A118801 -id:A118801 - cp's OEIS frontend

A038207 Triangle whose (i,j)-th entry is binomial(i,j)*2^(i-j).

1, 2, 1, 4, 4, 1, 8, 12, 6, 1, 16, 32, 24, 8, 1, 32, 80, 80, 40, 10, 1, 64, 192, 240, 160, 60, 12, 1, 128, 448, 672, 560, 280, 84, 14, 1, 256, 1024, 1792, 1792, 1120, 448, 112, 16, 1, 512, 2304, 4608, 5376, 4032, 2016, 672, 144, 18, 1, 1024, 5120, 11520, 15360, 13440, 8064, 3360, 960, 180, 20, 1

Offset: 0

N. J. A. Sloane

This infinite matrix is the square of the Pascal matrix (A007318) whose rows are [ 1,0,... ], [ 1,1,0,... ], [ 1,2,1,0,... ], ...
As an upper right triangle, table rows give number of points, edges, faces, cubes,
4D hypercubes etc. in hypercubes of increasing dimension by column. - Henry Bottomley, Apr 14 2000. More precisely, the (i,j)-th entry is the number of j-dimensional subspaces of an i-dimensional hypercube (see the Coxeter reference). - Christof Weber, May 08 2009
Number of different partial sums of 1+[1,1,2]+[2,2,3]+[3,3,4]+[4,4,5]+... with entries that are zero removed. - Jon Perry, Jan 01 2004
Row sums are powers of 3 (A000244), antidiagonal sums are Pell numbers (A000129). - Gerald McGarvey, May 17 2005
Riordan array (1/(1-2x), x/(1-2x)). - Paul Barry, Jul 28 2005
T(n,k) is the number of elements of the Coxeter group B_n with descent set contained in {s_k}, 0<=k<=n-1. For T(n,n), we interpret this as the number of elements of B_n with empty descent set (since s_n does not exist). - Elizabeth Morris (epmorris(AT)math.washington.edu), Mar 01 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then T(n,k) = the number of elements (x,y) of S for which y has exactly k more elements than x. - Ross La Haye, Oct 12 2007
T(n,k) is number of paths in the first quadrant going from (0,0) to (n,k) using only steps B=(1,0) colored blue, R=(1,0) colored red and U=(1,1). Example: T(3,2)=6 because we have BUU, RUU, UBU, URU, UUB and UUR. - Emeric Deutsch, Nov 04 2007
T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), and two kinds of step (1,0). - Joerg Arndt, Jul 01 2011
T(i,j) is the number of i-permutations of {1,2,3} containing j 1's. Example: T(2,1)=4 because we have 12, 13, 21 and 31; T(3,2)=6 because we have 112, 113, 121, 131, 211 and 311. - Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (2+x)^n. - N-E. Fahssi, Apr 13 2008
Sum of diagonals are Jacobsthal-numbers: A001045. - Mark Dols, Aug 31 2009
Triangle T(n,k), read by rows, given by [2,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 15 2009
Eigensequence of the triangle = A004211: (1, 3, 11, 49, 257, 1539, ...). - Gary W. Adamson, Feb 07 2010
f-vectors ("face"-vectors) for n-dimensional cubes [see e.g., Hoare]. (This is a restatement of Bottomley's above.) - Tom Copeland, Oct 19 2012
With P = Pascal matrix, the sequence of matrices I, A007318, A038207, A027465, A038231, A038243, A038255, A027466 ... = P^0, P^1, P^2, ... are related by Copeland's formula below to the evolution at integral time steps n= 0, 1, 2, ... of an exponential distribution exp(-x*z) governed by the Fokker-Planck equation as given in the Dattoli et al. ref. below. - Tom Copeland, Oct 26 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Unsigned diagonals of A133156 are rows of this array. - Tom Copeland, Oct 11 2014
Omitting the first row, this is the production matrix for A039683, where an equivalent differential operator can be found. - Tom Copeland, Oct 11 2016
T(n,k) is the number of functions f:[n]->[3] with exactly k elements mapped to 3. Note that there are C(n,k) ways to choose the k elements mapped to 3, and there are 2^(n-k) ways to map the other (n-k) elements to {1,2}. Hence, by summing T(n,k) as k runs from 0 to n, we obtain 3^n = Sum_{k=0..n} T(n,k). - Dennis P. Walsh, Sep 26 2017
Since this array is the square of the Pascal lower triangular matrix, the row polynomials of this array are obtained as the umbral composition of the row polynomials P_n(x) of the Pascal matrix with themselves. E.g., P_3(P.(x)) = 1 P_3(x) + 3 P_2(x) + 3 P_1(x) + 1 = (x^3 + 3 x^2 + 3 x + 1) + 3 (x^2 + 2 x + 1) + 3 (x + 1) + 1 = x^3 + 6 x^2 + 12 x + 8. - Tom Copeland, Nov 12 2018
T(n,k) is the number of 2-compositions of n+1 with some zeros allowed that have k zeros; see the Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020
Also the convolution triangle of A000079. - Peter Luschny, Oct 09 2022

			Triangle begins with T(0,0):
   1;
   2,  1;
   4,  4,  1;
   8, 12,  6,  1;
  16, 32, 24,  8,  1;
  32, 80, 80, 40, 10,  1;
  ... -  corrected by _Clark Kimberling_, Aug 05 2011
Seen as an array read by descending antidiagonals:
[0] 1, 2,  4,   8,    16,    32,    64,     128,     256, ...     [A000079]
[1] 1, 4,  12,  32,   80,    192,   448,    1024,    2304, ...    [A001787]
[2] 1, 6,  24,  80,   240,   672,   1792,   4608,    11520, ...   [A001788]
[3] 1, 8,  40,  160,  560,   1792,  5376,   15360,   42240, ...   [A001789]
[4] 1, 10, 60,  280,  1120,  4032,  13440,  42240,   126720, ...  [A003472]
[5] 1, 12, 84,  448,  2016,  8064,  29568,  101376,  329472, ...  [A054849]
[6] 1, 14, 112, 672,  3360,  14784, 59136,  219648,  768768, ...  [A002409]
[7] 1, 16, 144, 960,  5280,  25344, 109824, 439296,  1647360, ... [A054851]
[8] 1, 18, 180, 1320, 7920,  41184, 192192, 823680,  3294720, ... [A140325]
[9] 1, 20, 220, 1760, 11440, 64064, 320320, 1464320, 6223360, ... [A140354]

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 155.
H. S. M. Coxeter, Regular Polytopes, Dover Publications, New York (1973), p. 122.

T. D. Noe, Rows n=0..100 of triangle, flattened
Peter Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, On the f-Matrices of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1805.02274 [math.CO], 2018.
Jhon J. Bravo, Jose L. Herrera, and José L. Ramírez, Combinatorial Interpretation of Generalized Pell Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.2.1.
John Cartan, Starmaze: Cartan's Triangle.
Tom Copeland, Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras.
B. N. Cyvin, J. Brunvoll, and S. J. Cyvin, Isomer enumeration of unbranched catacondensed polygonal systems with pentagons and heptagons, Match, No. 34 (Oct 1996), 109-121.
S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757-774.
S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Isomer enumeration of some polygonal systems representing polycyclic conjugated hydrocarbons, Journal of Molecular Structure 376 (1996), 495-505.
G. Dattoli, A. Mancho, M. Quattromini and A. Torre, Exponential operators, generalized polynomials and evolution problems, Radiation Physics and Chemistry 61 (2001), 99-108. [From _Tom Copeland_, Oct 25 2012]
Filippo Disanto, Some Statistics on the Hypercubes of Catalan Permutations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.2.
Shishuo Fu and Yaling Wang, Bijective recurrences concerning two Schröder triangles, arXiv:1908.03912 [math.CO], 2019.
W. G. Harter, Representations of multidimensional symmetries in networks, J. Math. Phys., 15 (1974), 2016-2021.
Russell Jay Hendel, A Method for Uniformly Proving a Family of Identities, arXiv:2107.03549 [math.CO], 2021.
Graham Hoare, Hypercubes and Chebyshev, Math. Gaz. 74 (470) (1990), 375-377.
Brian Hopkins and Stéphane Ouvry, Combinatorics of Multicompositions, arXiv:2008.04937 [math.CO], 2020.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Marin Knežević, Vedran Krčadinac, and Lucija Relić, Matrix products of binomial coefficients and unsigned Stirling numbers, arXiv:2012.15307 [math.CO], 2020.
Katarzyna Kril and Wojciech Mlotkowski, Permutations of Type B with Fixed Number of Descents and Minus Signs, The Electronic Journal of Combinatorics, Vol. 26(1) (2019), Article P1.27.
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
Thomas Selig and Haoyue Zhu, Complete non-ambiguous trees and associated permutations: connections through the Abelian sandpile model, arXiv:2303.15756 [math.CO], 2023, see p. 27.
Wikipedia, Hypercube.

See A065109 for a signed version.
Columns 0-10 are A000079, A001787, A001788(n-1), A001789, A003472, A054849, A002409(n-6), A054851, A140325(n-8), A140354(n-9), A172242.
T(2n,n) gives A059304.
Cf. A007318, A013609, A013610, A062715, A099089, A004211, A135387.
Cf. A001861, A008277, A027465, A027466, A038231, A038243, A038255, A039683, A112857, A118801, A132440, A133156, A133314, A167374, A218272, A238385, A323939.

Flat(List([0..15], n->List([0..n], k->Binomial(n, k)*2^(n-k)))); # Stefano Spezia, Nov 21 2018

a038207 n = a038207_list !! n
a038207_list = concat $ iterate ([2,1] *) [1]
instance Num a => Num [a] where
   fromInteger k = [fromInteger k]
   (p:ps) + (q:qs) = p + q : ps + qs
   ps + qs         = ps ++ qs
   (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
    *                = []
-- Reinhard Zumkeller, Apr 02 2011

a038207' n k = a038207_tabl !! n !! k
a038207_row n = a038207_tabl !! n
a038207_tabl = iterate f [1] where
   f row = zipWith (+) ([0] ++ row) (map (* 2) row ++ [0])
-- Reinhard Zumkeller, Feb 27 2013

/* As triangle */ [[(&+[Binomial(n,i)*Binomial(i,k): i in [k..n]]): k in [0..n]]: n in [0..15]]; // Vincenzo Librandi, Nov 16 2018

for i from 0 to 12 do seq(binomial(i, j)*2^(i-j), j = 0 .. i) end do; # yields sequence in triangular form - Emeric Deutsch, Nov 04 2007
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, n -> 2^(n-1)); # Peter Luschny, Oct 09 2022

Table[CoefficientList[Expand[(y + x + x^2)^n], y] /. x -> 1, {n, 0,10}] // TableForm (* Geoffrey Critzer, Nov 20 2011 *)
Table[Binomial[n,k]2^(n-k),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, May 22 2020 *)

{T(n, k) = polcoeff((x+2)^n, k)}; /* Michael Somos, Apr 27 2000 */

def A038207_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in range(dim): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1]+2*M[n-1,k]
    return M
A038207_triangle(9)  # Peter Luschny, Sep 20 2012

T(n, k) = Sum_{i=0..n} binomial(n,i)*binomial(i,k).
T(n, k) = (-1)^k*A065109(n,k).
G.f.: 1/(1-2*z-t*z). - Emeric Deutsch, Nov 04 2007
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
From the formalism of A133314, the e.g.f. for the row polynomials of A038207 is exp(x*t)*exp(2x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-2x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*2x). The results generalize for 2 replaced by any number. - Tom Copeland, Aug 18 2008
Sum_{k=0..n} T(n,k)*x^k = (2+x)^n. - Philippe Deléham, Dec 15 2009
n-th row is obtained by taking pairwise sums of triangle A112857 terms starting from the right. - Gary W. Adamson, Feb 06 2012
T(n,n) = 1 and T(n,k) = T(n-1,k-1) + 2*T(n-1,k) for kJon Perry, Oct 11 2012
The e.g.f. for the n-th row is given by umbral composition of the normalized Laguerre polynomials A021009 as p(n,x) = L(n, -L(.,-x))/n! = 2^n L(n, -x/2)/n!. E.g., L(2,x) = 2 -4*x +x^2, so p(2,x)= (1/2)*L(2, -L(.,-x)) = (1/2)*(2*L(0,-x) + 4*L(1,-x) + L(2,-x)) = (1/2)*(2 + 4*(1+x) + (2+4*x+x^2)) = 4 + 4*x + x^2/2. - Tom Copeland, Oct 20 2012
From Tom Copeland, Oct 26 2012: (Start)
From the formalism of A132440 and A218272:
Let P and P^T be the Pascal matrix and its transpose and H= P^2= A038207.
Then with D the derivative operator,
   exp(x*z/(1-2*z))/(1-2*z)= exp(2*z D_z z) e^(x*z)= exp(2*D_x (x D_x)) e^(z*x)
   = (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
   = (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T
   = Sum_{n>=0} z^n * 2^n Lag_n(-x/2)= exp[z*EF(.,x)], an o.g.f. for the f-vectors (rows) of A038207 where EF(n,x) is an e.g.f. for the n-th f-vector. (Lag_n(x) are the un-normalized Laguerre polynomials.)
Conversely,
   exp(z*(2+x))= exp(2D_x) exp(x*z)= exp(2x) exp(x*z)
   = (1 x x^2 x^3 ...) H^T (1 z z^2/2! z^3/3! ...)^T
   = (1 z z^2/2! z^3/3! ...) H (1 x x^2 x^3 ...)^T
   = exp(z*OF(.,x)), an e.g.f for the f-vectors of A038207 where
     OF(n,x)= (2+x)^n is an o.g.f. for the n-th f-vector.
(End)
G.f.: R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ (1+y))*x/((2*k+2+ (1+y))*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
A038207 = exp[M*B(.,2)] where M = A238385-I and (B(.,x))^n = B(n,x) are the Bell polynomials (cf. A008277). B(n,2) = A001861(n). - Tom Copeland, Apr 17 2014
T = (A007318)^2 = A112857*|A167374| = |A118801|*|A167374| = |A118801*A167374| = |P*A167374*P^(-1)*A167374| = |P*NpdP*A167374|. Cf. A118801. - Tom Copeland, Nov 17 2016
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial 2^n*Sum_{k = 0..n} binomial(n,k)*x^k/k!. For example, the e.g.f. for the third subdiagonal is exp(x)*(8 + 24*x + 12*x^2 + 4*x^3/3) = 8 + 32*x + 80*x^2/2! + 160*x^3/3! + .... - Peter Bala, Mar 05 2017
T(3*k+2,k) = T(3*k+2,k+1), T(2*k+1,k) = 2*T(2*k+1,k+1). - Yuchun Ji, May 26 2020
From Robert A. Russell, Aug 05 2020: (Start)
G.f. for column k: x^k / (1-2*x)^(k+1).
E.g.f. for column k: exp(2*x) * x^k / k!. (End)
Also the array A(n, k) read by descending antidiagonals, where A(n, k) = (-1)^n*Sum_{j= 0..n+k} binomial(n + k, j)*hypergeom([-n, j+1], [1], 1). - Peter Luschny, Nov 09 2021

A119258 Triangle read by rows: T(n,0) = T(n,n) = 1 and for 0

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 7, 17, 15, 1, 1, 9, 31, 49, 31, 1, 1, 11, 49, 111, 129, 63, 1, 1, 13, 71, 209, 351, 321, 127, 1, 1, 15, 97, 351, 769, 1023, 769, 255, 1, 1, 17, 127, 545, 1471, 2561, 2815, 1793, 511, 1, 1, 19, 161, 799, 2561, 5503, 7937, 7423, 4097, 1023, 1

Offset: 0

Reinhard Zumkeller, May 11 2006

From Richard M. Green, Jul 26 2011: (Start)
T(n,n-k) is the (k-1)-st Betti number of the subcomplex of the n-dimensional half cube obtained by deleting the interiors of all half-cube shaped faces of dimension at least k.
T(n,n-k) is the (k-2)-nd Betti number of the complement of the k-equal real hyperplane arrangement in R^n.
T(n,n-k) gives a lower bound for the complexity of the problem of determining, given n real numbers, whether some k of them are equal.
T(n,n-k) is the number of nodes used by the Kronrod-Patterson-Smolyak cubature formula in numerical analysis. (End)

			Triangle begins as:
  1;
  1, 1;
  1, 3,  1;
  1, 5,  7,  1;
  1, 7, 17, 15,  1;
  1, 9, 31, 49, 31, 1;

Reinhard Zumkeller, Rows n=0..120 of triangle, flattened
A. Björner and V. Welker, The homology of "k-equal" manifolds and related partition lattices, Adv. Math., 110 (1995), 277-313.
Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO], 2023.
R. M. Green, Homology representations arising from the half cube, Adv. Math., 222 (2009), 216-239.
R. M. Green, Homology representations arising from the half cube, II, J. Combin. Theory Ser. A, 117 (2010), 1037-1048.
R. M. Green, Homology representations arising from the half cube, II, arXiv:0812.1208 [math.RT], 2008.
R. M. Green and Jacob T. Harper, Morse matchings on polytopes, arXiv preprint arXiv:1107.4993 [math.GT], 2011.
OEIS Wiki, Sequence of the Day for November 3.
K. Petras, On the Smolyak cubature error for analytic functions, Adv. Comput. Math., 12 (2000), 71-93.
M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Q., 48 (2010), 290-297.
M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, arXiv:1107.1063 [math.CO], 2010.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A119259, A007318, A007051, A005408, A056220, A027608, A055580, A000337, A000225, A112857 (row reverse).

A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011

a119258 n k = a119258_tabl !! n !! k
a119258_row n = a119258_tabl !! n
a119258_list = concat a119258_tabl
a119258_tabl = iterate (\row -> zipWith (+)
   ([0] ++ init row ++ [0]) $ zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Reinhard Zumkeller, Nov 15 2011

function T(n,k)
  if k eq 0 or k eq n then return 1;
  else return 2*T(n-1,k-1) + T(n-1,k);
  end if;
  return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 18 2019

# Case m = 2 of the more general:
A119258 := (n,k,m) -> (1-m)^(-n+k)-m^(k+1)*pochhammer(n-k, k+1) *hypergeom([1,n+1],[k+2],m)/(k+1)!;
seq(seq(round(evalf(A119258(n,k,2))),k=0..n), n=0..10); # Peter Luschny, Jul 25 2014

T[n_, k_] := Binomial[n, k] Hypergeometric2F1[-k, n-k, n-k+1, -1];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 10 2017 *)

T(n,k) = if(k==0 || k==n, 1, 2*T(n-1, k-1) + T(n-1,k) ); \\ G. C. Greubel, Nov 18 2019

@CachedFunction
def T(n, k):
    if (k==0 or k==n): return 1
    else: return 2*T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 18 2019

T(2*n,n-1) = T(2*n-1,n) for n > 0;
central terms give A119259; row sums give A007051;
T(n,0) = T(n,n) = 1;
T(n,1) = A005408(n-1) for n > 0;
T(n,2) = A056220(n-1) for n > 1;
T(n,n-4) = A027608(n-4) for n > 3;
T(n,n-3) = A055580(n-3) for n > 2;
T(n,n-2) = A000337(n-1) for n > 1;
T(n,n-1) = A000225(n) for n > 0.
T(n,k) = [k<=n]*(-1)^k*Sum_{i=0..k} (-1)^i*C(k-n,k-i)*C(n,i). - Paul Barry, Sep 28 2007
From Richard M. Green, Jul 26 2011: (Start)
T(n,k) = [k<=n] Sum_{i=n-k..n} (-1)^(n-k-i)*2^(n-i)*C(n,i).
T(n,k) = [k<=n] Sum_{i=n-k..n} C(n,i)*C(i-1,n-k-1).
G.f. for T(n,n-k): x^k/(((1-2x)^k)*(1-x)). (End)
T(n,k) = R(n,k,2) where R(n, k, m) = (1-m)^(-n+k)-m^(k+1)*Pochhammer(n-k,k+1)* hyper2F1([1,n+1], [k+2], m)/(k+1)!. - Peter Luschny, Jul 25 2014
From Peter Bala, Mar 05 2018: (Start)
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = 1 + 7*x + 17*x^2 + 15*x^3 + x^4 + O(x^5).
Row reverse of A112857. (End)

A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 2, -3, 1, 4, -8, 5, -1, 8, -20, 18, -7, 1, 16, -48, 56, -32, 9, -1, 32, -112, 160, -120, 50, -11, 1, 64, -256, 432, -400, 220, -72, 13, -1, 128, -576, 1120, -1232, 840, -364, 98, -15, 1, 256, -1280, 2816, -3584, 2912, -1568, 560, -128, 17, -1, 512, -2816, 6912, -9984, 9408, -6048, 2688, -816, 162, -19, 1

Offset: 0

Paul D. Hanna, May 02 2006

The matrix square, T^2, consists of columns that are all the same.
Matrix inverse is triangle A118801. Row sums form {0^n, n>=0}.
Unsigned row sums equal A025192(n) = 2*3^(n-1), n>=1.
Row squared sums equal A051708.
Antidiagonal sums equals all 1's.
Unsigned antidiagonal sums form A078057 (with offset).
Antidiagonal squared sums form A002002(n) = Sum_{k=0..n-1} C(n,k+1)*C(n+k,k), n>=1.
From Paul Barry, Nov 10 2008: (Start)
T is [1,1,0,0,0,...] DELTA [ -1,0,0,0,0,...] or C(1,n) DELTA -C(0,n). (DELTA defined in A084938).
The positive matrix T_p is [1,1,0,0,0,...] DELTA [1,0,0,0,0,...]. T_p*C^-1 is
[0,1,0,0,0,....] DELTA [1,0,0,0,0,...] which is C(n-1,k-1) for n,k>=1. (End)
The triangle formed by deleting the minus signs is the mirror of the self-fusion of Pascal's triangle; see Comments at A081277 and A193722. - Clark Kimberling, Aug 04 2011
Riordan array ( (1 - x)/(1 - 2*x), -x/(1 - 2*x) ). Cf. A209149. The matrix square is the Riordan array ( (1 - x)^2/(1 - 2*x), x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013
From Peter Bala, Feb 23 2019: (Start)
There is a 1-parameter family of solutions to the simultaneous equations C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle. Let T(k) denote the Riordan array ( (1 - k*x)/(1 - (k + 1)*x), -x/(1 - (k + 1)*x) ) so that T(1) = T. Then C*T(k)*C = T(k)^-1 and T(k)*C*T(k) = C^-1, for arbitrary k. For arbitrary m, the Riordan arrays (T(k)*C^m)^2 and (C^m*T(k))^2 both belong to the Appell subgroup of the Riordan group.
More generally, given a fixed m, we can ask for a lower triangular array X solving the simultaneous equations (C^m)*X*(C^m) = X^-1 and X*(C^m)*X = C^(-m). A 1-parameter family of solutions is given by the Riordan arrays X = ( (1 - m*k*x)/(1 - m*(k + 1)*x), -x/(1 - m*(k + 1)*x) ). The Riordan arrays X^2 , (X*C^n)^2 and (C^n*X)^2, for arbitrary n, all belong to the Appell subgroup of the Riordan group. (End)

			Triangle begins:
     1;
     1,    -1;
     2,    -3,     1;
     4,    -8,     5,     -1;
     8,   -20,    18,     -7,     1;
    16,   -48,    56,    -32,     9,     -1;
    32,  -112,   160,   -120,    50,    -11,     1;
    64,  -256,   432,   -400,   220,    -72,    13,    -1;
   128,  -576,  1120,  -1232,   840,   -364,    98,   -15,    1;
   256, -1280,  2816,  -3584,  2912,  -1568,   560,  -128,   17,   -1;
   512, -2816,  6912,  -9984,  9408,  -6048,  2688,  -816,  162,  -19,  1;
  1024, -6144, 16640, -26880, 28800, -21504, 11424, -4320, 1140, -200, 21, -1;
  ...
The matrix square, T^2, equals:
   1;
   0,  1;
   1,  0,  1;
   2,  1,  0,  1;
   4,  2,  1,  0,  1;
   8,  4,  2,  1,  0,  1;
  16,  8,  4,  2,  1,  0,  1;
  32, 16,  8,  4,  2,  1,  0,  1;
  64, 32, 16,  8,  4,  2,  1,  0,  1; ...
where all columns are the same.

Paul D. Hanna, Rows 0..45 of triangle, flattened.
Florian Luca, Attila Pethő, and László Szalay, Duplications in the k-generalized Fibonacci sequences, New York J. Math. (2021) Vol. 27, 1115-1133.

Cf. A118801 (inverse), A025192 (unsigned row sums), A051708 (row squared sums), A078057 (unsigned antidiagonal sums), A002002 (antidiagonal squared sums).

Cf. A135387, A209149, A000012, A097805, A130595, A167374, A200139, A239473, A321620.

(* This program generates A118800 as the mirror of the self-fusion of Pascal's triangle. *)
z = 8; a = 1; b = 1; c = 1; d = 1;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n;
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, 0];
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, _] = 1;
g[n_] := CoefficientList[w[n, -x], x];
TableForm[Table[Reverse[Abs@g[n]], {n, -1, z}]]
Flatten[Table[Reverse[Abs@g[n]], {n, -1, z}]] (* A081277 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A118800 *)
(* Clark Kimberling, Aug 04 2011 *)
T[ n_, k_] := If[ n<0 || k<0, 0, (-1)^k 2^(n-k) (Binomial[ n, k] + Binomial[ n-1, n-k]) / 2]; (* Michael Somos, Nov 25 2016 *)

{T(n,k)=if(n==0&k==0,1,(-1)^k*2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

/* Chebyshev Polynomials as Antidiagonals: */
{T(n,k)=local(Ox=x*O(x^(2*k))); polcoeff(((1+sqrt(1-x^2+Ox))^(n+k)+(1-sqrt(1-x^2+Ox))^(n+k))/2,2*k,x)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

# uses[riordan_square from A321620]
# Computes the unsigned triangle.
riordan_square((1-x)/(1-2*x), 8) # Peter Luschny, Jan 03 2019

T(n,k) = (-1)^k * 2^(n-k) * ( C(n,k) + C(n-1,k-1) )/2 for n>=k>=0 with T(0,0) = 1. Antidiagonals form the coefficients of Chebyshev polynomials: T(n,k) = [x^(2*n)] [(1+sqrt(1-x^2))^(n+k) + (1-sqrt(1-x^2))^(n+k)]/2.
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
O.g.f.: (1 - t)/(1 + t*(x - 2)) = 1 + (1 - x)*t + (2 - 3*x + x^2)^t^2 + (4 - 8*x + 5*x^2 - x^3)*t^3 + ....  Row polynomial R(n,x) = (1 - x)*(2 - x)^(n-1) for n >= 1. - Peter Bala, Jul 17 2013
T(n,k)=2*T(n-1,k)-T(n-1,k-1) with T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 25 2013
G.f. for row n (n>=1): Sum_{k=0..n} T(n,k)*x^k = (1-x)*(2-x)^(n-1). - Philippe Deléham, Nov 25 2013
From Tom Copeland, Nov 15 2016: (Start)
E.g.f. is [1 + (1-x)e^((2-x)t)]/(2-x), so the row polynomials are p_n(x) = (1-q,(x))^n, umbrally, where (q.(x))^k = q_k(x) are the row polynomials of A239473, or, equivalently, T = M*A239473, where M is the inverse Pascal matrix C^(-1) = A130595 with the odd rows negated, i.e., M(n,k) = (-1)^n C^(-1)(n,k) with e.g.f. exp[(1-x)t]. Cf. A200139: A200139(n,k) = (-1)^k* A118800(n,k).
TCT = C^(-1) = A130595 and A239473 = A000012*C^(-1) = S*C^(-1) imply (M*S)^2 = Identity matrix, i.e., M*S = (M*S)^(-1) = S^(-1)*M^(-1) = A167374*M^(-1). Note that M = M^(-1). Cf. A097805. (End)

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Abdullahi Umar, Aug 25 2008

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Harry Crane, Left-right arrangements, set partitions, and pattern avoidance, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.

Cf. A007051, A118801, A135233, columns: A000225, A000337, A055580, A027608.
Cf. also A038207, A250118, A250119.
Cf. A007318, A167374, A145661, A029635.

A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011

Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 8, 20, 18, 7, 1, 16, 48, 56, 32, 9, 1, 32, 112, 160, 120, 50, 11, 1, 64, 256, 432, 400, 220, 72, 13, 1, 128, 576, 1120, 1232, 840, 364, 98, 15, 1, 256, 1280, 2816, 3584, 2912, 1568, 560, 128, 17, 1, 512, 2816, 6912, 9984, 9408, 6048, 2688, 816, 162, 19, 1

Offset: 0

Philippe Deléham, Nov 13 2011

Riordan array ((1-x)/(1-2x),x/(1-2x)).
Product A097805*A007318 as infinite lower triangular arrays.
Product A193723*A130595 as infinite lower triangular arrays.
T(n,k) is the number of ways to place n unlabeled objects into any number of labeled bins (with at least one object in each bin) and then designate k of the bins. - Geoffrey Critzer, Nov 18 2012
Apparently, rows of this array are unsigned diagonals of A028297. - Tom Copeland, Oct 11 2014
Unsigned A118800, so my conjecture above is true. - Tom Copeland, Nov 14 2016

			Triangle begins:
   1
   1,   1
   2,   3,   1
   4,   8,   5,   1
   8,  20,  18,   7,   1
  16,  48,  56,  32,   9,   1
  32, 112, 160, 120,  50,  11,   1

Cf. A118800 (signed version), A081277, A039991, A001333 (antidiagonal sums), A025192 (row sums); diagonals: A000012, A005408, A001105, A002492, A072819l; columns: A011782, A001792, A001793, A001794, A006974, A006975, A006976.

Cf. A007318, A028297, A059260, A097805, A118801, A130595.

nn=15;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[(1-x)/(1-2x-y x) ,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Nov 18 2012 *)

T(n,k) = 2*T(n-1,k)+T(n-1,k-1) with T(0,0)=T(1,0)=T(1,1)=1 and T(n,k)=0 for k<0 or for n

T(n,k) = A011782(n-k)*A135226(n,k) = 2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2.

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for n=-1,0,1,2,3,4,5,6,7,8,9,10 respectively.

G.f.: (1-x)/(1-(2+y)*x).

T(n,k) = Sum_j>=0 T(n-1-j,k-1)*2^j.

T = A007318*A059260, so the row polynomials of this entry are given umbrally by p_n(x) = (1 + q.(x))^n, where q_n(x) are the row polynomials of A059260 and (q.(x))^k = q_k(x). Consequently, the e.g.f. is exp[tp.(x)] = exp[t(1+q.(x))] = e^t exp(tq.(x)) = [1 + (x+1)e^((x+2)t)]/(x+2), and p_n(x) = (x+1)(x+2)^(n-1) for n > 0. - Tom Copeland, Nov 15 2016

T^(-1) = A130595*(padded A130595), differently signed A118801. Cf. A097805. - Tom Copeland, Nov 17 2016

The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)/(1 + 2*x) * (1 + 2*x)^n about 0. For example, for n = 4, (1 + x)/(1 + 2*x) * (1 + 2*x)^4 = (8*x^4 + 20*x*3 + 18*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 24 2018

A145661 Triangle T(n,k) = (-1)^k * A119258(n,k) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -5, 7, -1, 1, -7, 17, -15, 1, 1, -9, 31, -49, 31, -1, 1, -11, 49, -111, 129, -63, 1, 1, -13, 71, -209, 351, -321, 127, -1, 1, -15, 97, -351, 769, -1023, 769, -255, 1, 1, -17, 127, -545, 1471, -2561, 2815, -1793, 511, -1, 1, -19, 161, -799, 2561

Offset: 0

Roger L. Bagula and Gary W. Adamson, Mar 16 2009

Row sums are (-1)^(n+1)*(n-1) for n >= 1.

A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011

			Triangle begins
  1;
  1,  -1;
  1,  -3,   1;
  1,  -5,   7,   -1;
  1,  -7,  17,  -15,    1;
  1,  -9,  31,  -49,   31,    -1;
  1, -11,  49, -111,  129,   -63,    1;
  1, -13,  71, -209,  351,  -321,  127,    -1;
  1, -15,  97, -351,  769, -1023,  769,  -255,    1;
  1, -17, 127, -545, 1471, -2561, 2815, -1793,  511,    -1;
  1, -19, 161, -799, 2561, -5503, 7937, -7423, 4097, -1023, 1;

J.-F. Chamayou, A Random Difference Equation with Dufresne Variables Revisited, arXiv preprint arXiv:1410.1708 [math.PR], 2014. See Table in Section XII.
M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Quart., 48 (2010), 290-297.

Cf. A135233, A112857.

A193844 is an essentially identical triangle.

A119258 := proc(n,k)
        if k=0 or k = n then
                1;
        elif k<0 or k> n then
                0;
        else
                2*procname(n-1,k-1)+procname(n-1,k) ;
        end if;
end proc:
seq(seq(A119258(n,k),k=0..n),n=0..10) ;
A145661 := proc(n,k)
        (-1)^k*A119258(n,k) ;
end proc: # R. J. Mathar, Oct 21 2011

Clear[M, T, d, a, x, a0];
T[n_, m_, d_] := If[ m == n + 1, 1, If[n == d, 1, 0]];
M[d_] := MatrixPower[Table[T[n, m, d], {n, 1, d}, {m, 1, d}], d];
Table[M[d], {d, 1, 10}];
Table[Det[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}]];

A135233 Triangle A007318 * A193554, read by rows.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 14, 7, 5, 1, 41, 15, 17, 7, 1, 122, 31, 49, 31, 9, 1, 365, 63, 129, 111, 49, 11, 1, 1094, 127, 321, 351, 209, 71, 13, 1, 3281, 255, 769, 1023, 769, 351, 97, 15, 1, 9842, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1

Offset: 0

Gary W. Adamson, Nov 23 2007

Row sums = 3^n.

Left column = A007051: (1, 2, 5, 14, 41, 122, ...).

			First few rows of the triangle:
   1;
   2,  1;
   5,  3,  1;
  14,  7,  5,  1;
  41, 15, 17,  7,  1;
...

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A007051, A007318, A118801, A119258, A193554.

function T(n,k)
  if k eq n then return 1;
  elif k eq 0 then return (3^n+1)/2;
  else return (&+[(-1)^(n-k+j)*2^j*Binomial(n, j): j in [0..n-k]]);
  end if; return T; end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 20 2019

T:= proc(n, k) option remember;
      if k=n then 1
    elif k=0 then (3^n_1)/2
    else add((-1)^(n-k+j)*binomial(n, j)*2^j, j=0..n-k)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Nov 20 2019

T[n_, k_]:= T[n, k]= If[k==n, 1, If[k==0, (3^n+1)/2, Sum [(-1)^(n-k+i)* Binomial[n, i]*2^i, {i, 0, n-k}]]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 20 2019 *)

T(n,k) = if(k==n, 1, if(k==0, (3^n+1)/2, sum(j=0, n-k, (-1)^(n-k+j)*binomial(n,j)*2^j) )); \\ G. C. Greubel, Nov 20 2019

@CachedFunction
def T(n, k):
    if (k==n): return 1
    elif (k==0): return (3^n+1)/2
    else: return sum((-1)^(n-k+j)*2^j*binomial(n, j) for j in (0..n-k))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 20 2019

Binomial transform of A193554, as infinite lower triangular matrices.

T(n,k) = Sum_{j=0..n-k} (-1)^(n-k+j)*binomial(n,j)*2^j, with T(n,n) = 1, and T(n,0) = (3^n + 1)/2. - G. C. Greubel, Nov 20 2019

Definition corrected by N. J. A. Sloane, Jul 30 2011

Showing 1-9 of 9 results.

cp's OEIS Frontend

A118802 Row squared sums of triangle A118801: a(n) = Sum_{k=0..n} A118801(n,k)^2.

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A118803 Self-convolution square-root of A118802, which is the row squared sums of triangle A118801.

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A038207 Triangle whose (i,j)-th entry is binomial(i,j)*2^(i-j).

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A119258 Triangle read by rows: T(n,0) = T(n,n) = 1 and for 0

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A118800 Triangle read by rows: T satisfies the matrix products: C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle.

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A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

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Maple

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A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

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A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.