A295555 -id:A295555 - cp's OEIS frontend

A169623 Generalized Pascal triangle read by rows: T(n,0) = T(0,n) = 1 for n >= 0, T(n,k) = 0 for k < 0 or k > n; otherwise T(n,k) = T(n-2,k-2) + T(n-2,k-1) + T(n-2,k) for 1 <= k <= n-1.

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 5, 14, 26, 35, 35, 26, 14, 5, 1, 1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1, 1, 6, 20, 45, 75, 96, 96, 75, 45, 20, 6, 1

Offset: 0

Roger L. Bagula and Gary W. Adamson, Dec 03 2009

The borders are all 1's, with zero entries outside. To get an internal entry, use the rule that D = A+B+C here:
         A   B   C
       *   *   *   *
     *   *   D   *   *
That is, add the three terms directly above you, two rows back.
This is the triangle er(n,k) defined in the Ehrenborg and Readdy link. See Proposition 2.4 and Table 1. - Michel Marcus, Sep 14 2016
If the offset is changed from 0 to 1, this is also the table U(n,k) of the coefficients [x^k] p_n(x) of the polynomials p_n(x) = (x + 1)*p_{n-1}(x) (if n even), p_n = (x^2 + x + 1)^floor(n/2) if n odd.
May be split into two triangles by taking the even-numbered and odd-numbered rows separately: the even-numbered rows give A027907.
From Peter Bala, Aug 19 2021: (Start)
Let M denote the lower unit triangular array A070909. 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 equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section below. The proof uses the hockey-stick identities from the Formula section. (End)

			Triangle begins:
                    1
                  1   1
                1   1   1
              1   2   2   1
            1   2   3   2   1
          1   3   5   5   3   1
        1   3   6   7   6   3   1
      1   4   9  13  13   9   4   1
    1   4  10  16  19  16  10   4   1
  ...
As a square array read by antidiagonals:
  1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 1, 1, 1, 1, ...
  1, 1, 2,  2,  3,  3,  4,  4,  5,  5,  6,  6, 7, 7, 8, 8, ...
  1, 2, 3,  5,  6,  9, 10, 14, 15, 20, 21, 27, ...
  1, 2, 5,  7, 13, 16, 26, 30, 45, ...
  1, 3, 6, 13, 19, 35, 45, 75, ...
  1, 3, 9, 16, 35, 51, 96, ...
  ...
From _Peter Bala_, Aug 19 2021: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
  /1        \/1        \/1       \ /1       \        /1         \
  |1 1      ||0 1      ||0 1      ||0 1      |       |1 1       |
  |1 0 1    ||0 1 1    ||0 0 1    ||0 0 1    |...  = |1 1 1     |
  |1 0 1 1  ||0 1 0 1  ||0 0 1 1  ||0 0 0 1  |       |1 2 2 1   |
  |1 0 1 0 1||0 1 0 1 1||0 0 1 0 1||0 0 0 1 1|       |1 2 3 2 1 |
  |...      ||...       |...      ||...      |       |...       |
(End)

Rémy Sigrist, Rows 0..199, flattened
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Richard Ehrenborg and Margaret A. Readdy, The Gaussian coefficient revisited, arXiv:1609.03216 [math.CO], 2016.
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (No. 2, 1998), 98-109. See Table 10.
Wikipedia, Hockey-stick identity.

A123149 is essentially the same triangle, except for a diagonal of zeros.
Row sums are in A182522 (essentially A038754).
Cf. A027907, A070909.
See A295555 for the next triangle in the series A007318, A169623 (this sequence).

T:=proc(n,k) option remember;
if n >= 0 and k = 0 then 1
elif n >= 0 and k = n then 1
elif (k < 0 or k > n) then 0
else T(n-2,k-2)+T(n-2,k-1)+T(n-2,k);
fi;
end;
for n from 0 to 14 do lprint([seq(T(n,k),k=0..n)]); od: # N. J. A. Sloane, Nov 23 2017

p[x, 1] := 1;
p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + x + 1)^Floor[n/2]]
a = Table[CoefficientList[p[x, n], x], {n, 1, 12}]
Flatten[a] (* This is for the same sequence but with offset 1 *)

From Peter Bala, Aug 19 2021: (Start)
T(2*n,k) = T(2*n-1,k-1) + T(2*n-2,k).
T(2*n,k) = T(2*n-1,k) + T(2*n-2,k-2).
T(2*n+1,k) = T(2*n,k) + T(2*n,k-1).
Hockey stick identities (relate row k entries to entries in row k-1):
T(2*n,k) = T(2*n-1,k-1) + T(2*n-3,k-1) + T(2*n-5,k-1) + ....
T(2*n+1,k) = T(2*n,k-1) + ( T(2*n-1,k-1) + T(2*n-3,k-1) + T(2*n-5,k-1) + ... ). (End)

Keyword:tabl added, notation standardized, formula added by the Assoc. Editors of the OEIS, Feb 02 2010

Entry revised by N. J. A. Sloane, Nov 23 2017

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A169623 Generalized Pascal triangle read by rows: T(n,0) = T(0,n) = 1 for n >= 0, T(n,k) = 0 for k < 0 or k > n; otherwise T(n,k) = T(n-2,k-2) + T(n-2,k-1) + T(n-2,k) for 1 <= k <= n-1.

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