A119467 -id:A119467 - cp's OEIS frontend

A131054 Triangle read by rows: A000012 * A119467 as infinite lower triangular matrices.

1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 3, 4, 7, 1, 1, 3, 9, 7, 11, 1, 1, 4, 9, 22, 11, 16, 1, 1, 4, 16, 22, 46, 16, 22, 1, 1, 5, 16, 50, 46, 86, 22, 29, 1, 1, 5, 25, 50, 130, 86, 148, 29, 37, 1, 1, 6, 25, 95, 130, 296, 148, 239, 37, 46, 1, 1

Offset: 0

Gary W. Adamson, Jun 12 2007

Row sums = 2^n.

			First few rows of the triangle are:
1;
1, 1;
2, 1, 1;
2, 4, 1, 1;
3, 4, 7, 1, 1;
3, 9, 7, 11, 1, 1;
4, 9, 22, 11, 16, 1, 1;
4, 16, 22, 46, 16, 22, 1, 1;
...

Cf. A000012, A119467.

a(46) = 5 inserted and more terms from Georg Fischer, May 29 2023

A136630 Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 4, 0, 1, 0, 1, 0, 10, 0, 1, 0, 0, 16, 0, 20, 0, 1, 0, 1, 0, 91, 0, 35, 0, 1, 0, 0, 64, 0, 336, 0, 56, 0, 1, 0, 1, 0, 820, 0, 966, 0, 84, 0, 1, 0, 0, 256, 0, 5440, 0, 2352, 0, 120, 0, 1, 0, 1, 0, 7381, 0, 24970, 0, 5082, 0, 165, 0, 1, 0, 0, 1024, 0, 87296, 0

Offset: 0

Paul D. Hanna, Jan 14 2008

For partitions into blocks of even size see A156289.
Essentially the unsigned matrix inverse of triangle A121408.
From Peter Bala, Jul 28 2014: (Start)
Define a polynomial sequence x_(n) by setting x_(0) = 1 and for n = 1,2,... setting x_(n) = x*(x + n - 2)*(x + n - 4)*...*(x + n - 2*(n - 1)). Then this table is the triangle of connection constants for expressing the monomial polynomials x^n in terms of the basis x_(k), that is, x^n = sum {k = 0..n} T(n,k)*x_(k) for n = 0,1,2,.... An example is given below.
Let M denote the lower unit triangular array A119467 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, omitting the first row and column, equals the infinite matrix product M(0)*M(1)*M(2)*.... (End)
Also the Bell transform of A000035(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

			Triangle begins:
  1;
  0, 1;
  0, 0,   1;
  0, 1,   0,    1;
  0, 0,   4,    0,    1;
  0, 1,   0,   10,    0,     1;
  0, 0,  16,    0,   20,     0,    1;
  0, 1,   0,   91,    0,    35,    0,    1;
  0, 0,  64,    0,  336,     0,   56,    0,   1;
  0, 1,   0,  820,    0,   966,    0,   84,   0,   1;
  0, 0, 256,    0, 5440,     0, 2352,    0, 120,   0, 1;
  0, 1,   0, 7381,    0, 24970,    0, 5082,   0, 165, 0, 1;
T(5,3) = 10. The ten partitions of the set [5] into 3 odd-sized blocks are
(1)(2)(345), (1)(3)(245), (1)(4)(235), (1)(5)(234), (2)(3)(145),
(2)(4)(135), (2)(5)(134), (3)(4)(125), (3)(5)(124), (4)(5)(123).
Connection constants: Row 5 = [0,1,0,10,0,1]. Hence, with the polynomial sequence x_(n) as defined in the Comments section we have x^5 = x_(1) + 10*x_(3) + x_(5) = x + 10*x*(x+1)*(x-1) + x*(x+3)*(x+1)*(x-1)*(x-3).

L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 225-226.

Ch. A. Charalambides, Central factorial numbers and related expansions, Fib. Quarterly, Vol. 19, No 5, Dec 1981, pp. 451-456.
Feng Qi and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Appl. Anal. Disc. Math. (2024) Vol. 18, No. 1, 1-24. See p. 13.

Cf. A121408; A136631 (antidiagonal sums), A003724 (row sums), A136632; A002452 (column 3), A002453 (column 5); A008958 (central factorial triangle), A156289. A185690, A196776.

A136630 := proc (n, k) option remember; if k < 0 or n < k then 0 elif k = n then 1 else procname(n-2, k-2) + k^2*procname(n-2, k) end if end proc: seq(seq(A136630(n, k), k = 1 .. n), n = 1 .. 12); # Peter Bala, Jul 27 2014
# The function BellMatrix is defined in A264428.
BellMatrix(n -> (n+1) mod 2, 9); # Peter Luschny, Jan 27 2016

t[n_, k_] := Coefficient[ x^k/Product[ 1 - (2*j + k - 2*Quotient[k, 2])^2*x^2, {j, 0, k/2}] + x*O[x]^n, x, n]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2013, after Pari *)
BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
rows = 13;
M = BellMatrix[Mod[#+1, 2]&, rows];
Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)

{T(n,k)=polcoeff(x^k/prod(j=0,k\2,1-(2*j+k-2*(k\2))^2*x^2 +x*O(x^n)),n)}

G.f. for column k: x^k/Product_{j=0..floor(k/2)} (1 - (2*j + k-2*floor(k/2))^2 * x^2).
G.f. for column 2*k: x^(2*k)/Product_{j=0..k} (1 - (2*j)^2*x^2).
G.f. for column 2*k+1: x^(2*k+1)/Product_{j=0..k} (1 - (2*j+1)^2*x^2).
From Peter Bala, Feb 21 2011 (Start)
T(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j-k)^n,
Recurrence relation T(n+2,k) = T(n,k-2) + k^2*T(n,k).
E.g.f.: F(x,z) = exp(x*sinh(z)) = Sum_{n>=0} R(n,x)*z^n/n! = 1 + x*z + x^2*z^2/2! + (x+x^3)*z^3/3! + ....
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x^2
R(3,x) = x+x^3.
The e.g.f. F(x,z) satisfies the partial differential equation d^2/dz^2(F) = x^2*F + x*F' + x^2*F'' where ' denotes differentiation w.r.t. x.
Hence the row polynomials satisfy the recurrence relation R(n+2,x) = x^2*R(n,x) + x*R'(n,x) + x^2*R''(n,x) with R(0,x) = 1.
The recurrence relation for T(n,k) given above follows from this.
(End)
For the corresponding triangle of ordered partitions into odd-sized blocks see A196776. Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of exp(t*M) lists the row polynomials for the present triangle. - Peter Bala, Oct 06 2011
Row generating polynomials equal D^n(exp(x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A196776. - Peter Bala, Dec 06 2011
From Peter Bala, Jul 28 2014: (Start)
E.g.f.: exp(t*sinh(x)) = 1 + t*x + t^2*x^2/2! + (t + t^3)*x^3/3! + ....
Hockey-stick recurrence: T(n+1,k+1) = Sum_{i = 0..floor((n-k)/2)} binomial(n,2*i)*T(n-2*i,k).
Recurrence equation for the row polynomials R(n,t):
R(n+1,t) = t*Sum_{k = 0..floor(n/2)} binomial(n,2*k)*R(n-2*k,t) with R(0,t) = 1. (End)

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

Paul Barry, May 26 2006

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

			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;

G. C. Greubel, Rows n=0..100 of triangle, flattened
Paul Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 28.
Tom Copeland, The Elliptic Lie Triad: Ricatti and KdV Equations, Infinigens, and Elliptic Genera, 2015.
Tom Copeland, Discussion of this sequence
Tom Copeland, SKipping over Dimensions, Juggling Zeros in the Matrix, 2020.
A. Hodges and C. V. Sukumar, Bernoulli, Euler, permutations and quantum algebras, Proc. R. Soc. A Oct. 2007 vol 463 no. 463 2086 2401-2414.
Peter Luschny, Additive decompositions of classical numbers via the Swiss-Knife polynomials [_Tom Copeland_, Oct 20 2015]
Miguel Méndez and Rafael Sánchez, On the combinatorics of Riordan arrays and Sheffer polynomials: monoids, operads and monops, arXiv:1707.00336 [math.CO], 2017, Section 4.3, Example 4.
Miguel A. Méndez and Rafael Sánchez Lamoneda, Monops, Monoids and Operads: The Combinatorics of Sheffer Polynomials, The Electronic Journal of Combinatorics 25(3) (2018), #P3.25.

Row sums are A155585. - Johannes W. Meijer, Apr 20 2011
Rows reversed: A081658.
Cf. A109449, A153641, A162660. - Philippe Deléham, Oct 26 2013
Cf. A000182, A046802, A119467, A133314.
Cf. A019538, A028246, A090582, A123125, A130850, A248727.

T := (n,k) -> binomial(n,k)*2^(n-k)*euler(n-k,1/2): # Peter Luschny, Jan 25 2009

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 *)

{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

@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:

# uses[riordan_array from A256893]
riordan_array(sech(x), x, 9, exp=true) # Peter Luschny, Apr 19 2015

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

A131047 (1/2) * (A007318 - A007318^(-1)).

Original entry on oeis.org

1, 0, 2, 1, 0, 3, 0, 4, 0, 4, 1, 0, 10, 0, 5, 0, 6, 0, 20, 0, 6, 1, 0, 21, 0, 35, 0, 7, 0, 8, 0, 56, 0, 56, 0, 8, 1, 0, 36, 0, 126, 0, 84, 0, 9

Offset: 1

Gary W. Adamson, Jun 12 2007

Row sums = (1, 2, 4, 8, ...). A131047 * (1,2,3, ...) = A087447 starting (1, 4, 10, 24, 56, ...). A generalized set of analogous triangles: (1/(Q+1)) * (P^Q - 1/P), Q an integer, generates triangles with row sums = powers of (Q+1). Cf. A131048, A131049, A131050, A131051 for triangles having Q = 2,3,4 and 5, respectively.

A007318, Pascal's triangle, = this triangle + A119467, since one triangle = the zeros or masks of the other. - Gary W. Adamson, Jun 12 2007

			First few rows of the triangle:
  1;
  0, 2;
  1, 0,  3;
  0, 4,  0,  4;
  1, 0, 10,  0,  5;
  0, 6,  0, 20,  0, 6;
  1, 0, 21,  0, 35, 0, 7;
  ...

Cf. A131048, A131049, A131050, A131051.

Cf. A119467.

Let A007318 (Pascal's triangle) = P, then A131047 = (1/2) * (P - 1/P); deleting the right border of zeros.

A119468 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(n,2j)*binomial(n-2j,k).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 6, 3, 1, 8, 16, 12, 4, 1, 16, 40, 40, 20, 5, 1, 32, 96, 120, 80, 30, 6, 1, 64, 224, 336, 280, 140, 42, 7, 1, 128, 512, 896, 896, 560, 224, 56, 8, 1, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 1, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1

Offset: 0

Paul Barry, May 21 2006

Product of Pascal's triangle A007318 and A119467. Row sums are A007051. Diagonal sums are A113225.
Variant of A080928, A115068 and A082137. - R. J. Mathar, Feb 09 2010
Matrix inverse of the Euler tangent triangle A081733. - Peter Luschny, Jul 18 2012
Central column: T(2*n,n) = A069723(n). - Peter Luschny, Jul 22 2012
Subtriangle of the triangle in A198792. - Philippe Deléham, Nov 10 2013

			Triangle begins
    1;
    1,    1;
    2,    2,    1;
    4,    6,    3,    1;
    8,   16,   12,    4,    1;
   16,   40,   40,   20,    5,    1;
   32,   96,  120,   80,   30,    6,    1;
   64,  224,  336,  280,  140,   42,    7,   1;
  128,  512,  896,  896,  560,  224,   56,   8,  1;
  256, 1152, 2304, 2688, 2016, 1008,  336,  72,  9,  1;
  512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1;

A082137 read as triangle with rows reversed.

Cf. A000182, A009006, A038207, A081733, A155585.

A119468_row := proc(n) local s,t,k;
  s := series(exp(z*x)/(1-tanh(x)),x,n+2);
  t := factorial(n)*coeff(s,x,n); seq(coeff(t,z,k), k=(0..n)) end:
for n from 0 to 7 do A119468_row(n) od; # Peter Luschny, Aug 01 2012
# Alternatively:
T := (n, k) -> 2^(n-k-1+0^(n-k))*binomial(n,k):
for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, Nov 10 2017

A[k_] := Table[If[m < n, 1, -1], {m, k}, {n, k}]; a = Join[{{1}}, Table[(-1)^n*CoefficientList[CharacteristicPolynomial[A[n], x], x], {n, 1, 10}]]; Flatten[a] (* Roger L. Bagula and Gary W. Adamson, Jan 25 2009 *)
Table[Sum[Binomial[n,2j]Binomial[n-2j,k],{j,0,n-k}],{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Dec 14 2022 *)

R = PolynomialRing(QQ, 'x')
def p(n,x) :
  return 1 if n==0 else add((-1)^n*binomial(n,k)*(x^(n-k)-1) for k in range(n))
def A119468_row(n):
    x = R.gen()
    return [abs(cf) for cf in list((p(n,x-1)-p(n,x+1))/2+x^n)]
for n in (0..8) : print(A119468_row(n)) # Peter Luschny, Jul 22 2012

G.f.: (1 - x - xy)/(1 - 2x - 2x*y + 2x^2*y + x^2*y^2).
Number triangle T(n,k) = Sum_{j=0..n} binomial(n,j)*binomial(j,k)*(1+(-1)^(j-k))/2.
Define matrix: A(n,m,k) = If[m < n, 1, -1];
  p(x,k) = CharacteristicPolynomial[A[n,m,k],x]; then t(n,m) = coefficients(p(x,n)). - Roger L. Bagula and Gary W. Adamson, Jan 25 2009
E.g.f.: exp(x*z)/(1-tanh(x)). - Peter Luschny, Aug 01 2012
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) - T(n-2,k-2) for n >= 2, T(0,0) = T(1,0) = T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 10 2013
E.g.f.: [(e^(2t)+1)/2] e^(tx) = e^(P.(x)t), so this is an Appell sequence with lowering operator D = d/dx and raising operator R = x + 2/(e^(-2D)+1), i.e., D P_n(x) = n P_{n-1}(x) and R p_n(x) = P_{n+1}(x) where P_n(x) = [(x+2)^n + x^n]/2. Also, (P.(x)+y)^n = P_n(x+y), umbrally. R = x + 1 + D - 2 D^3/3! + ... contains the e.g.f.(D) mod signs of A009006 and A155585 and signed, aerated A000182, the zag numbers, so the unsigned differential component 2/[e^(2D)+1] = 2 Sum_{n >= 0} Eta(-n) (-2D)^n/n!, where Eta(s) is the Dirichlet eta function, and 2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers. The polynomials PI_n(x) of A081733 are the umbral compositional inverses of this sequence, i.e., P_n(PI.(x)) = x^n = PI_n(P.(x)) under umbral composition. Aside from the signs and the main diagonals, multiplying this triangle by 2 gives the face-vectors of the hypercubes A038207. - Tom Copeland, Sep 27 2015
T(n,k) = 2^(n-k-1+0^(n-k))*binomial(n, k). - Peter Luschny, Nov 10 2017

A162590 Polynomials with e.g.f. exp(x*t)/csch(t), triangle of coefficients read by rows.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 1, 0, 3, 0, 0, 4, 0, 4, 0, 1, 0, 10, 0, 5, 0, 0, 6, 0, 20, 0, 6, 0, 1, 0, 21, 0, 35, 0, 7, 0, 0, 8, 0, 56, 0, 56, 0, 8, 0, 1, 0, 36, 0, 126, 0, 84, 0, 9, 0, 0, 10, 0, 120, 0, 252, 0, 120, 0, 10, 0, 1, 0, 55, 0, 330, 0, 462, 0, 165, 0, 11, 0, 0, 12, 0, 220, 0, 792, 0, 792, 0

Offset: 0

Peter Luschny, Jul 07 2009

Comment from Peter Bala (Dec 06 2011): "Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of (I-t*M)^-1 (apart from the initial 1) lists the row polynomials for" A196776(n,k), which gives the number of ordered partitions of an n set into k odd-sized blocks. - Peter Luschny, Dec 06 2011

The n-th row of the triangle is formed by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probability of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms in the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k) as the (2k)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k). - Luca Onnis, Oct 29 2023

			Triangle begins:
  0
  1,  0
  0,  2,  0
  1,  0,  3,  0
  0,  4,  0,  4,  0
  1,  0, 10,  0,  5,  0
  0,  6,  0, 20,  0,  6,  0
  1,  0, 21,  0, 35,  0,  7,  0
  ...
  p[0](x) = 0;
  p[1](x) = 1
  p[2](x) = 2*x
  p[3](x) = 3*x^2 +  1
  p[4](x) = 4*x^3 +  4*x
  p[5](x) = 5*x^4 + 10*x^2 +  1
  p[6](x) = 6*x^5 + 20*x^3 +  6*x
  p[7](x) = 7*x^6 + 35*x^4 + 21*x^2 + 1
  p[8](x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x
.
Cf. the triangle of odd-numbered terms in rows of Pascal's triangle (A034867).
p[n] (k), n=0,1,...
k=0:  0, 1,  0,   1,    0,     1, ... A000035, (A059841)
k=1:  0, 1,  2,   4,    8,    16, ... A131577, (A000079)
k=2:  0, 1,  4,  13,   40,   121, ... A003462
k=3:  0, 1,  6,  28,  120,   496, ... A006516
k=4:  0, 1,  8,  49,  272,  1441, ... A005059
k=5:  0, 1, 10,  76,  520,  3376, ... A081199, (A016149)
k=6:  0, 1, 12, 109,  888,  6841, ... A081200, (A016161)
k=7:  0, 1, 14, 148, 1400, 12496, ... A081201, (A016170)
k=8:  0, 1, 16, 193, 2080, 21121, ... A081202, (A016178)
k=9:  0, 1, 18, 244, 2952, 33616, ... A081203, (A016186)
k=10: 0, 1, 20, 301, 4040, 51001, ... ......., (A016190)
.
p[n] (k), k=0,1,...
p[0]: 0,  0,   0,    0,    0,     0, ... A000004
p[1]: 1,  1,   1,    1,    1,     1, ... A000012
p[2]: 0,  2,   4,    6,    8,    10, ... A005843
p[3]: 1,  4,  13,   28,   49,    76, ... A056107
p[4]: 0,  8,  40,  120,  272,   520, ... A105374
p[5]: 1, 16, 121,  496, 1441,  3376, ...
p[6]: 0, 32, 364, 2016, 7448, 21280, ...

Paul and Tatjana Ehrenfest, Über zwei bekannte Einwände gegen das Boltzmannsche H-Theorem, Physikalische Zeitschrift, vol. 8 (1907), pp. 311-314.
Luca Onnis, Animation of the Ehrenfest model.
Wikipedia, Ehrenfest model.

Cf. A119467.

# Polynomials: p_n(x)
p := proc(n,x) local k;
pow := (n,k) -> `if`(n=0 and k=0,1,n^k);
add((k mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;
# Coefficients: a(n)
seq(print(seq(coeff(i!*coeff(series(exp(x*t)/csch(t), t,16),t,i),x,n), n=0..i)), i=0..8);

p[n_, x_] := Sum[Binomial[n, 2*k-1]*x^(n-2*k+1), {k, 0, n+2}]; row[n_] := CoefficientList[p[n, x], x] // Append[#, 0]&; Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 28 2013 *)
n = 15; "n-th row"
mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];
mat[[1, 2]] = 1;
mat[[n + 1, n]] = 1;
For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];
For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];
mat // MatrixForm;
P2 = Dot[mat, mat];
R1 = Simplify[
  Eigenvectors[Transpose[P2]][[1]]/
   Total[Eigenvectors[Transpose[P2]][[1]]]]
R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]
even = R1*2^(n - 1) (* Luca Onnis, Oct 29 2023 *)

p_n(x) = Sum_{k=0..n} (k mod 2)*binomial(n,k)*x^(n-k).
E.g.f.: exp(x*t)/csch(t) = 0*(t^0/0!) + 1*(t^1/1!) + (2*x)*(t^2/2!) + (3*x^2+1)*(t^3/3!) + ...
The 'co'-polynomials with generating function exp(x*t)*sech(t) are the Swiss-Knife polynomials (A153641).

A193673 Triangle given by p(n,k)=(coefficient of x^(n-k) in (1/2) ((x+3)^n+(x+1)^n)), 0<=k<=n.

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 14, 15, 6, 1, 41, 56, 30, 8, 1, 122, 205, 140, 50, 10, 1, 365, 732, 615, 280, 75, 12, 1, 1094, 2555, 2562, 1435, 490, 105, 14, 1, 3281, 8752, 10220, 6832, 2870, 784, 140, 16, 1, 9842, 29529, 39384, 30660, 15372, 5166, 1176, 180, 18, 1, 29525

Offset: 0

Clark Kimberling, Aug 02 2011

			First five rows:
   1
   2   1
   5   4   1
  14  15   6   1
  41  56  30   8   1

Cf. A119467, A119468, A193661.

q[n_, k_] := 1; r[0] = 1;
 r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}]
p[n_, k_] := Coefficient[(1/2) ((x + 3)^n + (x + 1)^n), x, k] (* A193673 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 20}]    (* A193661 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 10}, {k, 0, n}]] (* A193673 as a triangle *)
Flatten[%] (* A193673 as a sequence *)

From Mélika Tebni, Dec 09 2023: (Start)
T(n,k) = binomial(n,k)*(1 + 3^(n-k)) / 2.
E.g.f. of column k: exp(2*x)*cosh(x)*x^k / k!. (End)
From  Peter Bala, Mar 07 2024: (Start)
Exponential Riordan array (exp(2*x)*cosh(x), x).
The zeros of the n-th row polynomial R(n,x) = ((1 + x)^n + (3 + x)^n)/2 lie on the vertical line Re(x) = -2 in the complex plane.
Triangle equals P * (I + P^2)/2 = P * A119468 = P^2 * A119467, where P denotes Pascal's triangle A007318. (End)

A141665 A signed half of Pascal's triangle A007318: p(x,n) = (1+I*x)^n; t(n,m) = real part of coefficients(p(x,n)).

Original entry on oeis.org

1, 1, 0, 1, 0, -1, 1, 0, -3, 0, 1, 0, -6, 0, 1, 1, 0, -10, 0, 5, 0, 1, 0, -15, 0, 15, 0, -1, 1, 0, -21, 0, 35, 0, -7, 0, 1, 0, -28, 0, 70, 0, -28, 0, 1, 1, 0, -36, 0, 126, 0, -84, 0, 9, 0, 1, 0, -45, 0, 210, 0, -210, 0, 45, 0, -1

Offset: 0

Roger L. Bagula and Gary W. Adamson, Sep 05 2008

Polynomials like these are seen in complex dynamics.
This method symmetrically breaks up Pascal's triangle A007318 into two parts as polynomial coefficient vectors. See the examples for the s(n,m) = imaginary part of coefficients(p(x,n)).
From Johannes W. Meijer, Mar 10 2012: (Start)
The row sums equal A146559 and the two antidiagonal sums lead to A104862 (minus a(0)) and A110161 (minus a(0)).
The mirror of this triangle (for the absolute values of the coefficients) is A119467. (End)

			s(n,m) = imaginary part of coefficients(p(x,n))
  {0},
  {0,   1},
  {0,   2,   0},
  {0,   3,   0,   -1},
  {0,   4,   0,   -4,   0},
  {0,   5,   0,  -10,   0,   1},
  {0,   6,   0,  -20,   0,   6,   0},
  {0,   7,   0,  -35,   0,  21,   0,   -1},
  {0,   8,   0,  -56,   0,  56,   0,   -8,   0},
  {0,   9,   0,  -84,   0, 126,   0,  -36,   0,   1},
  {0,  10,   0, -120,   0, 252,   0, -120,   0,  10,   0}

G. C. Greubel, Rows n=0..100 of triangle, flattened
Clark Kimberling, Polynomials associated with reciprocation, JIS 12 (2009) 09.3.4, section 5.

Cf. A007318, A146559, A104862, A110161, A119467.

From Johannes W. Meijer, Mar 10 2012: (Start)
nmax:=10: for n from 0 to nmax do p(x,n) := (1+I*x)^n: for m from 0 to n do t(n,m) := Re(coeff(p(x,n), x, m)) od: od: seq(seq(t(n,m), m=0..n), n=0..nmax);
nmax:=10: for n from 0 to nmax do for m from 0 to n do A119467(n,m) := binomial(n,m) * (1+(-1)^(n-m))/2: if (m mod 4 = 2) then x(n,m):= -1 else x(n,m):= 1 end if: od: od: for n from 0 to nmax do for m from 0 to n do t(n,m) := A119467(n,n-m)*x(n,m) od: od: seq(seq(t(n,m), m=0..n), n=0..nmax); # (End)

p[x_, n_] := If[n == 0, 1, Product[(1 + I*x), {i, 1, n}]]; Table[Expand[p[x, n]], {n, 0, 10}]; Table[Im[CoefficientList[p[x, n], x]], {n, 0, 10}]; Flatten[%] Table[Re[CoefficientList[p[x, n], x]], {n, 0, 10}]; Flatten[%]

p(x,n) = (1+I*x)^n
t(n,m) = real part of coefficients(p(x,n))
s(n,m) = imaginary part of coefficients(p(x,n))

Edited and information added by Johannes W. Meijer, Mar 10 2012

A158335 A triangle of matrix polynomials: m(n)=antisymmeticmatix(n).Transpose[antisymmeticmatix(n)].

Original entry on oeis.org

1, 0, -1, 1, -2, 1, 0, -9, 6, -1, 1, -12, 38, -12, 1, 0, -25, 100, -110, 20, -1, 1, -30, 255, -452, 255, -30, 1, 0, -49, 490, -1519, 1484, -511, 42, -1, 1, -56, 924, -3976, 6470, -3976, 924, -56, 1, 0, -81, 1512, -9324, 21816, -21942, 9240, -1548, 72, -1, 1, -90

Offset: 0

Roger L. Bagula, Mar 16 2009

Row sums are:
{1, -1, 0, -4, 16, -16, 0, -64, 256, -256, 0,...}. Unsigned row sums are:
{1, 1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144,...}.
Example matrix is:
M(3)={{2, -1, -1},
{-1, 2, -1},
{-1, -1, 2}}

			{1},
{0, -1},
{1, -2, 1},
{0, -9, 6, -1},
{1, -12, 38, -12, 1},
{0, -25, 100, -110, 20, -1},
{1, -30, 255, -452, 255, -30, 1},
{0, -49, 490, -1519, 1484, -511, 42, -1},
{1, -56, 924, -3976, 6470, -3976, 924, -56, 1},
{0, -81, 1512, -9324, 21816, -21942, 9240, -1548, 72, -1},
{1, -90, 2445, -19320, 63090, -92252, 63090, -19320, 2445, -90, 1}

A119467

Clear[M, T, d, a, x, a0];
T[n_, m_, d_] := If[ m < n, (-1)^(n + m), If[m > n, -(-1)^(n + m), 0]];
M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}].Transpose[Table[T[n, m, d], {n, 1, d}, {m, 1, d}]];
Table[Det[M[d]], {d, 1, 10}];
Table[M[d], {d, 1, 10}]
Table[CharacteristicPolynomial[M[d], x], {d, 1, 10}];
a = Join[{{1}}, Table[CoefficientList[Expand[CharacteristicPolynomial[M[ n], x]], x], {n, 1, 10}]];
Flatten[a]; Join[{1}, Table[Apply[Plus, CoefficientList[Expand[ CharacteristicPolynomial[M[n], x]], x]], {n, 1, 10}]];

m(n)=antisymmeticmatix(n).Transpose[antisymmeticmatix(n)];

out_(n,m)=coefficients(characteristicpolynomial(m(n),x),x).

A158336 A triangle of matrix polynomials: m(n)=antisymmeticmatix(n).pseudotranspose[antisymmeticmatix(n)].

Original entry on oeis.org

1, 0, -1, -1, 0, 1, 0, 9, 0, -1, 1, 0, -34, 0, 1, 0, -25, 0, 90, 0, -1, -1, 0, 195, 0, -195, 0, 1, 0, 49, 0, -931, 0, 371, 0, -1, 1, 0, -644, 0, 3334, 0, -644, 0, 1, 0, -81, 0, 4788, 0, -9846, 0, 1044, 0, -1, -1, 0, 1605, 0, -25290, 0, 25290, 0, -1605, 0, 1

Offset: 0

Roger L. Bagula, Mar 16 2009

The pseusotranspose operation used is: pseudotranspose[a(n)]=Reverse[I(n)].a(n).
Row sums are:
{1, -1, 0, 8, -32, 64, 0, -512, 2048, -4096, 0,...}. Unsigned row sums are:
{1, 1, 2, 10, 36, 116, 392, 1352, 4624, 15760, 53792,...}.
Example matrix is:
M(3)={{-1, -1, 2},
{-1, 2, -1},
{2, -1, -1}}

			{1},
{0, -1},
{-1, 0, 1},
{0, 9, 0, -1},
{1, 0, -34, 0, 1},
{0, -25, 0, 90, 0, -1},
{-1, 0, 195, 0, -195, 0, 1},
{0, 49, 0, -931, 0, 371, 0, -1},
{1, 0, -644, 0, 3334, 0, -644, 0, 1},
{0, -81, 0, 4788, 0, -9846, 0, 1044, 0, -1},
{-1, 0, 1605, 0, -25290, 0, 25290, 0, -1605, 0, 1}

A119467

Clear[M, T, d, a, x, a0];
pt[a_] := Reverse[IdentityMatrix[Length[a]]].a;
T[n_, m_, d_] := If[ m < n, (-1)^(n + m), If[m > n, -(-1)^(n + m), 0]];
M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}].pt[Table[T[ n, m, d], {n, 1, d}, {m, 1, d}]];
Table[Det[M[d]], {d, 1, 10}];
Table[M[d], {d, 1, 10}]
Table[CharacteristicPolynomial[M[d], x], {d, 1, 10}];
a = Join[{{1}}, Table[CoefficientList[Expand[CharacteristicPolynomial[M[ n], x]], x], {n, 1, 10}]];
Flatten[a]; Join[{1}, Table[Apply[Plus, CoefficientList[Expand[ CharacteristicPolynomial[M[n], x]], x]], {n, 1, 10}]];

m(n)=antisymmeticmatix(n).pseudotranspose[antisymmeticmatix(n)].;

out_(n,m)=coefficients(characteristicpolynomial(m(n),x),x).

cp's OEIS Frontend

A131054 Triangle read by rows: A000012 * A119467 as infinite lower triangular matrices.

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A136630 Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.

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

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A131047 (1/2) * (A007318 - A007318^(-1)).

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A119468 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(n,2j)*binomial(n-2j,k).

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A162590 Polynomials with e.g.f. exp(x*t)/csch(t), triangle of coefficients read by rows.

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A193673 Triangle given by p(n,k)=(coefficient of x^(n-k) in (1/2) ((x+3)^n+(x+1)^n)), 0<=k<=n.

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A141665 A signed half of Pascal's triangle A007318: p(x,n) = (1+I*x)^n; t(n,m) = real part of coefficients(p(x,n)).