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A263542 Triangle T(M, N): Number of M X N matrices where 1 all elements are distinct, all elements are at least 0 and at most M*N-1, and every 2 X 2 block of elements has the same sum.

%I A263542 #47 Sep 27 2024 20:55:00
%S A263542 24,112,376,768,2160,20352,5376,5904,86208,51840,64512,56736,1628352,
%T A263542 1342656,44084736
%N A263542 Triangle T(M, N): Number of M X N matrices where 1<N<=M, all elements are distinct, all elements are at least 0 and at most M*N-1, and every 2 X 2 block of elements has the same sum.
%C A263542 This sequence is given in this order: (2,2), (3,2), (3,3), (4,2), (4,3), (4,4), etc.
%C A263542 The idea of the program below is that the first row, first column, and the (1,1)th element uniquely determine the rest of the matrix. Hence, all permutations of m+n integers in the range 0..m*n-1 are generated to fill the first row, first column, and (1,1). Then the empty spots in the matrix are filled in and if at any point a condition is violated (duplicate, < 0, >= M*N), the program immediately moves on to the next permutation.
%C A263542 Much of the conversation in the main chat room of the Programming Puzzles and Code Golf Stack Exchange site - the Nineteenth Byte - following the linked message in the Links section deals with finding the terms of this sequence.
%C A263542 Observation: at least the first 15 terms are divisible by 8. - _Omar E. Pol_, Oct 20 2015, Nov 21 2015
%C A263542 When M and N are both even, the block sum is 2(MN-1). When one or both is odd the block sum can vary: e.g., for M=N=3, it varies from 12 to 20. - _Peter J. Taylor_, Oct 21 2015
%C A263542 When M and N are both even, all solutions are toroidal: the block sums created by wrapping from the last column to the first column or the last row to the first row also equal 2(MN-1). When one of M or N is even, all solutions are cylindrical, with wrapping in the even dimension, but they are toroidal only in the trivial case of Odd X 2. When both M and N are odd, except in the trivial case of 1 X 1, solutions do not wrap in either direction. - _Peter J. Taylor_, Oct 21 2015
%H A263542 The Nineteenth Byte, <a href="http://chat.stackexchange.com/transcript/message/24811363#24811363">Originating chat message</a>, ChatRoom.
%e A263542 One 3 X 3 solution (with a sum of 19) is:
%e A263542    0 4 2
%e A263542    8 7 6
%e A263542    3 1 5
%e A263542 One 4 X 4 solution (with a sum of 30) is:
%e A263542     0  3  4  7
%e A263542    12 15  8 11
%e A263542     1  2  5  6
%e A263542    13 14  9 10
%e A263542 One 5 X 5 solution (with a sum of 48) is:
%e A263542     0 24  1 23  2
%e A263542     9 15  8 16  7
%e A263542    10 14 11 13 12
%e A263542    19  5 18  6 17
%e A263542    20  4 21  3 22
%e A263542 The triangle T(M, N) begins:
%e A263542 M\N    2      3       4       5        6 ...
%e A263542 2:    24
%e A263542 3:   112    376
%e A263542 4:   768   2160   20352
%e A263542 5:  5376   5904   86208   51840
%e A263542 6: 64512  56736 1628352 1342656 44084736
%e A263542 ...reformatted. - _Wolfdieter Lang_, Dec 16 2015
%o A263542 (Python)
%o A263542 from itertools import permutations as P
%o A263542 n = 4; m = 4; permutes = P(range(m*n), m+n); counter = 0
%o A263542 for p in permutes:
%o A263542   grid = [p[:n]]
%o A263542   for i in range(m-1):
%o A263542     grid.append([p[n+i]]+[-1]*(n-1))
%o A263542   grid[1][1] = p[-1]
%o A263542   s = p[0]+p[1]+p[n]+p[-1]
%o A263542   has = list(p)
%o A263542   fail = 0
%o A263542   for y in range(1,m):
%o A263542     for x in range(1,n):
%o A263542       if x == y == 1: continue
%o A263542       r = s - (grid[y-1][x-1] + grid[y-1][x] + grid[y][x-1])
%o A263542       if r not in has and 0 <= r < m*n:
%o A263542         grid[y][x]=r
%o A263542         has.append(r)
%o A263542       else:
%o A263542        fail = 1
%o A263542        break
%o A263542     if fail: break
%o A263542   if not fail:
%o A263542     counter += 1
%o A263542 print(counter)
%K A263542 nonn,tabl,more
%O A263542 2,1
%A A263542 _Lee Burnette_, Oct 20 2015