A281726 - OEIS

%I #34 Dec 10 2023 11:10:56

%S 1,2,2,3,2,3,4,2,2,4,5,2,3,2,5,6,2,2,2,2,6,7,2,3,4,3,2,7,8,2,2,2,2,2,

%T 2,8,9,2,3,2,5,2,3,2,9,10,2,2,4,2,2,4,2,2,10,11,2,3,2,3,6,3,2,3,2,11,

%U 12,2,2,2,2,2,2,2,2,2,2,12,13,2,3,4,5,2,7,2,5,4,3,2,13

%N Triangular array read by rows: T(n,k) is the number 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.

%C From _Jon E. Schoenfield_, Dec 10 2023: (Start)

%C T(n,k) is also the number of lattice points that lie on a line segment from (0,0) to (n-k,k-1). Thus, row n of the triangle lists, for each of the n 1st-quadrant lattice points P whose Manhattan distance from the origin is n-1, the number of lattice points on a line segment from the origin to P.

%C E.g., for n = 5, the 5 1st-quadrant lattice points whose Manhattan distance from the origin is 4 are (0,4), (1,3), (2,2), (3,1), and (4,0), and a line segment drawn from the origin to each of these points will intersect 5, 2, 3, 2, and 5 lattice points, respectively; { 5, 2, 3, 2, 5 } is row 5 of the triangle. (End)

%H Alois P. Heinz, <a href="/A281726/b281726.txt">Rows n = 1..200, flattened</a>

%H Ana Rechtman, <a href="http://images.math.cnrs.fr/Janvier-2017-3e-defi.html">3ème Défi du Calendrier Mathématique</a>, Images des Mathématiques, CNRS, January 2017.

%F T(n,k) = 1 + gcd(n-1, k-1). - _Jon E. Schoenfield_, Dec 08 2023

%e For n=3 and k=2, the matrix will be

%e   1 2  and  1 4

%e   3 4       2 5

%e   5 6       3 6

%e and there are 2 identical terms (1 and 6).

%e Triangle begins:

%e   1;

%e   2, 2;

%e   3, 2, 3;

%e   4, 2, 2, 4;

%e   5, 2, 3, 2, 5;

%e   6, 2, 2, 2, 2, 6;

%e   ...

%p T:= (n, k)-> add(add(`if`(j+k*(i-1)=i+n*(j-1), 1, 0), i=1..n), j=1..k):

%p seq(seq(T(n,k), k=1..n), n=1..20);  # _Alois P. Heinz_, Jan 28 2017

%t Array[1+GCD[#,Range[0,#]]&,20,0] (* _Paolo Xausa_, Dec 08 2023 *)

%o (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] == mc[i,j]));}

%Y Cf. A109004, A281725.

%Y Main diagonal and column k=1 give A000027.

%K nonn,tabl

%O 1,2

%A _Michel Marcus_, Jan 28 2017