143), (12)(34), (13)(24), and (14)(23). - _Dennis P. Walsh_, Jan 25 2011
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
  / 1          \/1        \/1        \      / 1           \
  | 1  1       ||0 1      ||0 1      |      | 1  1        |
  | 2  2  1    ||0 1 1    ||0 0 1    |... = | 2  3  1     |
  | 6  6  3 1  ||0 2 2 1  ||0 0 1 1  |      | 6 11  6  1  |
  |24 24 12 4 1||0 6 6 3 1||0 0 2 2 1|      |24 50 35 10 1|
  |...         ||...      ||...      |      |...          |
(End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 93-94.
  • Sriram Pemmaraju and Steven Skiena, Computational Discrete Mathematics, Cambridge University Press, 2003, pp. 69-71. [Geoffrey Critzer, May 07 2010]

Crossrefs

See A008275, which is the main entry for these numbers; A094638 (reversed rows).
From Johannes W. Meijer, Oct 07 2009: (Start)
Row sums equal A000142.
The asymptotic expansions lead to A000142 (n=1), A000142(n=2; minus a(0)), A001710 (n=3), A001715 (n=4), A001720 (n=5), A001725 (n=6), A001730 (n=7), A049388 (n=8), A049389 (n=9), A049398 (n=10), A051431 (n=11), A008279 and A094587.
Cf. A163931 (E(x,m,n)), A028421 (m=2), A163932 (m=3), A163934 (m=4), A163936.
(End)
Cf. A136662.

Programs

  • Haskell
    a130534 n k = a130534_tabl !! n !! k
    a130534_row n = a130534_tabl !! n
    a130534_tabl = map (map abs) a008275_tabl
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    with(combinat): A130534 := proc(n,m): (-1)^(n+m)*stirling1(n+1,m+1) end proc: seq(seq(A130534(n,m), m=0..n), n=0..10); # Johannes W. Meijer, Oct 07 2009, revised Sep 11 2012
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0, ..) as column 0 (and shifts the enumeration).
    BellMatrix(n -> n!, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    Table[Table[ Length[Select[Map[ToInversionVector, Permutations[m]], Count[ #, 0] == n &]], {n, 0, m - 1}], {m, 0, 8}] // Grid (* Geoffrey Critzer, May 07 2010 *)
    rows = 10;
    t = Range[0, rows]!;
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)

Formula

T(0,0) = 1, T(n,k) = 0 if k > n or if n < 0, T(n,k) = T(n-1,k-1) + n*T(n-1,k). T(n,0) = n! = A000142(n). T(2*n,n) = A129505(n+1). Sum_{k=0..n} T(n,k) = (n+1)! = A000142(n+1). Sum_{k=0..n} T(n,k)^2 = A047796(n+1). T(n,k) = |Stirling1(n+1,k+1)|, see A008275. (x+1)(x+2)...(x+n) = Sum_{k=0..n} T(n,k)*x^k. [Corrected by Arie Bos, Jul 11 2008]
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. - Philippe Deléham, Nov 13 2007
For k=1..n, let A={a_1,a_2,...,a_k} denote a size-k subset of {1,2,...,n}. Then T(n,n-k) = Sum(Product_{i=1..k} a_i) where the sum is over all subsets A. For example, T(4,1)=50 since 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50. - Dennis P. Walsh, Jan 25 2011
The preceding formula means T(n,k) = sigma_{n-k}(1,2,3,..,n) with the (n-k)-th elementary symmetric function sigma with the indeterminates chosen as 1,2,...,n. See the Oct 24 2011 comment in A094638 with sigma called there a. - Wolfdieter Lang, Feb 06 2013
From Gary W. Adamson, Jul 08 2011: (Start)
n-th row of the triangle = top row of M^n, where M is the production matrix:
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 4, 6, 4, 1;
... (End)
Exponential Riordan array [1/(1 - x), log(1/(1 - x))]. Recurrence: T(n+1,k+1) = Sum_{i=0..n-k} (n + 1)!/(n + 1 - i)!*T(n-i,k). - Peter Bala, Jul 21 2014

A008517 Second-order Eulerian triangle T(n,k), 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 8, 6, 1, 22, 58, 24, 1, 52, 328, 444, 120, 1, 114, 1452, 4400, 3708, 720, 1, 240, 5610, 32120, 58140, 33984, 5040, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880
Offset: 1

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Author

Keywords

Comments

Second-order Eulerian numbers <> = T(n,k+1) count the permutations of the multiset {1,1,2,2,...,n,n} with k ascents with the restriction that for all m, all integers between the two copies of m are less than m. In particular, the two 1s are always next to each other.
When seen as coefficients of polynomials with descending exponents, evaluations are in A000311 (x=2) and A001662 (x=-1).
The row reversed triangle is A112007. There one can find comments on the o.g.f.s for the diagonals of the unsigned Stirling1 triangle |A008275|.
Stirling2(n,n-k) = Sum_{m=0..k-1} T(k,m+1)*binomial(n+k-1+m, 2*k), k>=1. See the Graham et al. reference p. 271 eq. (6.43).
This triangle is the coefficient triangle of the numerator polynomials appearing in the o.g.f. for the k-th diagonal (k >= 1) of the Stirling2 triangle A048993.
The o.g.f. for column k satisfies the recurrence G(k,x) = x*(2*x*(d/dx)G(k-1,x) + (2-k)*G(k-1,x))/(1-k*x), k >= 2, with G(1,x) = 1/(1-x). - Wolfdieter Lang, Oct 14 2005
This triangle is in some sense generated by the differential equation y' = 1 - 2/(1+x+y). (This is the differential equation satisfied by the function defined implicitly as x+y=exp(x-y).) If we take y = a(0) + a(1)x + a(2)x^2 + a(3)x^3 + ... and assume a(0)=c then all the a's may be calculated formally in terms of c and we have a(1) = (c-1)/(c+1) and, for n > 1, a(n) = 2^n/n! (1+c)^(1-2n)( T(n,1)c - T(n,2)c^2 + T(n,3)c^3 - ... + (-1)^(n-1) T(n,n)c^n ). - Moshe Shmuel Newman, Aug 08 2007
From the recurrence relation, the generating function F(x,y) := 1 + Sum_{n>=1, 1<=k<=n} [T(n,k)x^n/n!*y^k] satisfies the partial differential equation F = (1/y-2x)F_x + (y-1)F_y, with (non-elementary) solution F(x,y) = (1-y)/(1-Phi(w)) where w = y*exp(x(y-1)^2-y) and Phi(x) is defined by Phi(x) = x*exp(Phi(x)). By Lagrange inversion (see Wilf's book "generatingfunctionology", page 168, Example 1), Phi(x) = Sum_{n>=1} n^(n-1)*x^n/n!. Thus Phi(x) can alternatively be described as the e.g.f. for rooted labeled trees on n vertices A000169. - David Callan, Jul 25 2008
A method for solving PDEs such as the one above for F(x,y) is described in the Klazar reference (see pages 207-208). In his case, the auxiliary ODE dy/dx = b(x,y)/a(x,y) is exact; in this case it is not exact but has an integrating factor depending on y alone, namely y-1. The e.g.f. for the row sums A001147 is 1/sqrt(1-2*x) and the demonstration that F(x,1) = 1/sqrt(1-2*x) is interesting: two applications of l'Hopital's rule to lim_{y->1}F(x,y) yield F(x,1) = 1/(1-2x)*1/F(x,1). So l'Hopital's rule doesn't directly yield F(x,1) but rather an equation to be solved for F(x,1)!. - David Callan, Jul 25 2008
From Tom Copeland, Oct 12 2008; May 19 2010: (Start)
Let P(0,t)= 0, P(1,t)= 1, P(2,t)= t, P(3,t)= t + 2 t^2, P(4,t)= t + 8 t^2 + 6 t^3, ... be the row polynomials of the present array, then
exp(x*P(.,t)) = ( u + Tree(t*exp(u)) ) / (1-t) = WD(x*(1-t), t/(1-t)) / (1-t)
where u = x*(1-t)^2 - t, Tree(x) is the e.g.f. of A000169 and WD(x,t) is the e.g.f. for A134991, relating the Ward and 2-Eulerian polynomials by a simple transformation.
Note also apparently P(4,t) / (1-t)^3 = Ward Poly(4, t/(1-t)) = essentially an e.g.f. for A093500.
The compositional inverse of f(x,t) = exp(P(.,t)*x) about x=0 is
g(x,t) = ( x - (t/(1-t)^2)*(exp(x*(1-t))-x*(1-t)-1) )
= x - t*x^2/2! - t*(1-t)*x^3/3! - t*(1-t)^2*x^4/4! - t*(1-t)^3*x^5/5! - ... .
Can apply A134685 to these coefficients to generate f(x,t). (End)
Triangle A163936 is similar to the one given above except for an extra right hand column [1, 0, 0, 0, ... ] and that its row order is reversed. - Johannes W. Meijer, Oct 16 2009
From Tom Copeland, Sep 04 2011: (Start)
Let h(x,t) = 1/(1-(t/(1-t))*(exp(x*(1-t))-1)), an e.g.f. in x for row polynomials in t of A008292, then the n-th row polynomial in t of the table A008517 is given by ((h(x,t)*D_x)^(n+1))x with the derivative evaluated at x=0.
Also, df(x,t)/dx = h(f(x,t),t) where f(x,t) is an e.g.f. in x of the row polynomials in t of A008517, i.e., exp(x*P(.,t)) in Copeland's 2008 comment. (End)
The rows are the h-vectors of A134991. - Tom Copeland, Oct 03 2011
Hilbert series of the pre-WDVV ring, thus h-vectors of the Whitehouse simplicial complex (cf. Readdy, Table 1). - Tom Copeland, Sep 20 2014
Arises in Buckholtz's analysis of the error term in the series for exp(nz). - N. J. A. Sloane, Jul 05 2016

Examples

			Triangle begins:
  1;
  1,   2;
  1,   8,   6;
  1,  22,  58,  24;
  1,  52, 328, 444, 120; ...
Row 3: There are three plane increasing 0-1-2-3 trees on 3 vertices. The number of colors are shown to the right of a vertex.
.
    1o (2*t+1)         1o t*(t+2)      1o t*(t+2)
     |                 / \             / \
     |                /   \           /   \
    2o (2*t+1)      2o    3o        3o    2o
     |
     |
    3o
.
The total number of trees is (2*t+1)^2 + t*(t+2) + t*(t+2) = 1 + 8*t + 6*t^2.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270. [with offsets [0,0]: see A201637]

Crossrefs

Columns include A005803, A004301, A006260.
Right-hand columns include A000142, A002538, A002539.
Row sums are A001147.
For a (0,0) based version as used in 'Concrete Mathematics' and by Maple see A201637. For a (0,0) based version which has this triangle as a subtriangle see A340556.

Programs

  • Maple
    with(combinat): A008517 := proc(n, m) local k: add((-1)^(n+k)* binomial(2*n+1, k)* stirling1(2*n-m-k+1, n-m-k+1), k=0..n-m) end: seq(seq(A008517(n, m), m=1..n), n=1..8);
    # Johannes W. Meijer, Oct 16 2009, revised Nov 22 2012
    A008517 := proc(n,k) option remember; `if`(n=1,`if`(k=0,1,0), A008517(n-1,k)* (k+1) + A008517(n-1,k-1)*(2*n-k-1)) end: seq(print(seq(A008517(n,k), k=0..n-1)), n=1..9);
    # Peter Luschny, Apr 20 2011
  • Mathematica
    a[n_, m_] = Sum[(-1)^(n + k)*Binomial[2 n + 1, k]*StirlingS1[2n-m-k+1, n-m-k+1], {k, 0, n-m}]; Flatten[Table[a[n, m], {n, 1, 9}, {m, 1, n}]][[1 ;; 44]] (* Jean-François Alcover, May 18 2011, after Johannes W. Meijer *)
  • PARI
    {T(n, k) = my(z); if( n<1, 0, z = 1 + O(x); for( k=1, n, z = 1 + intformal( z^2 * (z+y-1))); n! * polcoeff( polcoeff(z, n),k))}; /* Michael Somos, Oct 13 2002 */
    
  • PARI
    {T(n,k)=polcoeff((1-x)^(2*n+1)*sum(j=0,2*n+1,j^(n+j)*x^j/j!*exp(-j*x +x*O(x^k))),k)} \\ Paul D. Hanna, Oct 31 2012
    for(n=1,10,for(k=1,n,print1(T(n,k),", "));print(""))
    
  • PARI
    T(n, m) = sum(k=0, n-m, (-1)^(n+k)*binomial(2*n+1, k)*stirling(2*n-m-k+1, n-m-k+1, 1)); \\ Michel Marcus, Dec 07 2021
    
  • Sage
    @CachedFunction
    def A008517(n, k):
        if n==1: return 1 if k==0 else 0
        return A008517(n-1,k)*(k+1)+A008517(n-1,k-1)*(2*n-k-1)
    for n in (1..9): [A008517(n, k) for k in(0..n-1)] # Peter Luschny, Oct 31 2012

Formula

T(n,k) = 0 if n < k, T(1,1) = 1, T(n,-1) = 0, T(n,k) = k*T(n-1,k) + (2*n-k)*T(n-1,k-1).
a(n,m) = Sum_{k=0..n-m} (-1)^(n+k)*binomial(2*n+1, k)*Stirling1(2*n-m-k+1, n-m-k+1). - Johannes W. Meijer, Oct 16 2009
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 = 1 and for the Eulerian numbers A008292 when k = 0.
Let v = -t*exp((1-t)^2*x-t) and let B(x,t) = -(1+1/t*LambertW(v))/(1+LambertW(v)). From the e.g.f. given by Copeland above we find B(x,t) = compositional inverse with respect to x of G(1,x,t) = Sum_{n>=1} R(n,t)*x^n/n! = x+(1+2*t)*x^2/2!+(1+8*t+6*t^2)*x^3/3!+.... The function B(x,t) satisfies the differential equation dB/dx = (1+B)*(1+t*B)^2 = 1 + (2*t+1)*B + t*(t+2)*B^2 + t^2*B^3.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees where each vertex has outdegree <= 3, the vertices of outdegree 1 come in 2*t+1 colors, the vertices of outdegree 2 come in t*(t+2) colors and the vertices of outdegree 3 come in t^2 colors. An example is given below. Cf. A008292. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x,t) = (1+x)*(1+t*x)^2 and let D be the operator f(x,t)*d/dx. Then R(n+1,t) = D^n(f(x,t)) evaluated at x = 0. (End)
From Tom Copeland, Oct 03 2011: (Start)
a(n,k) = Sum_{i=0..k} (-1)^(k-i) binomial(n-i,k-i) A134991(n,i), offsets 0.
P(n+1,t) = (1-t)^(2n+1) Sum_{k>=1} k^(n+k) [t*exp(-t)]^k / k! for n>0; consequently, Sum_{k>=1} (-1)^k k^(n+k) x^k/k!= [1+LW(x)]^(-(2n+1))P[n+1,-LW(x)] where LW(x) is the Lambert W-Function and P(n,t), for n > 0, are the row polynomials as given in Copeland's 2008 comment. (End)
The e.g.f. A(x,t) = -v * { Sum_{j>=1} D(j-1,u) (-z)^j / j! } where u=x*(1-t)^2-t, v=(1+u)/(1-t), z=(t+u)/[(1+u)^2] and D(j-1,u) are the polynomials of A042977. dA(x,t)/dx=(1-t)/[1+u-(1-t)A(x,t)]=(1-t)/{1+LW[-t exp(u)]}, (Copeland's e.g.f. in 2008 comment). - Tom Copeland, Oct 06 2011
A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=-t, and (a_n)=-P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 = 0. - Tom Copeland, Oct 08 2011
The compositional inverse (with respect to x) of y = y(t;x) = (x-t*(exp(x)-1)) is 1/(1-t)*y + t/(1-t)^3*y^2/2! + (t+2*t^2)/(1-t)^5*y^3/3! + (t+8*t^2+6*t^3)/(1-t)^7*y^4/4! + .... The numerator polynomials of the rational functions in t are the row polynomials of this triangle. As observed in the Comments section, the rational functions in t are the generating functions for the diagonals of the triangle of Stirling numbers of the second kind (A048993). See the Bala link for a proof. Cf. A112007 and A134991. - Peter Bala, Dec 04 2011
O.g.f. of row n: (1-x)^(2*n+1) * Sum_{k>=0} k^(n+k) * exp(-k*x) * x^k/k!. - Paul D. Hanna, Oct 31 2012
T(n, k) = n!*[x^n][t^k](egf) where egf = (1-t)/(1 + LambertW(-exp(t^2*x - 2*t*x - t + x)*t)) and after expansion W(-exp(-t)t) is substituted by (-t). - Shamil Shakirov, Feb 17 2025

A001498 Triangle a(n,k) (n >= 0, 0 <= k <= n) of coefficients of Bessel polynomials y_n(x) (exponents in increasing order).

Original entry on oeis.org

1, 1, 1, 1, 3, 3, 1, 6, 15, 15, 1, 10, 45, 105, 105, 1, 15, 105, 420, 945, 945, 1, 21, 210, 1260, 4725, 10395, 10395, 1, 28, 378, 3150, 17325, 62370, 135135, 135135, 1, 36, 630, 6930, 51975, 270270, 945945, 2027025, 2027025, 1, 45, 990, 13860, 135135, 945945, 4729725, 16216200, 34459425, 34459425
Offset: 0

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Author

Keywords

Comments

The row polynomials with exponents in increasing order (e.g., third row: 1+3x+3x^2) are Grosswald's y_{n}(x) polynomials, p. 18, Eq. (7).
Also called Bessel numbers of first kind.
The triangle a(n,k) has factorization [C(n,k)][C(k,n-k)]Diag((2n-1)!!) The triangle a(n-k,k) is A100861, which gives coefficients of scaled Hermite polynomials. - Paul Barry, May 21 2005
Related to k-matchings of the complete graph K_n by a(n,k)=A100861(n+k,k). Related to the Morgan-Voyce polynomials by a(n,k)=(2k-1)!!*A085478(n,k). - Paul Barry, Aug 17 2005
Related to Hermite polynomials by a(n,k)=(-1)^k*A060821(n+k, n-k)/2^n. - Paul Barry, Aug 28 2005
The row polynomials, the Bessel polynomials y(n,x):=Sum_{m=0..n} (a(n,m)*x^m) (called y_{n}(x) in the Grosswald reference) satisfy (x^2)*(d^2/dx^2)y(n,x) + 2*(x+1)*(d/dx)y(n,x) - n*(n+1)*y(n,x) = 0.
a(n-1, m-1), n >= m >= 1, enumerates unordered n-vertex forests composed of m plane (aka ordered) increasing (rooted) trees. Proof from the e.g.f. of the first column Y(z):=1-sqrt(1-2*z) (offset 1) and the Bergeron et al. eq. (8) Y'(z)= phi(Y(z)), Y(0)=0, with out-degree o.g.f. phi(w)=1/(1-w). See their remark on p. 28 on plane recursive trees. For m=1 see the D. Callan comment on A001147 from Oct 26 2006. - Wolfdieter Lang, Sep 14 2007
The asymptotic expansions of the higher order exponential integrals E(x,m,n), see A163931 for information, lead to the Bessel numbers of the first kind in an intriguing way. For the first four values of m these asymptotic expansions lead to the triangles A130534 (m=1), A028421 (m=2), A163932 (m=3) and A163934 (m=4). The o.g.f.s. of the right hand columns of these triangles in their turn lead to the triangles A163936 (m=1), A163937 (m=2), A163938 (m=3) and A163939 (m=4). The row sums of these four triangles lead to A001147, A001147 (minus a(0)), A001879 and A000457 which are the first four right hand columns of A001498. We checked this phenomenon for a few more values of m and found that this pattern persists: m = 5 leads to A001880, m=6 to A001881, m=7 to A038121 and m=8 to A130563 which are the next four right hand columns of A001498. So one by one all columns of the triangle of coefficients of Bessel polynomials appear. - Johannes W. Meijer, Oct 07 2009
a(n,k) also appear as coefficients of (n+1)st degree of the differential operator D:=1/t d/dt, namely D^{n+1}= Sum_{k=0..n} a(n,k) (-1)^{n-k} t^{1-(n+k)} (d^{n+1-k}/dt^{n+1-k}. - Leonid Bedratyuk, Aug 06 2010
a(n-1,k) are the coefficients when expanding (xI)^n in terms of powers of I. Let I(f)(x) := Integral_{a..x} f(t) dt, and (xI)^n := x Integral_{a..x} [ x_{n-1} Integral_{a..x_{n-1}} [ x_{n-2} Integral_{a..x_{n-2}} ... [ x_1 Integral_{a..x_1} f(t) dt ] dx_1 ] .. dx_{n-2} ] dx_{n-1}. Then: (xI)^n = Sum_{k=0..n-1} (-1)^k * a(n-1,k) * x^(n-k) * I^(n+k)(f)(x) where I^(n) denotes iterated integration. - Abdelhay Benmoussa, Apr 11 2025

Examples

			The triangle a(n, k), n >= 0, k = 0..n, begins:
  1
  1  1
  1  3   3
  1  6  15    15
  1 10  45   105    105
  1 15 105   420    945    945
  1 21 210  1260   4725  10395   10395
  1 28 378  3150  17325  62370  135135   135135
  1 36 630  6930  51975 270270  945945  2027025  2027025
  1 45 990 13860 135135 945945 4729725 16216200 34459425 34459425
  ...
And the first few Bessel polynomials are:
  y_0(x) = 1,
  y_1(x) = x + 1,
  y_2(x) = 3*x^2 + 3*x + 1,
  y_3(x) = 15*x^3 + 15*x^2 + 6*x + 1,
  y_4(x) = 105*x^4 + 105*x^3 + 45*x^2 + 10*x + 1,
  y_5(x) = 945*x^5 + 945*x^4 + 420*x^3 + 105*x^2 + 15*x + 1,
  ...
Tree counting: a(2,1)=3 for the unordered forest of m=2 plane increasing trees with n=3 vertices, namely one tree with one vertex (root) and another tree with two vertices (a root and a leaf), labeled increasingly as (1, 23), (2,13) and (3,12). - _Wolfdieter Lang_, Sep 14 2007
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 77.

Crossrefs

Cf. A001497 (same triangle but rows read in reverse order). Other versions of this same triangle are given in A144331, A144299, A111924 and A100861.
Columns from left edge include A000217, A050534.
Columns 1-6 from right edge are A001147, A001879, A000457, A001880, A001881, A038121.
Bessel polynomials evaluated at certain x are A001515 (x=1, row sums), A000806 (x=-1), A001517 (x=2), A002119 (x=-2), A001518 (x=3), A065923 (x=-3), A065919 (x=4). Cf. A043301, A003215.
Cf. A245066 (central terms). A113025 (y_n(2*x)).

Programs

  • Haskell
    a001498 n k = a001498_tabl !! n !! k
    a001498_row n = a001498_tabl !! n
    a001498_tabl = map reverse a001497_tabl
    -- Reinhard Zumkeller, Jul 11 2014
    
  • Magma
    /* As triangle: */ [[Factorial(n+k)/(2^k*Factorial(n-k)*Factorial(k)): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 15 2016
  • Maple
    Bessel := proc(n,x) add(binomial(n+k,2*k)*(2*k)!*x^k/(k!*2^k),k=0..n); end; # explicit Bessel polynomials
    Bessel := proc(n) option remember; if n <=1 then (1+x)^n else (2*n-1)*x*Bessel(n-1)+Bessel(n-2); fi; end; # recurrence for Bessel polynomials
    bessel := proc(n,x) add(binomial(n+k,2*k)*(2*k)!*x^k/(k!*2^k),k=0..n); end;
    f := proc(n) option remember; if n <=1 then (1+x)^n else (2*n-1)*x*f(n-1)+f(n-2); fi; end;
    # Alternative:
    T := (n,k) -> pochhammer(n+1,k)*binomial(n,k)/2^k:
    for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, May 11 2018
    T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)
    else (n - k + 1)* T(n, k - 1) + T(n - 1, k) fi fi end:
    for n from 0 to 9 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Oct 02 2023
  • Mathematica
    max=50; Flatten[Table[(n+k)!/(2^k*(n-k)!*k!), {n, 0, Sqrt[2 max]//Ceiling}, {k, 0, n}]][[1 ;; max]] (* Jean-François Alcover, Mar 20 2011 *)
  • PARI
    {T(n,k)=if(k<0||k>n, 0, binomial(n, k)*(n+k)!/2^k/n!)} /* Michael Somos, Oct 03 2006 */
    
  • PARI
    A001497_ser(N,t='t) = {
      my(x='x+O('x^(N+2)));
      serlaplace(deriv(exp((1-sqrt(1-2*t*x))/t),'x));
    };
    concat(apply(Vecrev, Vec(A001497_ser(9)))) \\ Gheorghe Coserea, Dec 27 2017
    

Formula

a(n, k) = (n+k)!/(2^k*(n-k)!*k!) (see Grosswald and Riordan). - Ralf Stephan, Apr 20 2004
a(n, 0)=1; a(0, k)=0, k > 0; a(n, k) = a(n-1, k) + (n-k+1) * a(n, k-1) = a(n-1, k) + (n+k-1) * a(n-1, k-1). - Len Smiley
a(n, m) = A001497(n, n-m) = A001147(m)*binomial(n+m, 2*m) for n >= m >= 0, otherwise 0.
G.f. for m-th column: (A001147(m)*x^m)/(1-x)^(2*m+1), m >= 0, where A001147(m) = double factorials (from explicit a(n, m) form).
Row polynomials y_n(x) are given by D^(n+1)(exp(t)) evaluated at t = 0, where D is the operator 1/(1-t*x)*d/dt. - Peter Bala, Nov 25 2011
G.f.: conjecture: T(0)/(1-x), where T(k) = 1 - x*y*(k+1)/(x*y*(k+1) - (1-x)^2/T(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 13 2013
Recurrence from Grosswald, p. 18, eq. (5), for the row polynomials: y_n(x) = (2*n-1)*x*y_{n-1} + y_{n-2}(x), y_{-1}(x) = 1 = y_{0} = 1, n >= 1. This becomes, for n >= 0, k = 0..n: a(n, k) = 0 for n < k (zeros not shown in the triangle), a(n, -1) = 0, a(0, 0) = 1 = a(1, 0) and otherwise a(n, k) = (2*n-1)*a(n-1, k-1) + a(n-2, k). Compare with the above given recurrences. - Wolfdieter Lang, May 11 2018
T(n, k) = Pochhammer(n+1,k)*binomial(n,k)/2^k = A113025(n,k)/2^k. - Peter Luschny, May 11 2018
a(n, k) = Sum_{i=0..min(n-1, k)} (n-i)(k-i) * a(n-1, i) where x(n) = x*(x-1)*...*(x-n+1) is the falling factorial, this equality follows directly from the operational formula we wrote in Apr 11 2025.- Abdelhay Benmoussa, May 18 2025

A005803 Second-order Eulerian numbers: a(n) = 2^n - 2*n.

Original entry on oeis.org

1, 0, 0, 2, 8, 22, 52, 114, 240, 494, 1004, 2026, 4072, 8166, 16356, 32738, 65504, 131038, 262108, 524250, 1048536, 2097110, 4194260, 8388562, 16777168, 33554382, 67108812, 134217674, 268435400, 536870854, 1073741764, 2147483586
Offset: 0

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Author

Keywords

Comments

Starting with n=2, a(n) is the second-order Eulerian number <> (see A008517).
Also, number of 3 X n binary matrices avoiding simultaneously the right-angled numbered polyomino patterns (ranpp) (00;1), (01;0) and (01;1). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1Sergey Kitaev, Nov 11 2004
This is the number of target DNA sequences of the correct length present after each successive cycle of the Polymerase Chain Reaction (PCR). The first two cycles produce 0 target sequences, there are 2 target sequences present after the third cycle, then 8 after the fourth cycle, and so forth. - A. Timothy Royappa, Jun 16 2012
a(n+2) = the row sums of A222403. - J. M. Bergot, Apr 04 2018

Examples

			G.f. = 1 + 2*x^3 + 8*x^4 + 22*x^5 + 52*x^6 + 114*x^7 + 240*x^8 + 494*x^9 + ...
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Equivalent to second column of A008517.
a(n) = A070313 + 1 = A052515 + 2. Bisection of A077866.
Equals for n =>3 the third right hand column of A163936.
Cf. A000918 (first differences).

Programs

  • Haskell
    a005803 n = 2 ^ n - 2 * n
    a005803_list = 1 : f 1 [0, 2 ..] where
       f x (z:zs@(z':_)) = y : f y zs  where y = (x + z) * 2 - z'
    -- Reinhard Zumkeller, Jan 19 2014
    
  • Magma
    [2^n-2*n: n in [0..30]]; // Wesley Ivan Hurt, Jun 04 2014
  • Maple
    A005803:=-2*z/(2*z-1)/(z-1)**2; # conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence except for three leading terms
  • Mathematica
    Table[2^n-2n,{n,0,50}] (* or *) LinearRecurrence[{4,-5,2},{1,0,0},51] (* Harvey P. Dale, May 21 2011 *)
  • PARI
    {a(n) = if( n<0, 0, 2^n - 2*n)}; /* Michael Somos, Oct 13 2002 */
    

Formula

G.f.: 1 + 2*x^3/((1-x)^2*(1-2*x)). a(n) = A008517(n-1, 2). - Michael Somos, Oct 13 2002
Equals binomial transform of [1, -1, 1, 1, 1, ...]. - Gary W. Adamson, Jul 14 2008
a(0) = 1 and a(n) = Sum_{k=0..n-3} ((-1)^(n+k+1)*binomial(2*n-1,k)*stirling1(2*n-k-3,n-k-2)), n=>1. - Johannes W. Meijer, Oct 16 2009
a(0)=1, a(1)=0, a(2)=0, a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3). - Harvey P. Dale, May 21 2011
a(n) = A000225(n+1) - A081494(n+1), n > 1. In other words, a(n) equals the sum of the elements in a Pascal triangle of depth n+1 minus the sum of the elements on its perimeter. - Ivan N. Ianakiev, Jun 01 2014
a(n) = A165900(n-1) + Sum_{i=0..n-1} a(i), for n > 0. - Ivan N. Ianakiev, Nov 24 2014
a(n) = A000225(n) - A005408(n-1). - Miquel Cerda, Nov 25 2016
E.g.f.: exp(x)*(exp(x) - 2*x). - Ilya Gutkovskiy, Nov 25 2016

A112007 Coefficient triangle for polynomials used for o.g.f.s for unsigned Stirling1 diagonals.

Original entry on oeis.org

1, 2, 1, 6, 8, 1, 24, 58, 22, 1, 120, 444, 328, 52, 1, 720, 3708, 4400, 1452, 114, 1, 5040, 33984, 58140, 32120, 5610, 240, 1, 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 362880, 3733920, 11026296, 12440064, 5765500, 1062500, 67260, 1004, 1
Offset: 0

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Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

This is the row reversed second-order Eulerian triangle A008517(k+1,k+1-m). For references see A008517.
The o.g.f. for the k-th diagonal, k >= 1, of the unsigned Stirling1 triangle |A008275| is G1(1,x)=1/(1-x) if k=1 and G1(k,x) = g1(k-2,x)/(1-x)^(2*k-1), if k >= 2, with the row polynomials g1(k;x):=Sum_{m=0..k} a(k,m)*x^m.
The recurrence eq. for the row polynomials is g1(k,x)=((k+1)+k*x)*g1(k-1,x) + x*(1-x)*(d/dx)g1(k-1,x), k >= 1, with input g1(0,x):=1.
The column sequences start with A000142 (factorials), A002538, A002539, A112008, A112485.
This o.g.f. computation was inspired by Bender et al. article where the Stirling polynomials have been rediscussed.
The A163936 triangle is identical to the triangle given above except for an extra right hand column [1, 0, 0, 0, ... ]. The A163936 triangle is related to the higher order exponential integrals E(x,m,n), see A163931 and A163932. - Johannes W. Meijer, Oct 16 2009

Examples

			Triangle begins:
    1;
    2,   1;
    6,   8,   1;
   24,  58,  22,   1;
  120, 444, 328,  52,   1;
  ...
G.f. for k=3 sequence A000914(n-1), [2,11,35,85,175,322,546,...], is G1(3,x)= g1(1,x)/(1-x)^5= (2+x)/(1-x)^5.
		

Crossrefs

Row sums give A001147(k+1) = (2*k+1)!!, k>=0.

Programs

  • Maple
    a:= proc(k,m) option remember; if m >= 0 and k >= 0 then (k+m+1)*procname(k-1,m)+(k-m+1)*procname(k-1,m-1) else 0 fi end proc:
    a(0,0):= 1:
    seq(seq(a(k,m),m=0..k),k=0..10); # Robert Israel, Jul 20 2017
  • Mathematica
    a[k_, m_] = Sum[(-1)^(k + n + 1)*Binomial[2k + 3, n]*StirlingS1[m + k - n + 2, m + 1 - n], {n, 0, m}]; Flatten[Table[a[k, m], {k, 0, 8}, {m, 0, k}]][[1 ;; 45]] (* Jean-François Alcover, Jun 01 2011, after Johannes W. Meijer *)
  • PARI
    a(k, m)=sum(n=0, m, (-1)^(k + n + 1)*binomial(2*k + 3, n)*stirling(m + k - n + 2, m + 1 - n, 1));
    for(k=0, 10, for(m=0, k, print1(a(k, m),", "))) \\ Indranil Ghosh, Jul 21 2017

Formula

a(k, m) = (k+m+1)*a(k-1, m) + (k-m+1)*a(k-1, m-1), if k >= m >= 0, a(0, 0)=1; a(k, -1):=0, otherwise 0.
a(k,m) = Sum_{n=0..m} (-1)^(k+n+1)*C(2*k+3,n)*Stirling1(m+k-n+2,m+1-n). - Johannes W. Meijer, Oct 16 2009
The compositional inverse (with respect to x) of y = y(t,x) = (x+t*log(1-x)) is x = x(t,y) = 1/(1-t)*y + t/(1-t)^3*y^2/2! + (2*t+t^2)/(1-t)^5*y^3/3! + (6*t+8*t^2+t^3)/(1-t)^7*y^4/4! + .... The numerator polynomials of the rational functions in t are the row polynomials of this triangle. As observed above, the rational functions in t are the generating functions for the diagonals of |A008275|. See the Bala link for a proof. Cf. A008517. - Peter Bala, Dec 02 2011

A340556 E2(n, k), the second-order Eulerian numbers with E2(0, k) = δ_{0, k}. Triangle read by rows, E2(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 8, 6, 0, 1, 22, 58, 24, 0, 1, 52, 328, 444, 120, 0, 1, 114, 1452, 4400, 3708, 720, 0, 1, 240, 5610, 32120, 58140, 33984, 5040, 0, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320, 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880
Offset: 0

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Author

Peter Luschny, Feb 05 2021

Keywords

Comments

The second-order Eulerian number E2(n, k) is the number of Stirling permutations of order n with exactly k descents; here the last index is defined to be a descent. More formally, let Q_n denote the set of permutations of the multiset {1,1,2,2, ..., n,n} in which, for all j, all entries between two occurrences of j are larger than j, then E2(n, k) = card({s in Q_n with des(s) = k}), where des(s) = card({j: s(j) > s(j+1)}) is the number of descents of s.
Also the number of Riordan trapezoidal words of length n with k distinct letters (see Riordan 1976, p. 9).
Also the number of rooted plane trees on n + 1 vertices with k leaves (see Janson 2008, p. 543).
Let b(n) = (1/2)*Sum_{k=0..n-1} (-1)^k*E2(n-1, k+1) / C(2*n-1, k+1). Apparently b(n) = Bernoulli(n, 1) = -n*Zeta(1 - n) = Integral_{x=0..1}F_n(x) for n >= 1. Here F_n(x) are the signed Fubini polynomials (A278075). (See Rzadkowski and Urlinska, example 4.)

Examples

			Triangle starts:
  [0] 1;
  [1] 0, 1;
  [2] 0, 1, 2;
  [3] 0, 1, 8,    6;
  [4] 0, 1, 22,   58,    24;
  [5] 0, 1, 52,   328,   444,     120;
  [6] 0, 1, 114,  1452,  4400,    3708,    720;
  [7] 0, 1, 240,  5610,  32120,   58140,   33984,    5040;
  [8] 0, 1, 494,  19950, 195800,  644020,  785304,   341136,   40320;
  [9] 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880.
To illustrate the generating function for row 3: The expansion of (1 - x)^7*(x*exp(-x) + 16*x^2*exp(-x)^2 + (243*x^3*exp(-x)^3)/2) gives the polynomial x + 8*x^2 + 6*x^3. The coefficients of this polynomial give row 3.
.
Stirling permutations of order 3 with exactly k descents: (When counting the descents one may assume an invisible '0' appended to the permutations.)
  T[3, k=0]:
  T[3, k=1]: 112233;
  T[3, k=2]: 331122; 223311; 221133; 133122; 122331; 122133; 113322; 112332;
  T[3, k=3]: 332211; 331221; 233211; 221331; 133221; 123321.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270.

Crossrefs

Indexing the second-order Eulerian numbers comes in three flavors: A008517 (following Riordan and Comtet), A201637 (following Graham, Knuth, and Patashnik) and this indexing, extending the definition of Gessel and Stanley. (A008517 is the main entry of the numbers.) The corresponding triangles of the first-order Eulerian numbers are A008292, A173018, and A123125.
Row reversed: A163936 (with offset = 0).
Values: E2poly(n, 1) = A001147(n), E2poly(n, -1) ~ -A001662(n+1), E2poly(n, 2) = A112487(n), 2^n*E2poly(n, 1/2) = A000311(n+1), 2^n*E2poly(n, -1/2) = A341106(n).

Programs

  • Maple
    # Using the recurrence:
    E2 := proc(n, k) option remember;
    if k = 0 and n = 0 then return 1 fi; if n < 0 then return 0 fi;
    E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) end: seq(seq(E2(n, k), k = 0..n), n = 0..9);
    # Using the row generating function:
    E2egf := n -> (1-x)^(2*n+1)*add(k^(n+k)/k!*(x*exp(-x))^k, k=0..n);
    T := (n, k) -> coeftayl(E2egf(n), x=0, k): seq(print(seq(T(n, j),j=0..n)), n=0..7);
    # Using the built-in function:
    E2 := (n, k) -> `if`(k=0, k^n, combinat:-eulerian2(n, k-1)):
    # Using the compositional inverse (series reversion):
    E2triangle := proc(N) local r, s, C; Order := N + 2;
    s := solve(y = series(x - t*(exp(x) - 1), x), x):
    r := n -> -n!*(t - 1)^(2*n - 1)*coeff(s, y, n); C := [seq(expand(r(n)), n = 1..N)];
    seq(print(seq(coeff(C[n+1], t, k), k = 0..n)), n = 0..N-1) end: E2triangle(10);
  • Mathematica
    T[n_, k_] := T[n, k] = If[k == 0, Boole[n == 0], If[n < 0, 0, k T[n - 1, k] + (2 n - k) T[n - 1, k - 1]]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
    (* Via row polynomials: *)
    E2poly[n_] := If[n == 0, 1,
      Expand[Simplify[x (x - 1)^(2 n) D[((1 - x)^(1 - 2 n) E2poly[n - 1]), x]]]];
    Table[CoefficientList[E2poly[n], x], {n, 0, 9}] // Flatten
    (* Series reversion *)
    Revert[gf_, len_] := Module[{S = InverseSeries[Series[gf, {x, 0, len + 1}], x]},
    Table[CoefficientList[(n + 1)! (1 - t)^(2 n + 1) Coefficient[S, x, n + 1], t],
    {n, 0, len}] // Flatten]; Revert[x + t - t Exp[x], 6]
  • PARI
    E2poly(n) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1)));
    { for(n = 0, 9, print(Vecrev(E2poly(n)))) }
    
  • PARI
    T(n, k) = sum(j=0, n-k, (-1)^(n-j)*binomial(2*n+1, j)*stirling(2*n-k-j+1, n-k-j+1, 1)); \\ Michel Marcus, Feb 11 2021
    
  • SageMath
    # See also link to notebook.
    @cached_function
    def E2(n, k):
        if n < 0: return 0
        if k == 0: return k^n
        return k * E2(n - 1, k) + (2*n - k) * E2(n - 1, k - 1)  # Peter Luschny, Mar 08 2025

Formula

E2(n, k) = E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) for n > 0 and 0 <= k <= n, and E2(0, 0) = 1; in all other cases E(n, k) = 0.
E2(n, k) = Sum_{j=0..n-k}(-1)^(n-j)*binomial(2*n+1, j)*Stirling1(2*n-k-j+1, n-k-j+1).
E2(n, k) = Sum_{j=0..k}(-1)^(k-j)*binomial(2*n + 1, k - j)*Stirling2(n + j, j).
Stirling1(x, x - n) = (-1)^n*Sum_{k=0..n} E2(n, k)*binomial(x + k - 1, 2*n).
Stirling2(x, x - n) = Sum_{k=0..n} E2(n, k)*binomial(x + n - k, 2*n).
E2poly(n, x) = Sum_{k=0..n} E2(n, k)*x^k, as row polynomials.
E2poly(n, x) = x*(x-1)^(2*n)*d_{x}((1-x)^(1-2*n)*E2poly(n-1)) for n>=1 and E2poly(0)=1.
E2poly(n, x) = (1 - x)^(2*n + 1)*Sum_{k=0..n}(k^(n + k)/k!)*(x*exp(-x))^k.
W(n, k) = [x^k] (1+x)^n*E2poly(n, x/(1 + x)) are the Ward numbers A269939.
E2(n, k) = [x^k] (1-x)^n*Wpoly(n, x/(1 - x)); Wpoly(n, x) = Sum_{k=0..n}W(n, k)*x^k.
W(n, k) = Sum_{j=0..k} E2(n, j)*binomial(n - j, n - k).
E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*W(n, j)*binomial(n - j, k - j).
The compositional inverse with respect to x of x - t*(exp(x) - 1) (see B. Drake):
T(n, k) = [t^k](n+1)!*(1-t)^(2*n+1)*[x^(n+1)] InverseSeries(x - t*(exp(x)-1), x).
AS1(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, j+1), where AS1(n, k) are the associated Stirling numbers of the first kind (A008306, A106828).
E2(n, k) = Sum_{j=0..n-k+1} (-1)^(n-k-j+1)*AS1(n+j, j)*binomial(n-j, n-k-j+1), for n >= 1.
AS2(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, n-k-j) for n >=1, where AS2(n, k) are the associated Stirling numbers of the second kind (A008299, A137375).
E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*AS2(n + j, j)*binomial(n - j, k - j).

A004301 Second-order Eulerian numbers <>.

Original entry on oeis.org

0, 6, 58, 328, 1452, 5610, 19950, 67260, 218848, 695038, 2170626, 6699696, 20507988, 62407890, 189123286, 571432036, 1722945672, 5187185766, 15600353130, 46882846680, 140820504700, 422822222266, 1269221639358, 3809241974028, 11431014253872, 34299887862990
Offset: 2

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Author

Keywords

Comments

See A008517 for the definition of second-order Eulerian numbers.

Examples

			G.f. = 6*x^3 + 58*x^4 + 328*x^5 + 1452*x^6 + 5610*x^7 + 19950*x^8 + ...
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd edition. Addison-Wesley, Reading, MA, 1994, p. 270.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

3rd column of A008517.
2nd column of A201637.
Equals the fourth right hand column of triangle A163936. - Johannes W. Meijer, Oct 16 2009

Programs

  • Mathematica
    LinearRecurrence[{10, -40, 82, -91, 52, -12}, {0, 6, 58, 328, 1452, 5610}, 26] (* Jean-François Alcover, Feb 27 2019 *)
  • PARI
    {a(n) = if( n<0, 0, (9*3^n - (12 + 8*n)*2^n + (3 + 6*n + 4*n^2))/2)}; /* Michael Somos, Oct 13 2002 */

Formula

From Michael Somos, Oct 13 2002: (Start)
G.f.: x^3(6-2x-12x^2)/((1-x)^3(1-2x)^2(1-3x)).
a(n) = A008517(n, 3) = (9*3^n - (12+8*n)*2^n + (3+6*n+4*n^2))/2. (End)
a(n) = Sum_{k=0..n-3} (-1)^(n+k)*binomial(2*n+1, k)*Stirling1(2*n-k-2, n-k-2). - Johannes W. Meijer, Oct 16 2009

Extensions

Edited by Olivier Gérard, Mar 28 2011

A006260 Second-order Eulerian numbers <>.

Original entry on oeis.org

0, 24, 444, 4400, 32120, 195800, 1062500, 5326160, 25243904, 114876376, 507259276, 2189829808, 9292526920, 38917528600, 161343812980, 663661077072, 2713224461136, 11039636532120, 44751359547420, 180880752056880
Offset: 3

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Author

Keywords

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd edition. Addison-Wesley, Reading, MA, 1994, p. 270.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A008517(n, 4).
3rd column of A201637.
Equals for n=>4 fifth right hand column of A163936. [Johannes W. Meijer, Oct 16 2009].

Programs

  • Maple
    G:=x^4*(24-36*x-280*x^2+652*x^3-168*x^4-288*x^5)/((1-x)^4*(1-2*x)^3*(1-3*x)^2*(1-4*x)): Gser:=series(G,x=0,27): seq(coeff(Gser,x^n),n=3..25);

Formula

G.f.: x^4(24-36x-280x^2+652x^3-168x^4-288x^5)/((1-x)^4(1-2x)^3(1-3x)^2(1-4x)). - Michael Somos, Oct 13 2002
a(n) = sum((-1)^(n+k+2)*binomial(2*n+1,k)*stirling1(2*n-k-3,n-k-3), k=0..n-4). [Johannes W. Meijer, Oct 16 2009].

Extensions

More terms from Emeric Deutsch, Dec 15 2004

A163938 Triangle related to the o.g.f.s. of the right hand columns of A163932 (E(x, m=3, n)).

Original entry on oeis.org

1, 3, 3, 11, 28, 6, 50, 225, 135, 10, 274, 1858, 2092, 486, 15, 1764, 16464, 29148, 13482, 1491, 21, 13068, 158352, 398640, 301220, 70485, 4152, 28, 109584, 1655172, 5552724, 6132780, 2432070, 322971, 10863, 36
Offset: 1

Views

Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

The asymptotic expansions of the higher order exponential integral E(x, m=3, n) lead to triangle A163932, see A163931 for information on the E(x,m,n). The o.g.f.s. of the right hand columns of triangle A163932 have a nice structure Gf(p) = W3(z,p)/(1-z)^(2*p+1) with p = 1 for the first right hand column, p = 2 for the second right hand column, etc. The coefficients of the W3(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. The row sums of this triangle lead to A001879, see A163936 for more information.

Examples

			The first few W3(z,p) polynomials are:
W3(z,p=1) = 1/(1-z)^3
W3(z,p=2) = (3 + 3*z)/(1-z)^5
W3(z,p=3) = (11 + 28*z + 6*z^2)/(1-z)^7
W3(z,p=4) = (50 + 225*z + 135*z^2 + 10*z^3)/(1-z)^9
		

Crossrefs

Row sums equal A001879.
A000254 equals the first left hand column.
A000217 equals the first right hand column.
Cf. A163931 (E(x,m,n)) and A163932.
Cf. A163936 (E(x,m=1,n)), A163937 (E(x,m=2,n)) and A163939 (E(x,m=4,n)).

Programs

  • Maple
    with(combinat): a := proc(n, m): add((-1)^(n+k+1)*((m-k+1)*(m-k)/2!)*binomial(2*n+1, k)*stirling1(m+n-k, m-k+1), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 27 2012
  • Mathematica
    Table[Sum[(-1)^(n + k + 1)*Binomial[m - k + 1, 2]*Binomial[2*n + 1, k]*StirlingS1[m + n - k, m - k + 1], {k, 0, m - 1}], {n, 1, 50}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)
  • PARI
    for(n=1,10, for(m=1,n, print1(sum(k=0,m-1, (-1)^(n+k+1)* binomial(m-k+1,2)*binomial(2*n+1,k) *stirling(m+n-k,m-k+1, 1)) ,", "))) \\ G. C. Greubel, Aug 13 2017

Formula

a(n,m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(m-k+1,2) *binomial(2*n+1,k) *stirling1(m+n-k,m-k+1), for 1 <= m <= n.

A163939 Triangle related to the o.g.f.s. of the right hand columns of A163934 (E(x,m=4,n)).

Original entry on oeis.org

1, 6, 4, 35, 60, 10, 225, 690, 325, 20, 1624, 7588, 6762, 1316, 35, 13132, 85288, 120358, 46928, 4508, 56, 118124, 1004736, 2028660, 1298860, 265365, 13896, 84, 1172700, 12529400, 33896400, 31862400, 11077255, 1313610, 39915, 120
Offset: 1

Views

Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

The asymptotic expansions of the higher order exponential integral E(x,m=4,n) lead to triangle A163934, see A163931 for information on the E(x,m,n). The o.g.f.s. of the right hand columns of triangle A163934 have a nice structure Gf(p) = W4(z,p)/(1-z)^(2*p+2) with p = 1 for the first right hand column, p = 2 for the second right hand column, etc.. The coefficients of the W4(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. The row sums of this triangle lead to A000457, see A163936 for more information.

Examples

			The first few W4(z,p) polynomials are:
W4(z,p=1) = 1/(1-z)^4
W4(z,p=2) = (6+4*z)/(1-z)^6
W4(z,p=3) = (35+60*z+10*z^2)/(1-z)^8
W4(z,p=4) = (225+690*z+325*z^2+20*z^3)/(1-z)^10
		

Crossrefs

Row sums equal A000457.
A000399 equals the first left hand column.
A000292 equals the first right hand column.
Cf. A163931 (E(x,m,n)) and A163934.
Cf. A163936 (E(x,m=1,n)), A163937 (E(x,m=2,n)) and A163938 (E(x,m=3,n)).

Programs

  • Maple
    with(combinat): a := proc(n, m): add((-1)^(n+k+1)*((m-k+2)*(m-k+1)*(m-k)/3!)*binomial(2*n+2, k)*stirling1(m+n-k+1, m-k+2), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 27 2012
  • Mathematica
    Table[Sum[(-1)^(n + k + 1)*Binomial[m - k + 2, 3]*Binomial[2*n + 2, k]*StirlingS1[m + n - k + 1, m - k + 2], {k, 0, m - 1}], {n, 1, 50}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)
  • PARI
    for(n=1,10, for(m=1,n, print1(sum(k=0, m-1, (-1)^(n+k+1)* binomial(m-k+2,3)* binomial(2*n+2,k)*stirling(m+n-k+1,m-k+2,1)), ", "))) \\ G. C. Greubel, Aug 13 2017

Formula

a(n,m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(m-k+2,3)* binomial(2*n+2,k)*stirling1(m+n-k+1,m-k+2), for 1<= m <=n.
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