A112857 -id:A112857 - 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)

A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.

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

Paul D. Hanna, May 02 2006

Matrix inverse is triangle A118800. Row sums are: (1-n). Unsigned row sums equal A007051(n) = (3^n + 1)/2. Row squared sums equal A118802. Antidiagonal sums equal A080956(n) = (n+1)(2-n)/2. Unsigned antidiagonal sums form A024537 (with offset).
T = C^2*D^-1 where matrix product D = C^-1*T*C = T^-1*C^2 has only 2 nonzero diagonals: D(n,n)=-D(n+1,n)=(-1)^n, with zeros elsewhere. Also, [B^-1]*T*[B^-1] = B*[T^-1]*B forms a self-inverse matrix, where B^2 = C and B(n,k) = C(n,k)/2^(n-k). - Paul D. Hanna, May 04 2006
Riordan array ( 1/(1 - x), -x/(1 - 2*x) ) The matrix square is the Riordan array ( (1 - 2*x)/(1 - x)^2, x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013

			Formulas for initial columns are, for n>=0:
T(n+1,1) = 1 - 2^(n+1);
T(n+2,2) = 1 + 2^(n+1)*n;
T(n+3,3) = 1 - 2^(n+1)*(n*(n+1)/2 + 1);
T(n+4,4) = 1 + 2^(n+1)*(n*(n+1)*(n+2)/6 + n);
T(n+5,5) = 1 - 2^(n+1)*(n*(n+1)*(n+2)*(n+3)/24 + n*(n+1)/2 + 1).
Triangle 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; ...
The matrix square, T^2, starts:
1;
0,1;
-1,0,1;
-2,-1,0,1;
-3,-2,-1,0,1;
-4,-3,-2,-1,0,1; ...
where all columns are the same.
The matrix product C^-1*T*C = T^-1*C^2 is:
1;
-1,-1;
0, 1, 1;
0, 0,-1,-1;
0, 0, 0, 1, 1; ...
where C(n,k) = n!/(n-k)!/k!.

M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Q., 48 (2010), 290-297.

Cf. A118800 (inverse), A007051 (unsigned row sums), A118802 (Row squared sums), A080956 (antidiagonal sums), A024537 (unsigned antidiagonal sums).
A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011
Cf. A007318, A032807, A097805, A112857, A130595, A156644, A167374, A200139, A239473.

Table[(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)=if(n==0&k==0,1,1+(-1)^k*2^(n-k+1)*sum(j=0,k\2,binomial(n-2*j-2,k-2*j-1)))}

T(n,k) = 1 + (-1)^k*2^(n-k+1)*Sum_{j=0..[k/2]} C(n-2j-2,k-2j-1) for n>=k>=0 with T(0,0) = 1.
For k>0, T(n,k) = -T(n-1,k-1) + 2*T(n-1,k). - Gerald McGarvey, Aug 05 2006
O.g.f.: (1 - 2*t)/(1 - t) * 1/(1 + t*(x - 2)) = 1 + (1 - x)*t + (1 - 3*x + x^2)*t^2 + (1 - 7*x + 5*x^2 - x^3)*t^3 + .... - Peter Bala, Jul 17 2013
From Tom Copeland, Nov 17 2016: (Start)
Let M = A200139^(-1) = (unsigned A118800)^(-1) and NpdP be the signed padded Pascal matrix defined in A097805. Then T(n,k) = (-1)^n* M(n,k) and T = P*NpdP = (A239473)^(-1)*P^(-1) = P*A167374*P^(-1) = A156644*P^(-1), where P is the Pascal matrix A007318 with inverse A130595. Cf. A112857.
Signed P^2 = signed A032807 = T*A167374. (End)

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}]];

A185410 A decomposition of the double factorials A001147.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 10, 4, 0, 1, 36, 60, 8, 0, 1, 116, 516, 296, 16, 0, 1, 358, 3508, 5168, 1328, 32, 0, 1, 1086, 21120, 64240, 42960, 5664, 64, 0, 1, 3272, 118632, 660880, 900560, 320064, 23488, 128, 0, 1, 9832, 638968, 6049744, 14713840, 10725184, 2225728, 95872, 256, 0

Offset: 0

Paul Barry, Jan 26 2011

Row sums are A001147. Reversal of A185411.
From Peter Bala, Jul 24 2012: (Start)
This is the case k = 2 of the 1/k—Eulerian polynomials introduced by Savage and Viswanathan. They give a combinatorial interpretation of the triangle in terms of an ascent statistic on sets of inversion sequences and a geometric interpretation in terms of lecture hall polytopes.
Row reverse of A156919.
(End)
Triangle T(n,k), 0<=k<=n, given by (1, 0, 3, 0, 5, 0, 7, 0, 9, 0, ...) DELTA (0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 12 2013

			Triangle begins:
  1,
  1,    0,
  1,    2,      0,
  1,   10,      4,       0,
  1,   36,     60,       8,        0,
  1,  116,    516,     296,       16,        0,
  1,  358,   3508,    5168,     1328,       32,       0,
  1, 1086,  21120,   64240,    42960,     5664,      64,     0,
  1, 3272, 118632,  660880,   900560,   320064,   23488,   128,   0,
  1, 9832, 638968, 6049744, 14713840, 10725184, 2225728, 95872, 256, 0,
  ...
In the Savage-Viswanathan paper, the coefficients appear as
  1
  1    2
  1   10     4
  1   36    60     8
  1  116   516   296    16
  1  358  3508  5168  1328   32
  1 1086 21120 64240 42960 5664 64
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
S.-M. Ma and T. Mansour, The 1/k-Eulerian polynomials and k-Stirling permutations, arXiv preprint arXiv:1409.6525 [math.CO], 2014.
C. D. Savage and G. Viswanathan, The 1/k-Eulerian polynomials, Elec. J. of Comb., Vol. 19, Issue 1, #P9 (2012).

Cf. A156919, A001147 (row sums), A112857, A173018, A186695, A202038 (alt. row sums).

T[0, 0] := 1;  T[n_, -1] := 0;  T[n_, n_] := 0; T[n_, k_] := T[n, k] = (n - k)*T[n - 1, k - 1] + (2*k + 1)*T[n - 1, k]; Join[{1}, Table[If[k < 0, 0, If[k >= n, 0, 2^k*T[n, k]]], {n, 1, 5}, {k, 0, n}] // Flatten] (* G. C. Greubel, Jun 30 2017 *)

G.f.: 1/(1-x/(1-2xy/(1-3x/(1-4xy/(1-5x/(1-6xy/(1-7x/(1-8xy/(1- .... (continued fraction).
From Peter Bala, Jul 24 2012: (Start)
T(n,k) = sum {j=0..k}(-1)^(k-j)/4^j*C(n+1/2,k-j)*C(2*j,j)*(2*j+1)^n.
Recurrence equation: T(n+1,k) = (2*k+1)*T(n,k) + 2*(n-k+1)*T(n,k-1).
E.g.f.: sqrt(E(x,2*z)) = 1 + z + (1+2*x)*z^2/2! + (1+10*x+4*x^2)*z^3/3! + ..., where E(x,z) = (1-x)/(exp(z*(x-1)) - x) is the e.g.f. for the Eulerian numbers (version A173018). Cf. A156919.
Row polynomial R(n,x) = sum {k = 1..n} 2^(n-2*k)*C(2*k,k)*k!*Stirling2(n,k)*(x-1)^(n-k). R(n,4*x)/(1-4*x)^(n+1/2) = sum {k>=0} C(2*k,k)*(2*k+1)^n*x^k. The sequence of rational functions x*R(n,x)/(1-x)^(n+1) conjecturally occurs in the first column of (I - x*A112857)^(-1). (1+x)^(n-1)*R(n,x/(x+1)) gives the n-th row polynomial of A186695.
Row sums A001147. Alt. row sums A202038. (End)
T(n,k) = 2^k*A102365(n,k). - Philippe Deléham, Feb 12 2013

A193845 Mirror of the triangle A193844.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Aug 07 2011

This triangle is obtained by reversing the rows of the triangle A193844.
From Philippe Deléham, Jan 17 2014: (Start)
Subtriangle of the triangle in A112857.
T(n,0)   = A000225(n+1).
T(n,1)   = A000337(n).
T(n+2,2) = A055580(n).
T(n+3,3) = A027608(n).
T(n+4,4) = A211386(n).
T(n+5,5) = A211388(n).
T(n,n)   = A000012(n).
T(n+1,n) = A005408(n).
T(n+2,n) = A056220(n+2).
T(n+3,n) = A199899(n+1).
Row sums are A003462(n+1).
Diagonal sums are A048739(n).
Riordan array (1/((1-2*x)*(1-x)), x/(1-2*x)). (End)
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-2)^0 + A_1*(x-2)^1 + A_2*(x-2)^2 + ... + A_n*(x-2)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
The n-th row lists the coefficients of the polynomial sum_{k=0..n} (X+2)^k, in order of increasing powers. - M. F. Hasler, Oct 15 2014

			First six rows:
1
3....1
7....5....1
15...17...7....1
31...49...31...9...1
63...129..111..49..11..1

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
Russell Jay Hendel, A Method for Uniformly Proving a Family of Identities, arXiv:2107.03549 [math.CO], 2021.

Cf. A193844.
Cf. Columns: A000225, A000337, A055580, A027608, A211386, A211388
Cf. Diagonals: A000012, A005408, A056220, A199899

z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 1)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193844 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)
Table[2^k*Binomial[n + 1, k]*Hypergeometric2F1[1, -k, -k + n + 2, 1/2], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Nov 09 2021 *)

for(n=0,20,for(k=0,n,print1(1/k!*sum(i=0,n,(2^(i-k)*prod(j=0,k-1,i-j))),", "))) \\ Derek Orr, Oct 14 2014

T(n,k) = A193844(n,n-k).

T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014

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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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A118801 Triangle T that satisfies the matrix products: C*[T^-1]*C = T and T*[C^-1]*T = C, where C is Pascal's triangle.

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

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A185410 A decomposition of the double factorials A001147.

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A193845 Mirror of the triangle A193844.

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A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.