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A169623
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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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11
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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
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OFFSET
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0,8
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COMMENTS
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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.
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)
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LINKS
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FORMULA
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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)
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EXAMPLE
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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, ...
...
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)
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MAPLE
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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
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MATHEMATICA
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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 */
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CROSSREFS
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A123149 is essentially the same triangle, except for a diagonal of zeros.
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KEYWORD
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AUTHOR
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EXTENSIONS
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Keyword:tabl added, notation standardized, formula added by the Assoc. Editors of the OEIS, Feb 02 2010
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STATUS
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approved
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