%I #36 Feb 29 2024 06:56:50
%S 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,
%T 63,1,1,127,2226,8526,8526,2226,127,1,1,255,8533,60588,113832,60588,
%U 8533,255,1,1,511,32587,424117,1444608,1444608,424117,32587,511,1
%N 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).
%C Apparently, the sum of the entries in row n is A005157(n).
%H Alois P. Heinz, <a href="/A184173/b184173.txt">Rows n = 0..12, flattened</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Minor_(linear_algebra)">Minor (linear algebra)</a>
%F The triangle is symmetric: T(n,k) = T(n,n-k).
%e 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.
%e Triangle starts:
%e 1;
%e 1, 1;
%e 1, 3, 1;
%e 1, 7, 7, 1;
%e 1, 15, 34, 15, 1;
%e 1, 31, 144, 144, 31, 1;
%e 1, 63, 574, 1155, 574, 63, 1;
%e 1, 127, 2226, 8526, 8526, 2226, 127, 1;
%e ...
%p with(combinat): with(LinearAlgebra):
%p T:= proc(n, k) option remember; `if`(n-k<k, T(n, n-k), (l-> add(add(
%p Determinant(SubMatrix(Matrix(n, (i, j)-> binomial(i-1, j-1)),
%p i, j)), j in l), i in l))(choose([$1..n], k)))
%p end:
%p seq(seq(T(n, k), k=0..n), n=0..7); # _Alois P. Heinz_, Feb 11 2019
%t T[n_, k_] := T[n, k] = If[k == 0 || k == n, 1, Module[{l, M},
%t l = Subsets[Range[n], {k}];
%t M = Table[Binomial[i-1, j-1], {i, n}, {j, n}];
%t Total[Det /@ Flatten[Table[M[[i, j]], {i, l}, {j, l}], 1]]]];
%t Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 09 2019 updated Feb 29 2024 *)
%Y Columns k=0-2 give: A000012, A000225, A306376.
%Y Cf. A005157, A007318.
%K nonn,tabl
%O 0,5
%A _Emeric Deutsch_, Jan 12 2011
%E Typo corrected by _Alois P. Heinz_, Feb 11 2019