0,8

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)

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.

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)

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)

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 *)

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).

Sequence in context: A176298 A259575 A370062 * A245558 A011847 A091325

Adjacent sequences: A169620 A169621 A169622 * A169624 A169625 A169626

nonn,easy,tabl

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

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

approved