A051708 -id:A051708 - cp's OEIS frontend

A035002 Square array read by antidiagonals: T(m,n) = Sum_{k=1..m-1} T(m-k,n) + Sum_{k=1..n-1} T(m,n-k).

1, 1, 1, 2, 2, 2, 4, 5, 5, 4, 8, 12, 14, 12, 8, 16, 28, 37, 37, 28, 16, 32, 64, 94, 106, 94, 64, 32, 64, 144, 232, 289, 289, 232, 144, 64, 128, 320, 560, 760, 838, 760, 560, 320, 128, 256, 704, 1328, 1944, 2329, 2329, 1944, 1328, 704, 256, 512, 1536, 3104, 4864, 6266

Offset: 1

Erich Friedman

T(m,n) is the sum of all the entries above it plus the sum of all the entries to the left of it.
T(m,n) equals the number of ways to move a chess rook from the lower left corner to square (m,n), with the rook moving only up or right. - Francisco Santos, Oct 20 2005
T(m+1,n+1) is the number of nim games that start with two piles of stones of sizes m and n. - Martin J. Erickson (erickson(AT)truman.edu), Dec 05 2008
The same sequences arises from reading the following triangle by rows: Start with 1, then use a Pascal-like rule, where each new entry is the sum of all terms in the two diagonals that converge at that point. See example below. - J. M. Bergot, Jun 08 2013
T(n,k) is odd iff (n,k) = (1,1), k = n-1, or k = n+1. - Peter Kagey, Apr 20 2020

			Table begins:
  1  1  2   4   8  16   32   64 ...
  1  2  5  12  28  64  144  320 ...
  2  5 14  37  94 232  560 1328 ...
  4 12 37 106 289 760 1944 4864 ...
Alternative construction as a triangle:
               1
             1   1
           2   2   2
         4   5   5   4
       8  12  14  12   8
    16  28  37  37  28  16

Peter Kagey, Table of n, a(n) for n = 1..10011 (first 141 antidiagonals, flattened)
C. Coker, Enumerating a class of lattice paths, Discrete Math., 271 (2003), 13-28.
M. Erickson, S. Fernando, and K. Tran, Enumerating Rook and Queen Paths, Bulletin of the Institute for Combinatorics and Its Applications, Volume 60 (2010), 37-48.

Cf. A035001, A051708, A025192 (antidiagonal sums).

A035002 := proc(m,n)
    option remember;
    if n = 1 and m= 1 then
        1;
    elif m = 1 then
        2^(n-2) ;
    elif n = 1 then
        2^(m-2) ;
    else
        add( procname(m-k,n),k=1..m-1) + add( procname(m,n-k),k=1..n-1) ;
    end if;
end proc: # R. J. Mathar, Jun 06 2013

T[n_, 1] = 2^(n-2); T[1, n_] = 2^(n-2); T[1, 1] = 1; T[m_, n_] := T[m, n] = Sum[T[m-k, n], {k, 1, m-1}] + Sum[T[m, n-k], {k, 1, n-1}]; Flatten[Table[T[m-n+1 , n], {m, 1, 11}, {n, 1, m}]] (* Jean-François Alcover, Nov 04 2011 *)
nMax = 11; T = (((x - 1)*y - x + 1)/((3*x - 2)*y - 2*x + 1) + O[x]^nMax // Normal // Expand) + O[y]^nMax // Normal // Expand // CoefficientList[#, {x, y}]&; Table[T[[n - k + 1, k]], {n, 1, nMax}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 18 2018, after Vladimir Kruchinin *)
T[ n_, m_] := SeriesCoefficient[ (1 - x)*(1 - y)/( 1 - 2*x - 2*y + 3*x*y), {x, 0, n}, {y, 0, m}]; (* Michael Somos, Oct 05 2023 *)

T(n,m):=sum(binomial(m-1,m-i)*sum(binomial(k+i,i)*binomial(n-1,n-k),k,0,n),i,0,m); /* Vladimir Kruchinin, Apr 14 2015 */

G.f. T(n; x) for n-th row satisfies: T(n; x) = Sum_{k=1..n} (1+x^k)*T(n-k; x), T(0; x) = 1. - Vladeta Jovovic, Sep 03 2002
T(m+1,n+1) = 2*T(m+1,n) + 2*T(m,n+1) - 3*T(m,n); T(n,1) = T(1,n) = A011782(n). - Francisco Santos, Oct 20 2005
G.f.: ((x-1)*y-x+1)/((3*x-2)*y-2*x+1). - Vladimir Kruchinin, Apr 14 2015
T(n,m) = Sum_{i=0..m} C(m-1,m-i)*Sum_{k=0..n} C(k+i,i)*C(n-1,n-k). - Vladimir Kruchinin, Apr 14 2015
T(n,m) = T(m,n) for all n and m. - Michael Somos, Oct 04 2023
T(n,2) = (n+2)*2^(n-3) for n>1; T(n,3) = (n^2+11*n+14)*2^(n-5) for n>1 - Erich Friedman, Jan 14 2025

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)

A181731 Table A(d,n) of the number of paths of a chess rook in a d-dimensional hypercube from (0...0) to (n...n) where the rook may move in steps that are multiples of (1,0..0), (0,1,0..0), ..., (0..0,1).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 6, 14, 4, 1, 24, 222, 106, 8, 1, 120, 6384, 9918, 838, 16, 1, 720, 291720, 2306904, 486924, 6802, 32, 1, 5040, 19445040, 1085674320, 964948464, 25267236, 56190, 64, 1, 40320, 1781750880, 906140159280, 4927561419120, 439331916888, 1359631776, 470010, 128, 1, 362880, 214899027840, 1224777388630320, 54259623434853360

Offset: 1

Manuel Kauers, Nov 16 2010

The table is enumerated along antidiagonals: A(1,0), A(2,0), A(1,1), A(3,0), A(2,1), A(1,2), A(4,0), A(3,1), A(2,2), A(1,3), ... .

			A(3,1) = 6 because there are 6 rook paths on 3D chessboards from (0,0,0) to (1,1,1).
Square table A(d,n) begins:
  1,   1,      2,          4,             8, ...
  1,   2,     14,        106,           838, ...
  1,   6,    222,       9918,        486924, ...
  1,  24,   6384,    2306904,     964948464, ...
  1, 120, 291720, 1085674320, 4927561419120, ...

Alois P. Heinz, Antidiagonals n = 1..20
M. Kauers and D. Zeilberger, The Computational Challenge of Enumerating High-Dimensional Rook Walks, arXiv:1011.4671 [math.CO], 2010.

Rows d=1-12 give: A011782, A051708 (from [1,1]), A144045 (from [1,1,1]), A181749, A181750, A181751, A181752, A181724, A181725, A181726, A181727, A181728.
Columns n=0-2 give: A000012, A000142, A105749.
Main diagonal gives A246623.

b:= proc(l) option remember; `if`({l[]} minus {0}={}, 1, add(add
       (b(sort(subsop(i=l[i]-j, l))), j=1..l[i]), i=1..nops(l)))
    end:
A:= (d, n)-> b([n$d]):
seq(seq(A(h-n, n), n=0..h-1), h=1..10); # Alois P. Heinz, Jul 21 2012

b[l_List] := b[l] = If[Union[l] ~Complement~ {0} == {}, 1, Sum[ Sum[ b[ Sort[ ReplacePart[l, i -> l[[i]] - j]]], {j, 1, l[[i]]}], {i, 1, Length[l]}]]; A[d_, n_] := b[Array[n&, d]]; Table[Table[A[h-n, n], {n, 0, h-1}], {h, 1, 10}] // Flatten (* Jean-François Alcover, Feb 25 2015, after Alois P. Heinz *)

Edited by Alois P. Heinz, Jul 21 2012

Minor edits by Vaclav Kotesovec, Sep 03 2014

A144045 Number of paths of a chess Rook in a cube, from (1,1,1) to (n,n,n), where the rook may move in steps that are multiples of (1,0,0), (0,0,1), or (0,0,1).

Original entry on oeis.org

1, 6, 222, 9918, 486924, 25267236, 1359631776, 75059524392, 4223303759148, 241144782230124, 13930829740017132, 812470444305924300, 47760356825349969600, 2826309951801018736800, 168207011284961649886800, 10060178088232285063542768, 604273284101165691102038556

Offset: 1

Martin J. Erickson (erickson(AT)truman.edu), Sep 08 2008

nonn

			a(2)=6 because there are 6 Rook paths from (1,1,1) to (2,2,2).
G.f. = x + 6*x^2 + 222*x^3 + 9918*x^4 + 486924*x^5 + 25267236*x^6 + ...

Alin Bostan, Calcul Formel pour la Combinatoire des Marches [The text is in English], Habilitation à Diriger des Recherches, Laboratoire d'Informatique de Paris Nord, Université Paris 13, December 2017.
M. Kauers and D. Zeilberger, The Computational Challenge of Enumerating High-Dimensional Rook Walks, arXiv:1011.4671 [math.CO], 2010.

Cf. A051708.

Row d=3 of A181731.

a(n) satisfies the recurrence relation a(1) = 1; a(2) = 6; a(3) = 222; a(4) = 9918; a(n) = ((-121 n^3 + 575n^2 - 872n + 412)a(n - 1) + (-475n^3 + 4887n^2 - 16202n + 17448)a(n - 2) + (1746n^3 - 19818n^2 + 75060n - 94896)a(n - 3) + (-1152n^3 + 16128n^2 - 74880n + 115200)a(n - 4))/(-(2n^3 - 8n^2 + 10n - 4)), n>= 5.
G.f.: 1+int(6*hypergeom([1/3, 2/3],[2],27*x*(3*x-2)/(4*x-1)^3)/((4*x-1)*(64*x-1)),x). [Mark van Hoeij, Dec 10 2009]
Asymptotics: a(n) ~ 9*sqrt(3)/(40*Pi*n)*64^(n-1). - Frederic Chyzak, 2010

A307584 Triangle read by rows: T(n,k) is the number of non-backtracking walks on Z^2 of length n that are active for k steps, where the walk is initially active and turns in the walk toggle the activity.

Original entry on oeis.org

1, 2, 1, 2, 6, 1, 2, 14, 10, 1, 2, 22, 42, 14, 1, 2, 30, 106, 86, 18, 1, 2, 38, 202, 318, 146, 22, 1, 2, 46, 330, 838, 722, 222, 26, 1, 2, 54, 490, 1774, 2514, 1382, 314, 30, 1, 2, 62, 682, 3254, 6802, 6062, 2362, 422, 34, 1, 2, 70, 906, 5406, 15378, 20406, 12570, 20406, 12570, 3726, 546, 38, 1

Offset: 1

Matthew Fahrbach, Apr 15 2019

			Triangle begins:
1;
2,  1;
2,  6,   1;
2, 14,  10,    1;
2, 22,  42,   14,     1;
2, 30, 106,   86,    18,     1;
2, 38, 202,  318,   146,    22,     1;
2, 46, 330,  838,   722,   222,    26,    1;
2, 54, 490, 1774,  2514,  1382,   314,   30,   1;
2, 62, 682, 3254,  6802,  6062,  2362,  422,  34,  1;
2, 70, 906, 5406, 15378, 20406, 12570, 3726, 546, 38, 1;
...

M. Fahrbach and D. Randall, Slow mixing of Glauber dynamics for the six-vertex model in the ferroelectric and antiferroelectric phases, arXiv:1904.01495 [cs.DS], 2019
R. J. Mathar, Walks of up and right steps in the square lattice with blocked squares (2022) Table 2.

Row sums give A000244. Cf. A051708 (subdiagonal T(2n,n)).

cp's OEIS Frontend

A035002 Square array read by antidiagonals: T(m,n) = Sum_{k=1..m-1} T(m-k,n) + Sum_{k=1..n-1} T(m,n-k).

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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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A181731 Table A(d,n) of the number of paths of a chess rook in a d-dimensional hypercube from (0...0) to (n...n) where the rook may move in steps that are multiples of (1,0..0), (0,1,0..0), ..., (0..0,1).

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A144045 Number of paths of a chess Rook in a cube, from (1,1,1) to (n,n,n), where the rook may move in steps that are multiples of (1,0,0), (0,0,1), or (0,0,1).

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A307584 Triangle read by rows: T(n,k) is the number of non-backtracking walks on Z^2 of length n that are active for k steps, where the walk is initially active and turns in the walk toggle the activity.

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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.