A323228 -id:A323228 - cp's OEIS frontend

A349841 Triangle T(n,k) built by placing all ones on the left edge, [1,0,0,0,0] repeated on the right edge, and filling the body using the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 1, 1, 5, 10, 10, 5, 2, 0, 1, 6, 15, 20, 15, 7, 2, 0, 1, 7, 21, 35, 35, 22, 9, 2, 0, 1, 8, 28, 56, 70, 57, 31, 11, 2, 0, 1, 9, 36, 84, 126, 127, 88, 42, 13, 2, 1

Offset: 0

Michael A. Allen, Dec 13 2021

This is the m=5 member in the sequence of triangles A007318, A059259, A118923, A349839, A349841 which have all ones on the left side, ones separated by m-1 zeros on the other side, and whose interiors obey Pascal's recurrence.
T(n,k) is the (n,n-k)-th entry of the (1/(1-x^5),x/(1-x)) Riordan array.
For n>0, T(n,n-1) = A002266(n+4).
For n>1, T(n,n-2) = A008732(n-2).
For n>2, T(n,n-3) = A122047(n-1).
Sums of rows give A349842.
Sums of antidiagonals give A349843.

			Triangle begins:
  1;
  1,   0;
  1,   1,   0;
  1,   2,   1,   0;
  1,   3,   3,   1,   0;
  1,   4,   6,   4,   1,   1;
  1,   5,  10,  10,   5,   2,   0;
  1,   6,  15,  20,  15,   7,   2,   0;
  1,   7,  21,  35,  35,  22,   9,   2,   0;
  1,   8,  28,  56,  70,  57,  31,  11,   2,   0;
  1,   9,  36,  84, 126, 127,  88,  42,  13,   2,   1;

Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Other members of sequence of triangles: A007318, A059259, A118923, A349839.
Columns: A000012, A001477, A000217, A000292, A000332, A323228.
Diagonals: A079998, A002266, A008732, A122047.

Flatten[Table[CoefficientList[Series[(1 - x*y)/((1 - (x*y)^5)(1 - x - x*y)), {x, 0, 20}, {y, 0, 10}], {x, y}][[n+1,k+1]],{n,0,10},{k,0,n}]]

G.f.: (1-x*y)/((1-(x*y)^5)(1-x-x*y)) in the sense that T(n,k) is the coefficient of x^n*y^k in the series expansion of the g.f.
T(n,0) = 1.
T(n,n) = delta(n mod 5,0).
T(n,1) = n-1 for n>0.
T(n,2) = (n-1)*(n-2)/2 for n>1.
T(n,3) = (n-1)*(n-2)*(n-3)/6 for n>2.
T(n,4) = (n-1)*(n-2)*(n-3)*(n-4)/24 for n>3.
T(n,5) = C(n-1,5) + 1 for n>4.
T(n,6) = C(n-1,6) + n - 6 for n>5.
For 0 <= k < n, T(n,k) = (n-k)*Sum_{j=0..floor(k/5)} binomial(n-5*j,n-k)/(n-5*j).
The g.f. of the n-th subdiagonal is 1/((1-x^5)(1-x)^n).

A323231 A(n, k) = [x^k] (1/(1-x) + x/(1-x)^n), square array read by descending antidiagonals for n, k >= 0.

Original entry on oeis.org

1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 2, 4, 4, 2, 1, 1, 2, 5, 7, 5, 2, 1, 1, 2, 6, 11, 11, 6, 2, 1, 1, 2, 7, 16, 21, 16, 7, 2, 1, 1, 2, 8, 22, 36, 36, 22, 8, 2, 1, 1, 2, 9, 29, 57, 71, 57, 29, 9, 2, 1, 1, 2, 10, 37, 85, 127, 127, 85, 37, 10, 2, 1

Offset: 0

Peter Luschny, Feb 10 2019

			Array starts:
[0] 1, 2,  1,  1,   1,   1,    1,    1,    1,     1,     1, ...
[1] 1, 2,  2,  2,   2,   2,    2,    2,    2,     2,     2, ... A040000
[2] 1, 2,  3,  4,   5,   6,    7,    8,    9,    10,    11, ... A000027
[3] 1, 2,  4,  7,  11,  16,   22,   29,   37,    46,    56, ... A000124
[4] 1, 2,  5, 11,  21,  36,   57,   85,  121,   166,   221, ... A050407
[5] 1, 2,  6, 16,  36,  71,  127,  211,  331,   496,   716, ... A145126
[6] 1, 2,  7, 22,  57, 127,  253,  463,  793,  1288,  2003, ... A323228
[7] 1, 2,  8, 29,  85, 211,  463,  925, 1717,  3004,  5006, ...
[8] 1, 2,  9, 37, 121, 331,  793, 1717, 3433,  6436, 11441, ...
[9] 1, 2, 10, 46, 166, 496, 1288, 3004, 6436, 12871, 24311, ...
.
Read as a triangle (by descending antidiagonals):
                                     1
                                  2,   1
                                1,   2,   1
                             1,   2,   2,   1
                           1,   2,   3,   2,   1
                        1,   2,   4,   4,   2,   1
                      1,   2,   5,   7,   5,   2,  1
                    1,  2,   6,  11,  11,   6,   2,  1
                  1,  2,   7,  16,  21,  16,   7,  2,  1
                1,  2,  8,  22,  36,  36,  22,   8,  2,  1
              1,  2,  9,  29,  57,  71,  57,  29,  9,  2,  1
.
A(0, 1) = C(-1, 0) + 1 = 2 because C(-1, 0) = 1. A(1, 0) = C(-1, -1) + 1 = 1 because C(-1, -1) = 0. Warning: Some computer algebra programs (for example Maple and Mathematica) return C(n, n) = 1 for n < 0. This contradicts the definition given by Graham et al. (see reference). On the other hand this definition preserves symmetry.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 154.

G. C. Greubel, Antidiagonals n = 0..50, flattened

Differs from A323211 only in the second term.
Rows include: A040000, A000027, A000124, A050407, A145126, A323228.
Diagonals A(n, n+d): A323230 (d=0), A260878 (d=1), A323229 (d=2).
Antidiagonal sums are A323227(n) if n!=1.
Cf. A007318 (Pascal's triangle).

using AbstractAlgebra
function Arow(n, len)
    R, x = PowerSeriesRing(ZZ, len+2, "x")
    gf = inv(1-x) + divexact(x, (1-x)^n)
    [coeff(gf, k) for k in 0:len-1] end
for n in 0:9 println(Arow(n, 11)) end

Binomial := (n, k) -> `if`(n < 0 and n = k, 0, binomial(n,k)):
A := (n, k) -> Binomial(n + k - 2, k - 1) + 1:
seq(lprint(seq(A(n, k), k=0..10)), n=0..10);

T[n_, k_]:= If[k==0, 1 + Boole[n==1], If[k==n, 1, Binomial[n-2, k-1] + 1]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 27 2021 *)

def Arow(n):
    R. = PowerSeriesRing(ZZ, 20)
    gf = 1/(1-x) + x/(1-x)^n
    return gf.padded_list(10)
for n in (0..9): print(Arow(n))

A(n, k) = binomial(n + k - 2, k - 1) + 1. Note that binomial(n, n) = 0 if n < 0.
A(n, k) = A(k, n) with the exception A(1,0) != A(0,1).
A(n, n) = binomial(2*n-2, n-1) + 1 = A323230(n).
From G. C. Greubel, Dec 27 2021: (Start)
T(n, k) = binomial(n-2, k-1) + 1 with T(n, 0) = 1 + [n=1], T(n, n) = 1.
T(2*n, n) = A323230(n).
Sum_{k=0..n} T(n,k) = n + 1 + 2^(n-2) - [n=0]/4 + [n=1]/2. (End)

cp's OEIS Frontend

A349841 Triangle T(n,k) built by placing all ones on the left edge, [1,0,0,0,0] repeated on the right edge, and filling the body using the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).

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