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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.

%I A169623 #57 Aug 04 2025 03:37:37
%S A169623 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,
%T A169623 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,
%U A169623 51,45,30,15,5,1,1,6,20,45,75,96,96,75,45,20,6,1
%N 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.
%C A169623 The borders are all 1's, with zero entries outside. To get an internal entry, use the rule that D = A+B+C here:
%C A169623          A   B   C
%C A169623        *   *   *   *
%C A169623      *   *   D   *   *
%C A169623 That is, add the three terms directly above you, two rows back.
%C A169623 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
%C A169623 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.
%C A169623 May be split into two triangles by taking the even-numbered and odd-numbered rows separately: the even-numbered rows give A027907.
%C A169623 From _Peter Bala_, Aug 19 2021: (Start)
%C A169623 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
%C A169623   /I_k 0\
%C A169623   \ 0  M/
%C A169623   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)
%H A169623 Rémy Sigrist, <a href="/A169623/b169623.txt">Rows 0..199, flattened</a>
%H A169623 Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Barry/barry321.html">Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices</a>, Journal of Integer Sequences, 19, 2016, #16.3.5.
%H A169623 Richard Ehrenborg and Margaret A. Readdy, <a href="http://arxiv.org/abs/1609.03216">The Gaussian coefficient revisited</a>, arXiv:1609.03216 [math.CO], 2016.
%H A169623 Richard L. Ollerton and Anthony G. Shannon, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/36-2/ollerton.pdf">Some properties of generalized Pascal squares and triangles</a>, Fib. Q., 36 (No. 2, 1998), 98-109. See Table 10.
%H A169623 Wikipedia, <a href="https://en.wikipedia.org/wiki/Hockey-stick_identity">Hockey-stick identity</a>.
%F A169623 From _Peter Bala_, Aug 19 2021: (Start)
%F A169623 T(2*n,k) = T(2*n-1,k-1) + T(2*n-2,k).
%F A169623 T(2*n,k) = T(2*n-1,k) + T(2*n-2,k-2).
%F A169623 T(2*n+1,k) = T(2*n,k) + T(2*n,k-1).
%F A169623 Hockey stick identities (relate row k entries to entries in row k-1):
%F A169623 T(2*n,k) = T(2*n-1,k-1) + T(2*n-3,k-1) + T(2*n-5,k-1) + ....
%F A169623 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)
%e A169623 Triangle begins:
%e A169623                     1
%e A169623                   1   1
%e A169623                 1   1   1
%e A169623               1   2   2   1
%e A169623             1   2   3   2   1
%e A169623           1   3   5   5   3   1
%e A169623         1   3   6   7   6   3   1
%e A169623       1   4   9  13  13   9   4   1
%e A169623     1   4  10  16  19  16  10   4   1
%e A169623   ...
%e A169623 As a square array read by antidiagonals:
%e A169623   1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 1, 1, 1, 1, ...
%e A169623   1, 1, 2,  2,  3,  3,  4,  4,  5,  5,  6,  6, 7, 7, 8, 8, ...
%e A169623   1, 2, 3,  5,  6,  9, 10, 14, 15, 20, 21, 27, ...
%e A169623   1, 2, 5,  7, 13, 16, 26, 30, 45, ...
%e A169623   1, 3, 6, 13, 19, 35, 45, 75, ...
%e A169623   1, 3, 9, 16, 35, 51, 96, ...
%e A169623   ...
%e A169623 From _Peter Bala_, Aug 19 2021: (Start)
%e A169623 With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
%e A169623   /1        \/1        \/1       \ /1       \        /1         \
%e A169623   |1 1      ||0 1      ||0 1      ||0 1      |       |1 1       |
%e A169623   |1 0 1    ||0 1 1    ||0 0 1    ||0 0 1    |...  = |1 1 1     |
%e A169623   |1 0 1 1  ||0 1 0 1  ||0 0 1 1  ||0 0 0 1  |       |1 2 2 1   |
%e A169623   |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 |
%e A169623   |...      ||...       |...      ||...      |       |...       |
%e A169623 (End)
%p A169623 T:=proc(n,k) option remember;
%p A169623 if n >= 0 and k = 0 then 1
%p A169623 elif n >= 0 and k = n then 1
%p A169623 elif (k < 0 or k > n) then 0
%p A169623 else T(n-2,k-2)+T(n-2,k-1)+T(n-2,k);
%p A169623 fi;
%p A169623 end;
%p A169623 for n from 0 to 14 do lprint([seq(T(n,k),k=0..n)]); od: # _N. J. A. Sloane_, Nov 23 2017
%t A169623 p[x, 1] := 1;
%t A169623 p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + x + 1)^Floor[n/2]]
%t A169623 a = Table[CoefficientList[p[x, n], x], {n, 1, 12}]
%t A169623 Flatten[a] (* This is for the same sequence but with offset 1 *)
%Y A169623 A123149 is essentially the same triangle, except for a diagonal of zeros.
%Y A169623 Row sums are in A182522 (essentially A038754).
%Y A169623 Cf. A027907, A070909.
%Y A169623 See A295555 for the next triangle in the series A007318, A169623 (this sequence).
%K A169623 nonn,easy,tabl
%O A169623 0,8
%A A169623 _Roger L. Bagula_ and _Gary W. Adamson_, Dec 03 2009
%E A169623 Keyword:tabl added, notation standardized, formula added by the Assoc. Editors of the OEIS, Feb 02 2010
%E A169623 Entry revised by _N. J. A. Sloane_, Nov 23 2017