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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A039755 Triangle of B-analogs of Stirling numbers of the second kind.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 13, 9, 1, 1, 40, 58, 16, 1, 1, 121, 330, 170, 25, 1, 1, 364, 1771, 1520, 395, 36, 1, 1, 1093, 9219, 12411, 5075, 791, 49, 1, 1, 3280, 47188, 96096, 58086, 13776, 1428, 64, 1, 1, 9841, 239220, 719860, 618870, 209622, 32340, 2388, 81, 1, 1
Offset: 0

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Author

Ruedi Suter (suter(AT)math.ethz.ch)

Keywords

Comments

Let M be an infinite lower triangular bidiagonal matrix with (1,3,5,7,...) in the main diagonal and (1,1,1,...) in the subdiagonal. n-th row = M^n * [1,0,0,0,...]. - Gary W. Adamson, Apr 13 2009
From Peter Bala, Aug 08 2011: (Start)
A type B_n set partition is a partition P of the set {1, 2, ..., n, -1, -2, ..., -n} such that for any block B of P, -B is also a block of P, and there is at most one block, called a zero-block, satisfying B = -B. We call (B, -B) a block pair of P if B is not a zero-block. Then T(n,k) is the number of type B_n set partitions with k block pairs. See [Wang].
For example, T(2,1) = 4 since the B_2 set partitions with 1 block pair are {1,2}{-1,-2}, {1,-2}{-1,2}, {1,-1}{2}{-2} and {2,-2}{1}{-1} (the last two partitions contain a zero block).
(End)
Exponential Riordan array [exp(x), (1/2)*(exp(2*x) - 1)]. Triangle of connection constants for expressing the monomial polynomials x^n as a linear combination of the basis polynomials (x-1)*(x-3)*...*(x-(2*k-1)) of A039757. An example is given below. Inverse array is A039757. Equals matrix product A008277 * A122848. - Peter Bala, Jun 23 2014
T(n, k) also gives the (dimensionless) volume of the multichoose(k+1, n-k) = binomial(n, k) polytopes of dimension n-k with side lengths from the set {1, 3, ..., 1+2*k}. See the column g.f.s and the complete homogeneous symmetric function formula for T(n, k) below. - Wolfdieter Lang, May 26 2017
T(n, k) is the number of k-dimensional subspaces (i.e., sets of fixed points like rotation axes and symmetry planes) of the n-cube. See "Sets of fixed points..." in LINKS section. - Tilman Piesk, Oct 26 2019

Examples

			Triangle T(n,k) begins:
  n\k 0     1       2        3       4       5      6     7    8   9 10 ...
  0:  1
  1:  1     1
  2:  1     4       1
  3:  1    13       9        1
  4:  1    40      58       16       1
  5:  1   121     330      170      25       1
  6:  1   364    1771     1520     395      36      1
  7:  1  1093    9219    12411    5075     791     49     1
  8:  1  3280   47188    96096   58086   13776   1428    64    1
  9:  1  9841  239220   719860  618870  209622  32340  2388   81   1
 10:  1 29524 1205941  5278240 6289690 2924712 630042 68160 3765 100  1
 ... reformatted and extended by _Wolfdieter Lang_, May 26 2017
The sequence of row polynomials of A214406 begins [1, 1+x, 1+8*x+3*x^2, ...]. The o.g.f.'s for the diagonals of this triangle thus begin
1/(1-x) = 1 + x + x^2 + x^3 + ...
(1+x)/(1-x)^3 = 1 + 4*x + 9*x^2 + 16*x^3 + ...
(1+8*x+3*x^2)/(1-x)^5 = 1 + 13*x + 58*x^2 + 170*x^3 + ... . - _Peter Bala_, Jul 20 2012
Connection constants: x^3 = 1 + 13*(x-1) + 9*(x-1)*(x-3) + (x-1)*(x-3)*(x-5). Hence row 3 = [1,13,9,1]. - _Peter Bala_, Jun 23 2014
Complete homogeneous symmetric functions: T(3, 1) = h^{(2)}_2 = 1^2 + 3^2 + 1^1*3^1 = 13. The three 2D polytopes are two squares and a rectangle. T(3, 2) = h^{(3)}_1 = 1^1 + 3^1 + 5^1 = 9. The 1D polytopes are three lines. - _Wolfdieter Lang_, May 26 2017
T(4, 3) = 16 is the number of 3-dimensional subspaces (mirror hyperplanes) of the 4-cube. (These are 4 cubes and 12 cuboids.) See "Sets of fixed points..." in LINKS section. - _Tilman Piesk_, Oct 26 2019
		

Crossrefs

Programs

  • Magma
    [[(&+[(-1)^(k-j)*(2*j+1)^n*Binomial(k, j): j in [0..k]])/( 2^k*Factorial(k)): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 14 2019
    
  • Maple
    A039755 := proc(n,k) if k < 0 or k > n then 0 ; elif n <= 1 then 1; else procname(n-1,k-1)+(2*k+1)*procname(n-1,k) ; end if; end proc:
    seq(seq(A039755(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Oct 30 2009
  • Mathematica
    t[n_, k_] = Sum[(-1)^(k-j)*(2j+1)^n*Binomial[k, j], {j, 0, k}]/(2^k*k!); Flatten[Table[t[n, k], {n, 0, 10}, {k, 0, n}]][[1 ;; 56]]
    (* Jean-François Alcover, Jun 09 2011, after Peter Bala *)
  • PARI
    T(n,k)=if(k<0 || k>n,0,n!*polcoeff(polcoeff(exp(x+y/2*(exp(2*x+x*O(x^n))-1)),n),k))
    
  • Sage
    [[sum((-1)^(k-j)*(2*j+1)^n*binomial(k, j) for j in (0..k))/( 2^k*factorial(k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 14 2019

Formula

E.g.f. row polynomials: exp(x + y/2 * (exp(2*x) - 1)).
T(n,k) = T(n-1,k-1) + (2*k+1)*T(n-1,k) with T(0,k) = 1 if k=0 and 0 otherwise. Sum_{k=0..n} T(n,k) = A007405(n). - R. J. Mathar, Oct 30 2009; corrected by Joshua Swanson, Feb 14 2019
T(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^(k-j)*C(k,j)*(2*j+1)^n.
T(n,k) = (1/(2^k*k!)) * A145901(n,k). - Peter Bala
The row polynomials R(n,x) satisfy the Dobinski-type identity:
R(n,x) = exp(-x/2)*Sum_{k >= 0} (2*k+1)^n*(x/2)^k/k!, as well as the recurrence equation R(n+1,x) = (1+x)*R(n,x)+2*x*R'(n,x). The polynomial R(n,x) has all real zeros (apply [Liu et al., Theorem 1.1] with f(x) = R(n,x) and g(x) = R'(n,x)). The polynomials R(n,2*x) are the row polynomials of A154537. - Peter Bala, Oct 28 2011
Let f(x) = exp((1/2)*exp(2*x)+x). Then the row polynomials R(n,x) are given by R(n,exp(2*x)) = (1/f(x))*(d/dx)^n(f(x)). Similar formulas hold for A008277, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
From Peter Bala, Jul 20 2012: (Start)
The o.g.f. for the n-th diagonal (with interpolated zeros) is the rational function D^n(x), where D is the operator x/(1-x^2)*d/dx. For example, D^3(x) = x*(1+8*x^2+3*x^4)/(1-x^2)^5 = x + 13*x^3 + 58*x^5 + 170*x^7 + ... . See A214406 for further details.
An alternative formula for the o.g.f. of the n-th diagonal is exp(-x/2)*(Sum_{k >= 0} (2*k+1)^(k+n-1)*(x/2*exp(-x))^k/k!).
(End)
From Tom Copeland, Dec 31 2015: (Start)
T(n,m) = Sum_{i=0..n-m} 2^(n-m-i)*binomial(n,i)*St2(n-i,m), where St2(n,k) are the Stirling numbers of the second kind, A048993 (also A008277). See p. 755 of Dolgachev and Lunts.
The relation of this entry's e.g.f. above to that of the Bell polynomials, Bell_n(y), of A048993 establishes this formula from a binomial transform of the normalized Bell polynomials, NB_n(y) = 2^n Bell_n(y/2); that is, e^x exp[(y/2)(e^(2x)-1)] = e^x exp[x*2*Bell.(y/2)] = exp[x(1+NB.(y))] = exp(x*P.(y)), so the row polynomials of this entry are given by P_n(y) = [1+NB.(y)]^n = Sum_{k=0..n} C(n,k) NB_k(y) = Sum_{k=0..n} 2^k C(n,k) Bell_k(y/2).
The umbral compositional inverses of the Bell polynomials are the falling factorials Fct_n(y) = y! / (y-n)!; i.e., Bell_n(Fct.(y)) = y^n = Fct_n(Bell.(y)). Since P_n(y) = [1+2Bell.(y/2)]^n, the umbral inverses are determined by [1 + 2 Bell.[ 2 Fct.[(y-1)/2] / 2 ] ]^n = [1 + 2 Bell.[ Fct.[(y-1)/2] ] ]^n = [1+y-1]^n = y^n. Therefore, the umbral inverse sequence of this entry's row polynomials is the sequence IP_n( y) = 2^n Fct_n[(y-1)/2] = (y-1)(y-3) .. (y-2n+1) with IP_0(y) = 1 and, from the binomial theorem, with e.g.f. exp[x IP.(y)]= exp[ x 2Fct.[(y-1)/2] ] = (1+2x)^[(y-1)/2] = exp[ [(y-1)/2] log(1+2x) ].
(End)
Let B(n,k) = T(n,k)*((2*k)!)/(2^k*k!) and P(n,x) = Sum_{k=0..n} B(n,k)*x^(2*k+1). Then (1) P(n+1,x) = (x+x^3)*P'(n,x) for n >= 0, and (2) Sum_{n>=0} B(n,k)/(n!)*t^n = binomial(2*k,k)*exp(t)*(exp(2*t)-1)^k/4^k for k >= 0, and (3) Sum_{n>=0} t^n* P(n,x)/(n!) = x*exp(t)/sqrt(1+x^2-x^2*exp(2*t)). - Werner Schulte, Dec 12 2016
From Wolfdieter Lang, May 26 2017: (Start)
G.f. column k: x^k/Product_{j=0..k} (1 - (1+2*j)*x), k >= 0.
T(n, k) = h^{(k+1)}_{n-k}, the complete homogeneous symmetric function of degree n-k of the k+1 symbols a_j = 1 + 2*j, j = 0, 1, ..., k. (End)
With p(n, x) = Sum_{k=0..n} A001147(k) * T(n, k) * x^k for n >= 0 holds:
(1) Sum_{i=0..n} p(i, x)*p(n-i, x) = 2^n*(Sum_{k=0..n} A028246(n+1, k+1)*x^k);
(2) p(n, -1/2) = (n!) * ([t^n] sqrt(2 / (1 + exp(-2*t)))). - Werner Schulte, Feb 16 2024

A163626 Triangle read by rows: The n-th derivative of the logistic function written in terms of y, where y = 1/(1 + exp(-x)).

Original entry on oeis.org

1, 1, -1, 1, -3, 2, 1, -7, 12, -6, 1, -15, 50, -60, 24, 1, -31, 180, -390, 360, -120, 1, -63, 602, -2100, 3360, -2520, 720, 1, -127, 1932, -10206, 25200, -31920, 20160, -5040, 1, -255, 6050, -46620, 166824, -317520, 332640, -181440, 40320, 1, -511, 18660
Offset: 0

Views

Author

Keywords

Comments

Apart from signs and offset, same as A028246. - Joerg Arndt, Nov 06 2016
Triangle T(n,k), read by rows, given by (1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,...) DELTA (-1,-1,-2,-2,-3,-3,-4,-4,-5,-5,-6,-6,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 05 2011
The "Stirling-Bernoulli transform" maps a sequence b_0, b_1, b_2, ... to a sequence c_0, c_1, c_2, ..., where if B has o.g.f. B(x), c has e.g.f. exp(x)*B(1 - exp(x)). More explicity, c_n = Sum_{k = 0..n} A163626(n,k)*b_k. - Philippe Deléham, May 26 2015
Row sums of absolute values of terms give A000629. - Yahia DJEMMADA, Aug 16 2016
This is the triangle of connection constants for expressing the monomial polynomials (-x)^n as a linear combination of the basis polynomials {binomial(x+n,n)}n>=0, that is, (-x)^n = Sum_{k = 0..n} T(n,k)*binomial(x+k,k). Cf. A145901. - Peter Bala, Jun 06 2019
Row sums for n > 0 are zero. - Shel Kaphan, May 14 2024
The Akiyama-Tanigawa algorithm applied to a sequence yields the same result as the Stirling-Bernoulli Transform applied to the same sequence. See Philippe Deléham's comment of May 26 2015. - Shel Kaphan, May 16 2024

Examples

			y = 1/(1+exp(-x))
y^(0) = y
y^(1) = y-y^2
y^(2) = y-3*y^2+2*y^3
y^(3) = y-7*y^2+12*y^3-6*y^4
Triangle begins :
n\k 0     1     2     3     4     5    6
----------------------------------------
0:  1
1:  1    -1
2:  1    -3     2
3:  1    -7    12    -6
4:  1   -15    50   -60    24
5:  1   -31   180  -390   360  -120
6:  1   -63   602 -2100  3360 -2520  720
7:  1  -127 ... - Reformatted by _Philippe Deléham_, May 26 2015
Change of basis constants: x^4 = 1 - 15*binomial(x+1,1) + 50*binomial(x+2,2) - 60*binomial(x+3,3) + 24*binomial(x+4,4). - _Peter Bala_, Jun 06 2019
		

Crossrefs

Programs

  • Maple
    A163626 := (n, k) -> add((-1)^j*binomial(k, j)*(j+1)^n, j = 0..k):
    for n from 0 to 6 do seq(A163626(n, k), k = 0..n) od; # Peter Luschny, Sep 21 2017
  • Mathematica
    Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1-y[x]);
    Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n-1][y][x], x];
    row[n_] := CoefficientList[Derivative[n][y][x], y[x]] // Rest;
    Table[row[n], {n, 0, 9}] // Flatten
    (* or *) Table[(-1)^k*k!*StirlingS2[n+1, k+1], {n, 0, 9}, {k, 0, n}] // Flatten
    (* Jean-François Alcover, Dec 16 2014 *)
  • Python
    from sympy.core.cache import cacheit
    @cacheit
    def T(n, k):return 1 if n==0 and k==0 else 0 if k>n or k<0 else (k + 1)*T(n - 1, k) - k*T(n - 1, k - 1)
    for n in range(51): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Sep 11 2017

Formula

T(n, k) = (-1)^k*k!*Stirling2(n+1, k+1). - Jean-François Alcover, Dec 16 2014
T(n, k) = (k+1)*T(n-1,k) - k*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k>n or if k<0. - Philippe Deléham, May 29 2015
Worpitzky's representation of the Bernoulli numbers B(n, 1) = Sum_{k = 0..n} T(n,k)/(k+1) = A164555(n)/A027642(n) (Bernoulli numbers). - Philippe Deléham, May 29 2015
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(j+1)^n. - Peter Luschny, Sep 21 2017
Let W_n(x) be the row polynomials of this sequence and F_n(x) the row polynomials of A278075. Then W_n(1 - x) = F_n(x). Also Integral_{x=0..1} U_n(x) = Bernoulli(n, 1) for U in {W, F}. - Peter Luschny, Aug 10 2021

A046802 T(n, k) = Sum_{j=k..n} binomial(n, j)*E1(j, j-k), where E1 are the Eulerian numbers A173018. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 33, 15, 1, 1, 31, 131, 131, 31, 1, 1, 63, 473, 883, 473, 63, 1, 1, 127, 1611, 5111, 5111, 1611, 127, 1, 1, 255, 5281, 26799, 44929, 26799, 5281, 255, 1, 1, 511, 16867, 131275, 344551, 344551, 131275, 16867, 511, 1, 1, 1023, 52905
Offset: 0

Views

Author

Keywords

Comments

T(n,k) is the number of positroid cells of the totally nonnegative Grassmannian G+(k,n) (cf. Postnikov/Williams). It is the triangle of the h-vectors of the stellahedra. - Tom Copeland, Oct 10 2014
See A248727 for a simple transformation of the row polynomials of this entry that produces the umbral compositional inverses of the polynomials of A074909, related to the face polynomials of the simplices. - Tom Copeland, Jan 21 2015
From Tom Copeland, Jan 24 2015: (Start)
The reciprocal of this entry's e.g.f. is [t e^(-xt) - e^(-x)] / (t-1) = 1 - (1+t) x + (1+t+t^2) x^2/2! - (1+t+t^2+t^3) x^3/3! + ... = e^(q.(0;t)x), giving the base sequence (q.(0;t))^n = q_n(0;t) = (-1)^n [1-t^(n+1)] / (1-t) for the umbral compositional inverses (q.(0;t)+z)^n = q_n(z;t) of the Appell polynomials associated with this entry, p_n(z;t) below, i.e., q_n(p.(z;t)) = z^n = p_n(q.(z;t)), in umbral notation. The relations in A133314 then apply between the two sets of base polynomials. (Inserted missing index in a formula - Mar 03 2016.)
The associated o.g.f. for the umbral inverses is Og(x) = x / (1-x q.(0:t)) = x / [(1+x)(1+tx)] = x / [1+(1+t)x+tx^2]. Applying A134264 to h(x) = x / Og(x) = 1 + (1+t) x + t x^2 leads to an o.g.f. for the Narayana polynomials A001263 as the comp. inverse Oginv(x) = [1-(1+t)x-sqrt[1-2(1+t)x+((t-1)x)^2]] / (2xt). Note that Og(x) gives the signed h-polynomials of the simplices and that Oginv(x) gives the h-polynomials of the simplicial duals of the Stasheff polynomials, or type A associahedra. Contrast this with A248727 = A046802 * A007318, which has o.g.f.s related to the corresponding f-polynomials. (End)
The Appell polynomials p_n(x;t) in the formulas below specialize to the Swiss-knife polynomials of A119879 for t = -1, so the Springer numbers A001586 are given by 2^n p_n(1/2;-1). - Tom Copeland, Oct 14 2015
The row polynomials are the h-polynomials associated to the stellahedra, whose f-polynomials are the row polynomials of A248727. Cf. page 60 of the Buchstaber and Panov link. - Tom Copeland, Nov 08 2016
The row polynomials are the h-polynomials of the stellohedra, which enumerate partial permutations according to descents. Cf. Section 10.4 of the Postnikov-Reiner-Williams reference. - Lauren Williams, Jul 05 2022
From p. 60 of the Buchstaber and Panov link, S = P * C / T where S, P, C, and T are the bivariate e.g.f.s of the h vectors of the stellahedra, permutahedra, hypercubes, and (n-1)-simplices, respectively. - Tom Copeland, Jan 09 2017
The number of Le-diagrams of type (k, n) this means the diagram uses the bounding box size k x (n-k), equivalently the number of Grassmann necklaces of type (k, n) and also the number of decorated permutations with k anti-exceedances. - Thomas Scheuerle, Dec 29 2024

Examples

			The triangle T(n, k) begins:
n\k 0   1     2      3      4      5      6     7
0:  1
1:  1   1
2:  1   3     1
3:  1   7     7      1
4:  1  15    33     15      1
5:  1  31   131    131     31      1
6:  1  63   473    883    473     63      1
7:  1 127  1611   5111   5111   1611    127     1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, Holland, 1974, page 245 [From Roger L. Bagula, Nov 21 2009]
  • D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.

Crossrefs

Programs

  • Maple
    T := (n, k) -> add(binomial(n, r)*combinat:-eulerian1(r, r-k), r = k .. n):
    for n from 0 to 8 do seq(T(n, k), k=0..n) od; # Peter Luschny, Jun 27 2018
  • Mathematica
    t[, 1] = 1; t[n, n_] = 1; t[n_, 2] = 2^(n-1)-1;
    t[n_, k_] = Sum[((i-k+1)^i*(k-i)^(n-i-1) - (i-k+2)^i*(k-i-1)^(n-i-1))*Binomial[n-1, i], {i, 0, k-1}];
    T[n_, k_] := t[n+1, k+1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten
    (* Jean-François Alcover, Jan 22 2015, after Tom Copeland *)
    T[ n_, k_] := Coefficient[n! SeriesCoefficient[(1-x) Exp[t] / (1 - x Exp[(1-x) t]), {t, 0, n}] // Simplify, x, k];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Michael Somos, Jan 22 2015 *)

Formula

E.g.f.: (y-1)*exp(x*y)/(y-exp((y-1)*x)). - Vladeta Jovovic, Sep 20 2003
p(t,x) = (1 - x)*exp(t)/(1 - x*exp(t*(1 - x))). - Roger L. Bagula, Nov 21 2009
With offset=0, T(n,0)=1 otherwise T(n,k) = sum_{i=0..k-1} C(n,i)((i-k)^i*(k-i+1)^(n-i) - (i-k+1)^i*(k-i)^(n-i)) (cf. Williams). - Tom Copeland, Oct 10 2014
With offset 0, T = A007318 * A123125. Second column is A000225; 3rd, appears to be A066810. - Tom Copeland, Jan 23 2015
A raising operator (with D = d/dx) associated with this entry's row polynomials is R = x + t + (1-t) / [1-t e^{(1-t)D}] = x + t + 1 + t D + (t+t^2) D^2/2! + (t+4t^2+t^3) D^3/3! + ... , containing the e.g.f. for the Eulerian polynomials of A123125. Then R^n 1 = (p.(0;t)+x)^n = p_n(x;t) are the Appell polynomials with this entry's row polynomials p_n(0;t) as the base sequence. Examples of this formalism are given in A028246 and A248727. - Tom Copeland, Jan 24 2015
With offset 0, T = A007318*(padded A090582)*(inverse of A097805) = A007318*(padded A090582)*(padded A130595) = A007318*A123125 = A007318*(padded A090582)*A007318*A097808*A130595, where padded matrices are of the form of padded A007318, which is A097805. Inverses of padded matrices are just the padded versions of inverses of the unpadded matrices. This relates the face vectors, or f-vectors, and h-vectors of the permutahedra / permutohedra to those of the stellahedra / stellohedra. - Tom Copeland, Nov 13 2016
Umbrally, the row polynomials (offset 0) are r_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A123125. - Tom Copeland, Nov 16 2016
From the previous umbral statement, OP(x,d/dy) y^n = (y + q.(x))^n, where OP(x,y) = exp[y * q.(x)] = (1-x)/(1-x*exp((1-x)y)), the e.g.f. of A123125, so OP(x,d/dy) y^n evaluated at y = 1 is r_n(x), the n-th row polynomial of this entry, with offset 0. - Tom Copeland, Jun 25 2018
Consolidating some formulas in this entry and A248727, in umbral notation for concision, with all offsets 0: Let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry (A046802, the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
From Peter Luschny, Apr 30 2021: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A122045(n).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A007047(n).
Sum_{k=0..n} T(n, n-k) = A000522(n).
Sum_{k=0..n} T(n-k, k) = Sum_{k=0..n} (n - k)^k = A026898(n-1) for n >= 1.
Sum_{k=0..n} k*T(n, k) = A036919(n) = floor(n*n!*e/2).
(End)

Extensions

More terms from Vladeta Jovovic, Sep 20 2003
First formula corrected by Wolfdieter Lang, Feb 14 2015
Offset set to 0 and edited by Peter Luschny, Apr 30 2021

A145901 Triangle of f-vectors of the simplicial complexes dual to the permutohedra of type B_n.

Original entry on oeis.org

1, 1, 2, 1, 8, 8, 1, 26, 72, 48, 1, 80, 464, 768, 384, 1, 242, 2640, 8160, 9600, 3840, 1, 728, 14168, 72960, 151680, 138240, 46080, 1, 2186, 73752, 595728, 1948800, 3037440, 2257920, 645120, 1, 6560, 377504, 4612608, 22305024, 52899840
Offset: 0

Views

Author

Peter Bala, Oct 26 2008

Keywords

Comments

The Coxeter group of type B_n may be realized as the group of n X n matrices with exactly one nonzero entry in each row and column, that entry being either +1 or -1. The order of the group is 2^n*n!. The orbit of the point (1,2,...,n) (or any sufficiently generic point (x_1,...,x_n)) under the action of this group is a set of 2^n*n! distinct points whose convex hull is defined to be the permutohedron of type B_n. The rows of this table are the f-vectors of the simplicial complexes dual to these type B permutohedra. Some examples are given in the Example section below. See A060187 for the corresponding table of h-vectors of type B permutohedra.
This is the (unsigned) triangle of connection constants between the polynomial sequences (2*x + 1)^n, n >= 0, and binomial(x+k,k), k >= 0. For example, (2*x + 1)^2 = 8*binomial(x+2,2) - 8*binomial(x+1,1) + 1 and (2*x + 1)^3 = 48*binomial(x+3,3) - 72*binomial(x+2,2) + 26*binomial(x+1,1) - 1. Cf. A163626. - Peter Bala, Jun 06 2019

Examples

			The triangle begins
n\k|..0.....1.....2.....3.....4.....5
=====================================
0..|..1
1..|..1.....2
2..|..1.....8.....8
3..|..1....26....72....48
4..|..1....80...464...768...384
5..|..1...242..2640..8160..9600..3840
...
Row 2: the permutohedron of type B_2 is an octagon with 8 vertices and 8 edges. Its dual, also an octagon, has f-vector (1,8,8) - row 3 of this triangle.
Row 3: for an appropriate choice of generic point in R_3, the permutohedron of type B_3 is realized as the great rhombicuboctahedron, also known as the truncated cuboctahedron, with 48 vertices, 72 edges and 26 faces (12 squares, 8 regular hexagons and 6 regular octagons). See the Wikipedia entry and also [Fomin and Reading p.22]. Its dual polyhedron is a simplicial polyhedron, the disdyakis dodecahedron, with 26 vertices, 72 edges and 48 triangular faces and so its f-vector is (1,26,72,48) - row 4 of this triangle.
From _Peter Bala_, Jun 06 2019: (Start)
Examples of falling factorials identities for odd numbered rows: Let (x)_n = x*(x - 1)*...*(x - n + 1) with (x)_0 = 1 denote the falling factorial power.
Row 1: 2*(x)_1 + (0 - 2*x)_1 = 0.
Row 3: 48*(x)_3 + 72*(x)_2 * (2 - 2*x)_1 + 26*(x)_1 * (2 - 2*x)_2 + (2 - 2*x)_3 = 0
Row 5: 3840*(x)_5 + 9600*(x)_4 * (4 - 2*x)_(1) + 8160*(x)_3 * (4 - 2*x)_2 + 2640*(x)_2 * (4 - 2*x)_3 + 242*(x)_1 * (4 - 2*x)_4 + (4 - 2*x)_5 = 0. (End)
		

Crossrefs

Cf. A019538 (f-vectors type A permutohedra), A060187 (h-vectors type B permutohedra), A080253 (row sums), A145905, A062715, A028246.

Programs

  • Maple
    with(combinat):
    T:= (n,k) -> add((-1)^(k-i)*binomial(k,i)*(2*i+1)^n,i = 0..k):
    for n from 0 to 9 do
    seq(T(n,k),k = 0..n);
    end do;
  • Mathematica
    T[n_, k_] := Sum[(-1)^(k - i)*Binomial[k, i]*(2*i + 1)^n, {i, 0, k}];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 02 2019 *)

Formula

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(2*i+1)^n.
Recurrence relation: T(n,k) = (2*k + 1)*T(n-1,k) + 2*k*T(n-1,k-1) with T(0,0) = 1 and T(0,k) = 0 for k >= 1.
Relation with type B Stirling numbers of the second kind: T(n,k) = 2^k*k!*A039755(n,k).
Row sums A080253. The matrix product A060187 * A007318 produces the mirror image of this triangle.
E.g.f.: exp(t)/(1 + x - x* exp(2*t)) = 1 + (1 + 2*x)*t + (1 + 8*x + 8*x^2 )*t^2/2! + ... .
From Peter Bala, Oct 13 2011: (Start)
The polynomials in the first column of the array ((1+t)*P^(-1)-t*P)^(-1), P Pascal's triangle and I the identity, are the row polynomials of this table.
The polynomials in the first column of the array ((1+t)*I-t*A062715)^(-1) are, apart from the initial 1, the row polynomials of this table with an extra factor of t. Cf. A060187. (End)
From Peter Bala, Jul 18 2013: (Start)
Integrating the above e.g.f. with respect to x from x = 0 to x = 1 gives Sum_{k = 0..n} (-1)^k*T(n,k)/(k + 1) = 2^n*Bernoulli(n,1/2), the n-th cosecant number.
The corresponding Type A result is considered in A028246 as Worpitzky's algorithm.
Also for n >= 0, Sum_{k = 0..2*n} (-1)^k*T(2*n,k)/((k + 1)*(k + 2)) = 1/2*2^(2*n)*Bernoulli(2*n,1/2) and for n >= 1, Sum_{k = 0..2*n-1} (-1)^k*T(2*n - 1,k)/((k + 1)*(k + 2)) = -1/2 * 2^(2*n)* Bernoulli(2*n,1/2).
The nonzero cosecant numbers are given by A001896/A001897. (End)
From Peter Bala, Jul 22 2014: (Start)
The row polynomials R(n,x) satisfy the recurrence equation R(n+1,x) = D(R(n,x)) with R(0,x) = 1, where D is the operator 1 + 2*x + 2*x(1 + x)*d/dx.
R(n,x) = 1/(1 + x)* Sum_{k = 0..inf} (2*k + 1)^n*(x/(1 + x))^k, valid for x in the open interval (-1/2, inf). Cf. A019538.
The shifted row polynomial x*R(n,x) = (1 + x)^n*P(n,x/(1 + x)) where P(n,x) denotes the n-th row polynomial of A060187.
The row polynomials R(n,x) have only real zeros.
Symmetry: R(n,x) = (-1)^n*R(n,-1 - x). Consequently the zeros of R(n,x) lie in the open interval (-1, 0). (End)
From Peter Bala, May 28 2015: (Start)
Recurrence for row polynomials: R(n,x) = 1 + x*Sum_{k = 0..n-1} binomial(n,k)2^(n-k)*R(k,x) with R(0,x) = 1.
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(k,z))) )^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). A(k,z) = A(-(k + 1),-z). Cf. A019538.
For cases see A258377 (k = 1), A258378(k = 2), A258379 (k = 3), A258380 (k = 4) and A258381 (k = 5). (End)
T(n,k) = A154537(n,k)*k! = A039755(n,k)*(2^k*k!), 0 <= k <= n. - Wolfdieter Lang, Apr 19 2017
From Peter Bala, Jan 12 2018: (Start)
n-th row polynomial R(n,x) = (1 + 2*x) o (1 + 2*x) o ... o (1 + 2*x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E13 in the Bala link.
R(n,x) = Sum_{k = 0..n} binomial(n,k)*2^k*F(k,x) where F(k,x) is the Fubini polynomial of order k, the k-th row polynomial of A019538. (End)

A090582 T(n, k) = Sum_{j=0..n-k} (-1)^j*binomial(n - k + 1, j)*(n - k + 1 - j)^n. Triangle read by rows, T(n, k) for 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 24, 36, 14, 1, 120, 240, 150, 30, 1, 720, 1800, 1560, 540, 62, 1, 5040, 15120, 16800, 8400, 1806, 126, 1, 40320, 141120, 191520, 126000, 40824, 5796, 254, 1, 362880, 1451520, 2328480, 1905120, 834120, 186480, 18150, 510, 1, 3628800, 16329600, 30240000, 29635200, 16435440, 5103000, 818520, 55980, 1022, 1
Offset: 1

Views

Author

Hugo Pfoertner, Jan 11 2004

Keywords

Comments

Let Q(m, n) = Sum_(k=0..n-1) (-1)^k * binomial(n, k) * (n-k)^m. Then Q(m,n) is the numerator of the probability P(m,n) = Q(m,n)/n^m of seeing each card at least once if m >= n cards are drawn with replacement from a deck of n cards, written in a two-dimensional array read by antidiagonals with Q(m,m) as first row and Q(m,1) as first column.
The sequence is given as a matrix with the first row containing the cases #draws = size_of_deck. The second row contains #draws = 1 + size_of_deck. If "mn" indicates m cards drawn from a deck with n cards then the locations in the matrix are:
11 22 33 44 55 66 77 ...
21 32 43 54 65 76 87 ...
31 42 53 64 75 86 97 ...
41 52 63 74 85 .. .. ...
read by antidiagonals ->:
11, 22, 21, 33, 32, 31, 44, 43, 42, 41, 55, 54, 53, 52, ....
The probabilities are given by Q(m,n)/n^m:
.(m,n):.....11..22..21..33..32..31..44..43..42..41...55...54..53..52..51
.....Q:......1...2...1...6...6...1..24..36..14...1..120..240.150..30...1
...n^m:......1...4...1..27...8...1.256..81..16...1.3125.1024.243..32...1
%.Success:.100..50.100..22..75.100...9..44..88.100....4...23..62..94.100
P(n,n) = n!/n^n which can be approximated by sqrt(Pi*(2n+1/3))/e^n (Gosper's approximation to n!).
Let P[n] be the set of all n-permutations. Build a superset Q[n] of P[n] composed of n-permutations in which some (possibly all or none) ascents have been designated. An ascent in a permutation s[1]s[2]...s[n] is a pair of consecutive elements s[i],s[i+1] such that s[i] < s[i+1]. As a triangular array read by rows T(n,k) is the number of elements in Q[n] that have exactly k distinguished ascents, n >= 1, 0 <= k <= n-1. Row sums are A000670. E.g.f. is y/(1+y-exp(y*x)). For example, T(3,1)=6 because there are four 3-permutations with one ascent, with these we would also count 1->2 3, and 1 2->3 where exactly one ascent is designated by "->". (After Flajolet and Sedgewick.) - Geoffrey Critzer, Nov 13 2012
Sum_(k=1..n) Q(n, k)*binomial(x, k) = x^n such that Sum_{k=1..n} Q(n, i)*binomial(x+1,i+1) = Sum_{k=1..x} k^n. - David A. Corneth, Feb 17 2014
A141618(n,n-k+1) = a(n,k) * C(n,k-1) / k. - Tom Copeland, Oct 25 2014
See A074909 and above g.f.s below for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
For connections to toric varieties and Eulerian polynomials (in addition to those noted below), see the Dolgachev and Lunts and the Stembridge links in A019538. - Tom Copeland, Dec 31 2015
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra (this entry) and stellahedra. - Tom Copeland, Nov 14 2016
From the Hasan and Franco and Hasan papers: The m-permutohedra for m=1,2,3,4 are the line segment, hexagon, truncated octahedron and omnitruncated 5-cell. The first three are well-known from the study of elliptic models, brane tilings and brane brick models. The m+1 torus can be tiled by a single (m+2)-permutohedron. Relations to toric Calabi-Yau Kahler manifolds are also discussed. - Tom Copeland, May 14 2020

Examples

			For m = 4, n = 2, we draw 4 times from a deck of two cards. Call the cards "a" and "b" - of the 16 possible combinations of draws (each of which is equally likely to occur), only two do not contain both a and b: a, a, a, a and b, b, b, b. So the probability of seeing both a and b is 14/16. Therefore Q(m, n) = 14.
Table starts:
  [1] 1;
  [2] 2,      1;
  [3] 6,      6,       1;
  [4] 24,     36,      14,      1;
  [5] 120,    240,     150,     30,      1;
  [6] 720,    1800,    1560,    540,     62,     1;
  [7] 5040,   15120,   16800,   8400,    1806,   126,    1;
  [8] 40320,  141120,  191520,  126000,  40824,  5796,   254,   1;
  [9] 362880, 1451520, 2328480, 1905120, 834120, 186480, 18150, 510, 1.
		

Crossrefs

Cf. A073593 first m >= n giving at least 50% probability, A085813 ditto for 95%, A055775 n^n/n!, A090583 Gosper's approximation to n!.
Reflected version of A019538.
Cf. A233734 (central terms).

Programs

  • Haskell
    a090582 n k = a090582_tabl !! (n-1) !! (k-1)
    a090582_row n = a090582_tabl !! (n-1)
    a090582_tabl = map reverse a019538_tabl
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    T := (n, k) -> add((-1)^j*binomial(n - k + 1, j)*(n - k + 1 - j)^n, j = 0..n-k):
    # Or:
    T := (n, k) -> (n - k + 1)!*Stirling2(n, n - k + 1):
    for n from 1 to 9 do seq( T(n, k), k = 1..n) od; # Peter Luschny, May 21 2021
  • Mathematica
    In[1]:= Table[Table[k! StirlingS2[n, k], {k, n, 1, -1}], {n, 1, 6}] (* Victor Adamchik, Oct 05 2005 *)
    nn=6; a=y/(1+y-Exp[y x]); Range[0,nn]! CoefficientList[Series[a, {x,0,nn}], {x,y}]//Grid (* Geoffrey Critzer, Nov 10 2012 *)
  • PARI
    a(n)={m=ceil((-1+sqrt(1+8*n))/2);k=m+1+binomial(m,2)-n;k*sum(i=1,k,(-1)^(i+k)*i^(m-1)*binomial(k-1,i-1))} \\ David A. Corneth, Feb 17 2014

Formula

T(n, k) = (n - k + 1)!*Stirling2(n, n - k + 1), generated by Stirling numbers of the second kind multiplied by a factorial. - Victor Adamchik, Oct 05 2005
Triangle T(n,k), 1 <= k <= n, read by rows given by [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] DELTA [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2006
From Tom Copeland, Oct 07 2008: (Start)
G(x,t) = 1/ (1 + (1-exp(x*t))/t) = 1 + 1*x + (2 + t)*x^2/2! + (6 + 6*t + t^2)*x^3/3! + ... gives row polynomials of A090582, the f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1 + t)*x^2/2! + (1 + 4*t + t^2)*x^3/3! + ... gives row polynomials for A008292, the h-polynomials of permutohedra.
G[(t+1)x,-1/(t+1)] = 1 + (1 + t)*x + (1 + 3*t + 2*t^2)*x^2/2! + ... gives row polynomials of A028246. (End)
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = G(x,t) - 1, the compositional inverse in x is
B(x,t) = log((t+1)-t/(1+x))/t. Let h(x,t) = 1/(dB/dx) = (1+x)*(1+(1+t)x), then the row polynomial P(n,t) is given by (1/n!)*(h(x,t)*d/dx)^n x, evaluated at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t). (End)
k <= 0 or k > n yields Q(n, k) = 0; Q(1,1) = 1; Q(n, k) = k * (Q(n-1, k) + Q(n-1, k-1)). - David A. Corneth, Feb 17 2014
T = A008292*A007318. - Tom Copeland, Nov 13 2016
With all offsets 0 for this entry, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125 with offsets -1 so that the array becomes A008292; i.e., we ignore the first row and first column of A123125. Then the row polynomials of this entry, A090582, are given by A_n(1 + x;0). Other specializations of A_n(x;y) give A028246, A046802, A119879, A130850, and A248727. - Tom Copeland, Jan 24 2020

A119879 Exponential Riordan array (sech(x),x).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 0, -3, 0, 1, 5, 0, -6, 0, 1, 0, 25, 0, -10, 0, 1, -61, 0, 75, 0, -15, 0, 1, 0, -427, 0, 175, 0, -21, 0, 1, 1385, 0, -1708, 0, 350, 0, -28, 0, 1, 0, 12465, 0, -5124, 0, 630, 0, -36, 0, 1, -50521, 0, 62325, 0, -12810, 0, 1050, 0, -45, 0, 1
Offset: 0

Views

Author

Paul Barry, May 26 2006

Keywords

Comments

Row sums have e.g.f. exp(x)*sech(x) (signed version of A009006). Inverse of masked Pascal triangle A119467. Transforms the sequence with e.g.f. g(x) to the sequence with e.g.f. g(x)*sech(x).
Coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent and Bernoulli number (triangle read by rows). Another version in A153641. - Philippe Deléham, Oct 26 2013
Relations to Green functions and raising/creation and lowering/annihilation/destruction operators are presented in Hodges and Sukumar and in Copeland's discussion of this sequence and 2020 pdf. - Tom Copeland, Jul 24 2020

Examples

			Triangle begins:
     1;
     0,    1;
    -1,    0,     1;
     0,   -3,     0,   1;
     5,    0,    -6,   0,   1;
     0,   25,     0, -10,   0,   1;
   -61,    0,    75,   0, -15,   0,   1;
     0, -427,     0, 175,   0, -21,   0,  1;
  1385,    0, -1708,   0, 350,   0, -28,  0,  1;
		

Crossrefs

Row sums are A155585. - Johannes W. Meijer, Apr 20 2011
Rows reversed: A081658.

Programs

  • Maple
    T := (n,k) -> binomial(n,k)*2^(n-k)*euler(n-k,1/2): # Peter Luschny, Jan 25 2009
  • Mathematica
    T[n_, k_] := Binomial[n, k] 2^(n-k) EulerE[n-k, 1/2];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, after Peter Luschny *)
  • PARI
    {T(n,k) = binomial(n,k)*2^(n-k)*(2/(n-k+1))*(subst(bernpol(n-k+1, x), x, 1/2) - 2^(n-k+1)*subst(bernpol(n-k+1, x), x, 1/4))};
    for(n=0,5, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 25 2019
  • Sage
    @CachedFunction
    def A119879_poly(n, x) :
        return 1 if n == 0  else add(A119879_poly(k, 0)*binomial(n, k)*(x^(n-k)-1+n%2) for k in range(n)[::2])
    def A119879_row(n) :
        R = PolynomialRing(ZZ, 'x')
        return R(A119879_poly(n,x)).coeffs()  # Peter Luschny, Jul 16 2012
    # Alternatively:
    
  • Sage
    # uses[riordan_array from A256893]
    riordan_array(sech(x), x, 9, exp=true) # Peter Luschny, Apr 19 2015
    

Formula

Number triangle whose k-th column has e.g.f. sech(x)*x^k/k!.
T(n,k) = C(n,k)*2^(n-k)*E_{n-k}(1/2) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1; k even} binomial(n,k)*p{k}(0)*((n mod 2) - 1 + x^(n-k)). - Peter Luschny, Jul 16 2012
E.g.f.: exp(x*z)/cosh(x). - Peter Luschny, Aug 01 2012
Sum_{k=0..n} T(n,k)*x^k = A122045(n), A155585(n), A119880(n), A119881(n) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 27 2013
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of this entry, A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Triangle equals P*((I + P^2)/2)^(-1), where P denotes Pascal's triangle A007318. - Peter Bala, Mar 07 2024

A371761 Array read by antidiagonals: The number of parades with n girls and k boys that begin with a girl and end with a boy.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 5, 1, 0, 0, 1, 13, 13, 1, 0, 0, 1, 29, 73, 29, 1, 0, 0, 1, 61, 301, 301, 61, 1, 0, 0, 1, 125, 1081, 2069, 1081, 125, 1, 0, 0, 1, 253, 3613, 11581, 11581, 3613, 253, 1, 0, 0, 1, 509, 11593, 57749, 95401, 57749, 11593, 509, 1, 0
Offset: 0

Views

Author

Peter Luschny, Apr 05 2024

Keywords

Comments

The name derives from a proposition by Donald Knuth, who describes the setup of a "girls and boys parade" as follows: "There are m girls {g_1, ..., g_m} and n boys {b_1, ..., b_n}, where g_i is younger than g_{i+1} and b_j is younger than b_{j+1}, but we know nothing about the relative ages of g_i and b_j. In how many ways can they all line up in a sequence such that no girl is directly preceded by an older girl and no boy is directly preceded by an older boy?" [Our notation: A <- D, n <- m, k <- n].
In A344920, the Worpitzky transform is defined as a sequence-to-sequence transformation WT := A -> B, where B(n) = Sum_{k=0..n} A163626(n, k)*A(k). (If A(n) = 1/(n + 1) then B(n) are the Bernoulli numbers (with B(1) = 1/2.)) The rows of the array are the Worpitzky transforms of the powers up to the sign (-1)^k.
The array rows are recursively generated by applying the Akiyama-Tanigawa algorithm to the powers (see the Python implementation below). In this way the array becomes the image of A004248 under the AT-transformation when applied to the rows of A004248. This makes the array closely linked to A344499, which is generated in the same way, but applied to the columns of A004248.
Conjecture: Row n + 1 is row 2^n in table A136301, where a probabilistic interpretation is given (see the link to Parsonnet's paper below).

Examples

			Array starts:
[0] 1, 0,   0,     0,      0,       0,        0,         0,          0, ...
[1] 0, 1,   1,     1,      1,       1,        1,         1,          1, ...
[2] 0, 1,   5,    13,     29,      61,      125,       253,        509, ...
[3] 0, 1,  13,    73,    301,    1081,     3613,     11593,      36301, ...
[4] 0, 1,  29,   301,   2069,   11581,    57749,    268381,    1191989, ...
[5] 0, 1,  61,  1081,  11581,   95401,   673261,   4306681,   25794781, ...
[6] 0, 1, 125,  3613,  57749,  673261,  6487445,  55213453,  431525429, ...
[7] 0, 1, 253, 11593, 268381, 4306681, 55213453, 610093513, 6077248381, ...
.
Seen as triangle T(n, k) = A(n - k, k):
  [0] 1;
  [1] 0, 0;
  [2] 0, 1,  0;
  [3] 0, 1,  1,   0;
  [4] 0, 1,  5,   1,   0;
  [5] 0, 1, 13,  13,   1,  0;
  [6] 0, 1, 29,  73,  29,  1, 0;
  [7] 0, 1, 61, 301, 301, 61, 1, 0;
.
A(n, k) as sum of powers:
  A(2, k) =  -3+   2*2^k;
  A(3, k) =   7-  12*2^k+    6*3^k;
  A(4, k) = -15+  50*2^k-   60*3^k+   24*4^k;
  A(5, k) =  31- 180*2^k+  390*3^k-  360*4^k+  120*5^k;
  A(6, k) = -63+ 602*2^k- 2100*3^k+ 3360*4^k- 2520*5^k+  720*6^k;
  A(7, k) = 127-1932*2^k+10206*3^k-25200*4^k+31920*5^k-20160*6^k+5040*7^k;
		

Crossrefs

Variant: A272644.
Rows include: A344920 (row 2, signed), A006230 (row 3).
Row sums of triangle (n>=2): A297195, alternating row sums: A226158.
Diagonal of array: A048144.

Programs

  • Maple
    egf := 1/(exp(w) + exp(z) - exp(w + z)): serw := n -> series(egf, w, n + 1):
    # Returns row n (>= 0) with length len (> 0):
    R := n -> len -> local k;
    seq(k!*coeff(series(n!*coeff(serw(n), w, n), z, len), z, k), k = 0..len - 1):
    seq(lprint(R(n)(9)), n = 0..7);
    # Explicit with Stirling2 :
    A := (n, k) -> local j; add(j!^2*Stirling2(n, j)*Stirling2(k, j), j = 0..min(n, k)): seq(lprint(seq(A(n, k), k = 0..8)), n = 0..7);
    # Using the unsigned Worpitzky transform.
    WT := (a, len) -> local n, k;
    seq(add((-1)^(n - k)*k!*Stirling2(n + 1, k + 1)*a(k), k=0..n), n = 0..len-1):
    Arow := n -> WT(x -> x^n, 8): seq(lprint(Arow(n)), n = 0..8);
    # Two recurrences:
    A := proc(n, k) option remember; local j; if k = 0 then return k^n fi;
    add(binomial(n, j)*(A(n-j, k-1) + A(n-j+1, k-1)), j = 1..n) end:
    A := proc(n, k) option remember; local j; if n = 0 then 0^k else
    add(binomial(k + `if`(j=0,0,1), j+1)*A(n-1, k-j), j = 0..k) fi end:
  • Mathematica
    (* Using the unsigned Worpitzky transform. *)
    Unprotect[Power]; Power[0, 0] = 1;
    W[n_, k_] := (-1)^(n - k) k! StirlingS2[n + 1, k + 1];
    WT[a_, len_] := Table[Sum[W[n, k] a[k], {k, 0, n}], {n, 0, len-1}];
    A371761row[n_, len_] := WT[#^n &, len];
    Table[A371761row[n, 9], {n, 0, 8}] // MatrixForm
    (* Row n >= 1 by linear recurrence: *)
    RowByLRec[n_, len_] := LinearRecurrence[Table[-StirlingS1[n+1, n+1-k], {k, 1, n}],
    A371761row[n, n+1], len]; Table[RowByLRec[n, 9], {n, 1, 8}] // MatrixForm
  • Python
    from functools import cache
    from math import comb as binomial
    @cache
    def A(n, k):
        if n == 0: return int(k == 0)
        return sum(binomial(k + int(j > 0), j + 1) * A(n - 1, k - j)
               for j in range(k + 1))
    for n in range(8): print([A(n, k) for k in range(8)])
    
  • Python
    # The Akiyama-Tanigawa algorithm for powers generates the rows.
    def ATPowList(n, len):
        A = [0] * len
        R = [0] * len
        for k in range(len):
            R[k] = k**n   # Changing this to R[k] = (n + 1)**k generates A344499.
            for j in range(k, 0, -1):
                R[j - 1] = j * (R[j] - R[j - 1])
            A[k] = R[0]
        return A
    for n in range(8): print([n], ATPowList(n, 9))
  • SageMath
    def A371761(n, k): return sum((-1)^(j - k) * factorial(j) * stirling_number2(k + 1, j + 1) * j^n for j in range(k + 1))
    for n in range(9): print([A371761(n, k) for k in range(8)])
    

Formula

A(n, k) = k! * [z^k] (n! * [w^n] 1/(exp(w) + exp(z) - exp(w + z))).
A(n, k) = k! * [w^k] (Sum_{j=0..n} A075263(n, n - j) * exp(j*w)).
A(n, k) = Sum_{j=0..k} (-1)^(j-k) * Stirling2(k + 1, j + 1) * j! * j^n.
A(n, k) = Sum_{j=0..min(n,k)} (j!)^2 * Stirling2(n, j) * Stirling2(k, j).
A(n, k) = Sum_{j=0..n} (-1)^(n-j)*A028246(n, j) * j^k; this is explicit:
A(n, k) = Sum_{j=0..n} Sum_{m=0..n} binomial(n-m, n-j) * Eulerian1(n, m) * j^k *(-1)^(n-j), where Eulerian1 = A173018.
A(n, k) = Sum_{j=0..k} binomial(k + [j>0], j+1)*A(n-1, k-j) for n > 0.
A(n, k) = Sum_{j=1..n} binomial(n, j)*(A(n-j, k-1) + A(n-j+1, k-1)) for n,k >= 1.
Row n (>=1) satisfies a linear recurrence:
A(n, k) = -Sum_{j=1..n} Stirling1(n + 1, n + 1 - j)*A(n, k - j) if k > n.
A(n, k) = [x^k] (Sum_{j=0..n} A371762(n, j)*x^j) / (Sum_{j=0..n} Stirling1(n + 1, n + 1 - j)*x^j).
A(n, k) = A(k, n). (From the symmetry of the bivariate exponential g.f.)
Let T(n, k) = A(n - k, k) and G(n) = Sum_{k=0..n} (-1)^k*T(n, k) the alternating row sums of the triangle. Then G(n) = (n + 2)*Euler(n + 1, 1) and as shifted Genocchi numbers G(n) = -2*(n + 2)*PolyLog(-n - 1, -1) = -A226158(n + 2).

A038719 Triangle T(n,k) (0 <= k <= n) giving number of chains of length k in partially ordered set formed from subsets of n-set by inclusion.

Original entry on oeis.org

1, 2, 1, 4, 5, 2, 8, 19, 18, 6, 16, 65, 110, 84, 24, 32, 211, 570, 750, 480, 120, 64, 665, 2702, 5460, 5880, 3240, 720, 128, 2059, 12138, 35406, 57120, 52080, 25200, 5040, 256, 6305, 52670, 213444, 484344, 650160, 514080, 221760, 40320, 512, 19171
Offset: 0

Views

Author

N. J. A. Sloane, May 02 2000

Keywords

Comments

The relation of this triangle to A143494 given in the Formula section leads to the following combinatorial interpretation: T(n,k) gives the number of partitions of the set {1,2,...,n+2} into k + 2 blocks where 1 and 2 belong to two distinct blocks and the remaining k blocks are labeled from a fixed set of k labels. - Peter Bala, Jul 10 2014
Also, the number of distinct k-level fuzzy subsets of a set consisting of n elements ordered by set inclusion. - Rajesh Kumar Mohapatra, Mar 16 2020

Examples

			Triangle begins
   1;
   2,   1;
   4,   5,   2;
   8,  19,  18,   6;
  16,  65, 110,  84,  24;
  ...
From _Peter Bala_, Feb 02 2022: (Start)
Table of successive differences of k^2 starting at k = 2
4   9   16
  5   7
    2
gives [4, 5, 2] as row 2 of this triangle.
Table of successive differences of k^3 starting at k = 2
8   27   64   125
  19   37   61
     18   24
        6
gives [8, 19, 8, 6] as row 3 of this triangle. (End)
		

Crossrefs

Row sums give A007047. Columns give A000079, A001047, A038721. Next-to-last diagonal gives A038720.
Diagonal gives A000142. - Rajesh Kumar Mohapatra, Mar 16 2020

Programs

  • Haskell
    a038719 n k = a038719_tabl !! n !! k
    a038719_row n = a038719_tabl !! n
    a038719_tabl = iterate f [1] where
       f row = zipWith (+) (zipWith (*) [0..] $ [0] ++ row)
                           (zipWith (*) [2..] $ row ++ [0])
    -- Reinhard Zumkeller, Jul 08 2012
  • Maple
    T:= proc(n, k) option remember;
          `if` (n=0, `if`(k=0, 1, 0), k*T(n-1, k-1) +(k+2)*T(n-1, k))
        end:
    seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Aug 02 2011
  • Mathematica
    t[n_, k_] := Sum[ (-1)^(k-i)*Binomial[k, i]*(2+i)^n, {i, 0, k}]; Flatten[ Table[ t[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, after Pari *)
  • PARI
    T(n,k)=sum(i=0,k,(-1)^(k-i)*binomial(k,i)*(2+i)^n)
    

Formula

T(n, k) = Sum_{j=0..k} (-1)^j*C(k, j)*(k+2-j)^n.
T(n+1, k) = k*T(n, k-1) + (k+2)*T(n, k), T(0,0) = 1, T(0,k) = 0 for k>0.
E.g.f.: exp(2*x)/(1+y*(1-exp(x))). - Vladeta Jovovic, Jul 21 2003
A038719 as a lower triangular matrix is the binomial transform of A028246. - Gary W. Adamson, May 15 2005
Binomial transform of n-th row = 2^n + 3^n + 4^n + ...; e.g., binomial transform of [8, 19, 18, 6] = 2^3 + 3^3 + 4^3 + 5^3 + ... = 8, 27, 64, 125, ... - Gary W. Adamson, May 15 2005
From Peter Bala, Jul 09 2014: (Start)
T(n,k) = k!*( Stirling2(n+2,k+2) - Stirling2(n+1,k+2) ).
T(n,k) = k!*A143494(n+2,k+2).
n-th row polynomial = 1/(1 + x)*( sum {k >= 0} (k + 2)^n*(x/(1 + x))^k ). Cf. A028246. (End)
The row polynomials have the form (2 + x) o (2 + x) o ... o (2 + x), where o denotes the black diamond multiplication operator of Dukes and White. See example E12 in the Bala link. - Peter Bala, Jan 18 2018
Z(P,m) = Sum_{k=0..n} T(n,k)Binomial(m-2,k) = m^n, the zeta polynomial of the poset B_n. Each length m multichain from 0 to 1 in B_n corresponds to a function from [n] into [m]. - Geoffrey Critzer, Dec 25 2020
The entries in row n are the first terms in a table of the successive differences of the sequence [2^n, 3^n, 4^n, ...]. Examples are given below. - Peter Bala, Feb 02 2022

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), May 09 2000

A263634 Irregular triangle read by rows: row n gives coefficients of n-th logarithmic polynomial L_n(x_1, x_2, ...) with monomials sorted into standard order.

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -6, 12, -4, -3, 1, 24, -60, 20, 30, -5, -10, 1, -120, 360, -120, -270, 30, 120, 30, -6, -15, -10, 1, 720, -2520, 840, 2520, -210, -1260, -630, 42, 210, 140, 210, -7, -21, -35, 1
Offset: 1

Views

Author

N. J. A. Sloane, Oct 29 2015

Keywords

Comments

"Standard order" here means as produced by Maple's "sort" command.
From Petros Hadjicostas, May 27 2020: (Start)
According to the Maple help files for the "sort" command, polynomials in multiple variables are "sorted in total degree with ties broken by lexicographic order (this is called graded lexicographic order)."
Thus for example, x_1^2*x_3 = x_1*x_1*x_3 > x_1*x_2*x_2 = x_1*x_2^2, while x_1^2*x_4 = x_1*x_1*x_4 > x_1*x_2*x_3. (End)
Row sums are 0 (for n > 1). Numbers of terms in rows are partition numbers A000041.
From Tom Copeland, Nov 06 2015: (Start)
With the formal Taylor series f(x) = 1 + x[1] x + x[2] x^2/2! + ... , the partition polynomials of this entry give d[log(f(x))]/dx = L_1(x[1]) + L_2(x[1], x[2]) x + L_3(...) x^2/2! + ..., and the coefficients of the reduced polynomials with x[n] = t are signed A028246.
The raising operator R = x + d[log(f(D)]/dD = x + L_1(x[1]) + L_2[x[1], x[2]) D + L_3(x[1], x[2], x[3]) D^2/2! + ... with D = d/dx generates an Appell sequence of polynomials, given umbrally by P_n(x[1], ..., x[n]; x) = (x[.] + x)^n = Sum_{k=0..n} binomial(n,k) x[k] * x^(n-k) = R^n 1 with the e.g.f. f(t)*e^(x*t) = exp[t P.(x[1], ..., x[.]; x)]. P_0 = x[0] = 1.
The umbral compositional inverse Appell sequence is generated by R = x - d[log(f(D))]/dD with e.g.f. e^(x*t)/f(t) = exp[t IP.(x[1], ..., x[.]; x)], so umbrally IP_n(x[1], ..., x[n]; P.(x[1], ..., x[n]; x)) = x^n = P_n(x[1], ..., x[n]; IP.(x[1], ..., x[n]; x)). An unsigned array for the reduced IP_n(x[1], ..., x[n]; x) polynomials with IP_0 = x[0] = 1 and x[n] = -1 for n > 0 is A154921, for which f(t) = 2 - e^t. (End)
From Tom Copeland, Sep 08 2016: (Start)
The Appell formalism allows a matrix representation in the power basis x^n of the raising operator R that incorporates this array's partition polynomials L_n(x[1], ..., x[n]):
VP_(n+1) = VP_n * R = VP_n * XPS^(-1) * MX * XPS, where XPS is the matrix formed from multiplying the n-th diagonal of the Pascal matrix PS of A007318 by the indeterminate x[n], with x[0] = 1 for the main diagonal of ones, i.e., XPS[n,k] = PS[n,k] * x[n-k]; the matrix MX is A129185; the matrix XPS^(-1) is the inverse of XPS, which can be formed by multiplying the diagonals of the Pascal matrix by the partition polynomials IPT(n, x[1], ..., x[n]) of A133314, i.e., XPS^(-1)[n,k] = PS[n,k] * IPT(n-k, x[1], ...); and VP_n is the row vector in the power basis representing the Appell polynomial P_n(x) formed from the basic sequence of moments 1, x[1], x[2], ..., i.e., umbrally P_n(x) = (x[.] + x)^n = Sum_{k=0..n} binomial(n,k) * x[k] * x^(n-k).
Then R = XPS^(-1) * MX * XPS is the Pascal matrix PS with an additional first superdiagonal of ones and the other lower diagonals multiplied by the partition polynomials of this array, i.e., R[n,k] = PS[n,k] * L_{n+1-k}(x[1], ..., x[n+1-k]) except for the first superdiagonal of ones.
Consistently, VP_n = (1, 0, 0, ...) * R^n = (1, 0, 0, ...) * XPS^(-1) * MX^n * XPS = (1, 0, 0, ...) * MX^n * XPS = the n-th row vector of XPS, which is the vector representation of P_n(x) = (x[.] + x)^n with x[0] = 1.
See the Copeland link for the umbral representation R = exp[g.*D] * x * exp[h.*D] that reflects the matrix representations.
The Stirling partition polynomials of the first kind St1_n(a[1], a[2], ..., a[n]) of A036039, the Stirling partition polynomials of the second kind St2_n(b[1], b[2], ..., b[n]) of A036040, and the refined Lah polynomials Lah_n[c[1], c[2], ..., c[n]) of A130561 are Appell sequences in the respective distinguished indeterminates a[1], b[1], and c[1]. Comparing the formulas for their raising operators with that in this entry, L_n(x[1], x[2], ..., x[n]) evaluates to
A) (n-1)! * a[n] for x[n] = St1_n(a[1], a[2], ..., a[n]);
B) b[n] for x[n] = St2_n(b[1], b[2], ..., b[n]);
C) n! * c[n] for x[n] = Lah_n(c[1], c[2], ..., c[n]).
Conversely, from the respective e.g.f.s (added Sep 12 2016)
D) x[n] = St1_n(L_1(x[1])/0!, ..., L_n(x[1], ..., x[n])/(n-1)!);
E) x[n] = St2_n(L_1(x[1]), ..., L_n(x[1], ..., x[n]));
F) x[n] = Lah_n(L_1(x[1])/1!, ..., L_n(x[1], ..., x[n])/n!).
Given only the Appell sequence with no closed form for the e.g.f., the raising operator can be generated using this formalism, as has been partially done for A134264. (End)
For the Appell sequences above, the raising operator is related to the recursion P_(n+1)(x) = x * P_n(x) + Sum_{k=0..n} binomial(n,k) * L_(n-k+1)(x[1], ..., x[n+k-1]) * P_k(x). For a derivation and connections to formal cumulants (c_n = L_n(x[1], ...)) and moments (m_n = x[n]), see the Copeland link on noncrossing partitions. With x = 0, the recursion reduces to x[n+1] = Sum_{k = 0..n} binomial(n,k) * L_(n-k+1)(x[1], ..., x[n+k-1]) * x[k] with x[0] = 1. This array is a differently ordered version of A127671. - Tom Copeland, Sep 13 2016
With x[n] = x^(n-1), a signed version of A130850 is obtained. - Tom Copeland, Nov 14 2016
See p. 2 of Getzler for a relation to stable graphs called necklaces used in computations for Deligne-Mumford-Knudsen moduli spaces of stable curves of genus 1. - Tom Copeland, Nov 15 2019
For a relation to a combinatorial Faa di Bruno Hopf algebra related to functional composition, as presented by Connes and Moscovici, see Figueroa et al. - Tom Copeland, Jan 17 2020
From Tom Copeland, May 17 2020: (Start)
The e.g.f. of an Appell sequence is f(t) e^(x*t) with f(0) = 1. Given the Laguerre-Polya class function f(t) = e^(-a*t^2 + b*t) Product_m (1 - t/z_m) e^(t/z_m) with a = 0 for simplicity (more generally a >= 0) and b real and where the product runs over all the zeros z_m of f(t) with all zeros real and Sum_m 1/(z_m)^2 convergent, the raising operator of the Appell polynomials is R = x + b - Sum_{k > 0} c_(k+1) D^k with c_k = Sum_m (1/(z_m)^k), i.e., traces of powers of the reciprocals of the zeros. From R in earlier comments, b = L_1(x_1) and otherwise c_k = -L_k(x_1, ..., x_k).
The Laguerre / Turan / de Gua inequalities (Csordas and Williamson and Skovgaard) imply that all the zeros of each Appell polynomial are real and simple and its extrema are local maxima above the x-axis and local minima below and are located above or below the zeros of the next lower degree Appell polynomial. (End)
From Tom Copeland, Oct 15 2020: (Start)
With a_n = n! * b_n = (n-1)! * c_n for n > 0, represent a function with f(0) = a_0 = b_0 = 1 as an
A) exponential generating function (e.g.f), or formal Taylor series: f(x) = e^{a.x} = 1 + Sum_{n > 0} a_n * x^n/n!
B) ordinary generating function (o.g.f.), or formal power series: f(x) = 1/(1-b.x) = 1 + Sum_{n > 0} b_n * x^n
C) logarithmic generating function (l.g.f): f(x) = 1 - log(1 - c.x) = 1 + Sum_{n > 0} c_n * x^n /n.
Expansions of log(f(x)) are given in
I) A127671 and A263634 for the e.g.f: log[ e^{a.*x} ] = e^{L.(a_1,a_2,...)x} = Sum_{n > 0} L_n(a_1,...,a_n) * x^n/n!, the logarithmic polynomials, cumulant expansion polynomials
II) A263916 for the o.g.f.: log[ 1/(1-b.x) ] = log[ 1 - F.(b_1,b_2,...)x ] = -Sum_{n > 0} F_n(b_1,...,b_n) * x^n/n, the Faber polynomials.
Expansions of exp(f(x)-1) are given in
III) A036040 for an e.g.f: exp[ e^{a.x} - 1 ] = e^{BELL.(a_1,...)x}, the Bell/Touchard/exponential partition polynomials, a.k.a. the Stirling partition polynomials of the second kind
IV) A130561 for an o.g.f.: exp[ b.x/(1-b.x) ] = e^{LAH.(b.,...)x}, the Lah partition polynomials
V) A036039 for an l.g.f.: exp[ -log(1-c.x) ] = e^{CIP.(c_1,...)x}, the cycle index polynomials of the symmetric groups S_n, a.k.a. the Stirling partition polynomials of the first kind.
Since exp and log are a compositional inverse pair, one can extract the indeterminates of the log set of partition polynomials from the exp set and vice versa. For a discussion of the relations among these polynomials and the combinatorics of connected and disconnected graphs/maps, see Novak and LaCroix on classical moments and cumulants and the two books on statistical mechanics referenced in A036040. (End)
Ignoring signs, these polynomials appear in Schröder in the set of equations (II) on p. 343 and in Stewart's translation on p. 31. - Tom Copeland, Aug 25 2021

Examples

			The first few polynomials are:
(1) x[1].
(2) -x[1]^2 + x[2].
(3) 2*x[1]^3 - 3*x[1]*x[2] + x[3].
(4) -6*x[1]^4 + 12*x[1]^2*x[2] - 4*x[1]*x[3] - 3*x[2]^2 + x[4].
(5) 24*x[1]^5 - 60*x[1]^3*x[2] + 20*x[1]^2*x[3] + 30*x[1]*x[2]^2 - 5*x[1]*x[4] - 10*x[2]*x[3] + x[5].
(6) -120*x[1]^6 + 360*x[1]^4*x[2] - 120*x[1]^3*x[3] - 270*x[1]^2*x[2]^2 + 30*x[1]^2*x[4] + 120*x[1]*x[2]*x[3] + 30*x[2]^3 - 6*x[1]*x[5] - 15*x[2]*x[4] - 10*x[3]^2 + x[6].
...
[1]    1
[2]   -1,    1
[3]    2,   -3,     1
[4]   -6,   12,    -4,    -3,   1
[5]   24,  -60,    20,    30,  -5,  -10,   1
[6] -120,  360,  -120,  -270,  30,  120,  30, -6, -15, -10, 1
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 140, 156, 308.

Crossrefs

Programs

  • Maple
    triangle := proc(numrows) local E, s, Q;
    E := add(x[i]*t^i/i!, i=1..numrows);
    s := series(log(1 + E), t, numrows+1);
    Q := k -> sort(expand(k!*coeff(s, t, k)));
    seq(print(coeffs(Q(k))), k=1..numrows) end:
    triangle(6); # updated by Peter Luschny, May 27 2020

Formula

G.f.: Log(1 + Sum_{i >= 1} x_i*t^i/i!) = Sum_{n >= 1} L_n(x_1, x_2, ...)*t^n/n!. [Comtet, p. 140, Eq. [5a]. - corrected by Tom Copeland, Sep 08 2016]
Conjecture: row polynomials are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} binomial(n-2,j-1)*R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = x_k for k > 0. - Mikhail Kurkov, Mar 30 2025

A112486 Coefficient triangle for polynomials used for e.g.f.s for unsigned Stirling1 diagonals.

Original entry on oeis.org

1, 1, 1, 2, 5, 3, 6, 26, 35, 15, 24, 154, 340, 315, 105, 120, 1044, 3304, 4900, 3465, 945, 720, 8028, 33740, 70532, 78750, 45045, 10395, 5040, 69264, 367884, 1008980, 1571570, 1406790, 675675, 135135, 40320, 663696, 4302216, 14777620, 29957620
Offset: 0

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Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

The k-th diagonal of |A008275| appears as the k-th column in |A008276| with k-1 leading zeros.
The recurrence, given below, is derived from (d/dx)g1(k,x) - g1(k,x)= x*(d/dx)g1(k-1,x) + g1(k-1,x), k >= 1, with input g(-1,x):=0 and initial condition g1(k,0)=1, k >= 0. This differential recurrence for the e.g.f. g1(k,x) follows from the one for unsigned Stirling1 numbers.
The column sequences start with A000142 (factorials), A001705, A112487- A112491, for m=0,...,5.
The main diagonal gives (2*k-1)!! = A001147(k), k >= 1.
This computation was inspired by the Bender article (see links), where the Stirling polynomials are discussed.
The e.g.f. for the k-th diagonal, k >= 1, of the unsigned Stirling1 triangle |A008275| with k-1 leading zeros is g1(k-1,x) = exp(x)*Sum_{m=0..k-1} a(k,m)*(x^(k-1+m))/(k-1+m)!.
a(k,n) = number of lists with entries from [n] such that (i) each element of [n] occurs at least once and at most twice, (ii) for each i that occurs twice, all entries between the two occurrences of i are > i, and (iii) exactly k elements of [n] occur twice. Example: a(1,2)=5 counts 112, 121, 122, 211, 221, and a(2,2)=3 counts 1122,1221,2211. - David Callan, Nov 21 2011

Examples

			Triangle begins:
    1;
    1,    1;
    2,    5,     3;
    6,   26,    35,    15;
   24,  154,   340,   315,   105;
  120, 1044,  3304,  4900,  3465,   945;
  720, 8028, 33740, 70532, 78750, 45045, 10395;
k=3 column of |A008276| is [0,0,2,11,35,85,175,...] (see A000914), its e.g.f. exp(x)*(2*x^2/2! + 5* x^3/3! + 3*x^4/4!).
		

Crossrefs

Cf. A112007 (triangle for o.g.f.s for unsigned Stirling1 diagonals). A112487 (row sums).

Programs

  • Maple
    A112486 := proc(n,k)
        if n < 0 or k<0 or  k> n then
            0 ;
        elif n = 0 then
            1 ;
        else
            (n+k)*procname(n-1,k)+(n+k-1)*procname(n-1,k-1) ;
        end if;
    end proc: # R. J. Mathar, Dec 19 2013
  • Mathematica
    A112486 [n_, k_] := A112486[n, k] = Which[n<0 || k<0 || k>n, 0, n == 0, 1, True, (n+k)*A112486[n-1, k]+(n+k-1)*A112486[n-1, k-1]]; Table[A112486[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2014, after R. J. Mathar *)

Formula

a(k, m) = (k+m)*a(k-1, m) + (k+m-1)*a(k-1, m-1) for k >= m >= 0, a(0, 0)=1, a(k, -1):=0, a(k, m)=0 if k < m.
From Tom Copeland, Oct 05 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = -(1 + t)
P(3,t) = 2 + 5 t + 3 t^2
P(4,t) = -( 6 + 26 t + 35 t^2 + 15 t^3)
P(5,t) = 24 + 154 t +340 t^2 + 315 t^3 + 105 t^4
Apparently, P(n,t) = (-1)^(n+1) PW[n,-(1+t)] where PW are the Ward polynomials A134991. If so, an e.g.f. for the polynomials is
A(x,t) = -(x+t+1)/t - LW{-((t+1)/t) exp[-(x+t+1)/t]}, where LW(x) is a suitable branch of the Lambert W Fct. (e.g., see A135338). The comp. inverse in x (about x = 0) is B(x) = x + (t+1) [exp(x) - x - 1]. See A112487 for special case t = 1. These results are a special case of A134685 with u(x) = B(x), i.e., u_1=1 and (u_n)=(1+t) for n>0.
Let h(x,t) = 1/(dB(x)/dx) = 1/[1+(1+t)*(exp(x)-1)], an e.g.f. in x for row polynomials in t of signed A028246 , then P(n,t), is given by
(h(x,t)*d/dx)^n x, evaluated at x=0, i.e., A(x,t)=exp(x*h(u,t)*d/du) u, evaluated at u=0. Also, dA(x,t)/dx = h(A(x,t),t).
The e.g.f. A(x,t) = -v * Sum_{j>=1} D(j-1,u) (-z)^j / j! where u=-(x+t+1)/t, v=1+u, z=(1+t*v)/(t*v^2) and D(j-1,u) are the polynomials of A042977. dA/dx = -1/[t*(v-A)].(End)
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+1, and (a_n)=t*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 row polynomials R(n,x) may be calculated using R(n,x) = 1/x^(n+1)*D^n(x), where D is the operator (x^2+x^3)*d/dx. - Peter Bala, Jul 23 2012
For n>0, Sum_{k=0..n} a(n,k)*(-1/(1+W(t)))^(n+k+1) = (t d/dt)^(n+1) W(t), where W(t) is Lambert W function. For t=-x, this gives Sum_{k>=1} k^(k+n)*x^k/k! = - Sum_{k=0..n} a(n,k)*(-1/(1+W(-x)))^(n+k+1). - Max Alekseyev, Nov 21 2019
Conjecture: row polynomials are R(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling2(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(n+j+k)/((n+j-k)!*(i-j)!*k!). - Mikhail Kurkov, Apr 21 2025
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