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%I A281725 #20 Oct 19 2024 15:57:32
%S A281725 1,3,5,6,7,15,10,9,13,34,15,11,24,21,65,21,13,19,25,31,111,28,15,33,
%T A281725 58,54,43,175,36,17,25,33,41,49,57,260,45,19,42,37,115,55,96,73,369,
%U A281725 55,21,31,82,51,61,142,81,91,505,66,23,51,45,84,201,117,89,150,111,671
%N A281725 Triangular array T(n,k) is the sum of elements in an n X k matrix that will be assigned the same value whether the integers from 1 to n*k are assigned to elements in row-major order or column-major order.
%H A281725 Alois P. Heinz, <a href="/A281725/b281725.txt">Antidiagonals n = 1..141, flattened</a>
%H A281725 Ana Rechtman, <a href="http://images-archive.math.cnrs.fr/Janvier-2017-3e-defi.html">3ème Défi du Calendrier Mathématique</a>, Images des Mathématiques, CNRS, January 2017.
%e A281725 For n=2 and k=3, the matrix will be
%e A281725 1 2 3  and  1 3 5
%e A281725 4 5 6       2 4 6
%e A281725 and there are 2 identical terms, 1 and 6, whose sum is 7.
%e A281725 The triangle begins
%e A281725    1;
%e A281725    3,  5;
%e A281725    6,  7, 15;
%e A281725   10,  9, 13, 34;
%e A281725   15, 11, 24, 21, 65;
%e A281725   21, 13, 19, 25, 31, 111;
%e A281725   ...
%p A281725 T:= (n, k)-> add(add(`if`(j+k*(i-1)=
%p A281725     i+n*(j-1), j+k*(i-1), 0), i=1..n), j=1..k):
%p A281725 seq(seq(T(n,k), k=1..n), n=1..20);  # _Alois P. Heinz_, Jan 28 2017
%o A281725 (PARI) a(n, k) = {ml = matrix(n, k, i, j, ((i-1)*k+j)); mc = matrix(n, k, i, j, ((j-1)*n+i)); sum(i=1, n, sum(j=1, k, ml[i,j]*(ml[i,j] == mc[i,j])));}
%Y A281725 Cf. A281726.
%Y A281725 Main diagonal gives A006003.
%Y A281725 Column k=1 gives A000217.
%Y A281725 T(2n,n) gives A058331 for n>0.
%Y A281725 T(2n+1,n+1) gives A081436.
%K A281725 nonn,tabl
%O A281725 1,2
%A A281725 _Michel Marcus_, Jan 28 2017

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