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A184173
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Triangle read by rows: T(n,k) is the sum of the k X k minors in the n X n Pascal matrix (0<=k<=n; the empty 0 X 0 minor is defined to be 1).
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4
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1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 34, 15, 1, 1, 31, 144, 144, 31, 1, 1, 63, 574, 1155, 574, 63, 1, 1, 127, 2226, 8526, 8526, 2226, 127, 1, 1, 255, 8533, 60588, 113832, 60588, 8533, 255, 1, 1, 511, 32587, 424117, 1444608, 1444608, 424117, 32587, 511, 1
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OFFSET
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0,5
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COMMENTS
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Apparently, the sum of the entries in row n is A005157(n).
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LINKS
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FORMULA
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The triangle is symmetric: T(n,k) = T(n,n-k).
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EXAMPLE
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T(3,1) = 7 because in the 3 X 3 Pascal matrix [1,0,0/1,1,0/1,2,1] the sum of the entries is 7.
Triangle starts:
1;
1, 1;
1, 3, 1;
1, 7, 7, 1;
1, 15, 34, 15, 1;
1, 31, 144, 144, 31, 1;
1, 63, 574, 1155, 574, 63, 1;
1, 127, 2226, 8526, 8526, 2226, 127, 1;
...
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MAPLE
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with(combinat): with(LinearAlgebra):
T:= proc(n, k) option remember; `if`(n-k<k, T(n, n-k), (l-> add(add(
Determinant(SubMatrix(Matrix(n, (i, j)-> binomial(i-1, j-1)),
i, j)), j in l), i in l))(choose([$1..n], k)))
end:
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MATHEMATICA
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T[n_, k_] := T[n, k] = If[k == 0 || k == n, 1, Module[{l, M},
l = Subsets[Range[n], {k}];
M = Table[Binomial[i-1, j-1], {i, n}, {j, n}];
Total[Det /@ Flatten[Table[M[[i, j]], {i, l}, {j, l}], 1]]]];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 09 2019 updated Feb 29 2024 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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