A292671 Upper right triangle A(m,n) = least number of symbols required to fill a grid of size n X n row by row in the greedy way such that in no row or column or m X m square a symbol occurs more than once.

1, 2, 4, 4, 6, 9, 4, 6, 11, 16, 8, 7, 13, 18, 25, 8, 8, 13, 18, 27, 36, 8, 10, 13, 20, 29, 38, 49, 8, 10, 13, 20, 32, 38, 51, 64, 16, 13, 14, 22, 33, 40, 53, 66, 81, 16, 15, 14, 22, 33, 40, 56, 66, 83, 100, 16, 16, 15, 23, 33, 41, 57, 68, 85, 102

Offset

1,2

Comments

Consider the symbols as positive integers. By the greedy way we mean to fill the grid row by row from left to right always with the least possible positive integer such that the three constraints (on rows, columns and rectangular blocks) are satisfied. In contrast to the sudoku case, the m X m rectangles have "floating" borders, so the constraint is actually equivalent to say that any element must be different from all neighbors in a Moore neighborhood of range m-1 (having up to (2m-1)^2 grid points). See A292673 for examples.

One can consider the infinite square array A(m,n) defined in the same way, but the lower triangular part of it is uninteresting: for m>n one has A(m,n) = n^2, i.e., the columns continue below the diagonal indefinitely with the same value, n^2.

Links

Michael S. Branicky, Table of n, a(n) for n = 1..5050 (first 100 rows)

Eric Weisstein's World of Mathematics, Moore Neighborhood.

Example

The infinite square array would look as follows: (but the sequence only lists the upper right triangle: 1; 2, 4; 4, 6, 9; 4, 6, 11, 16; ...):
   [1  2  4  4  8  8  8  8 16 ...]  m=1: A(1,n) = 2^ceil(log_2(n)) = A062383(n-1)
   [1\_4  6  6  7  8 10 10 13 ...]  m=2: A(2,n) = A292672(n)
   [1  4\_9 11 13 13 13 13 14 ...]  m=3: A(3,n) = A292673(n) : see here
   [1  4  9\16 18 18 20 20 22 ...]  m=4: A(4,n) = A292674(n) :  for examples
   [1  4  9 16\25 27 29 32 33 ...]  m=5: A(5,n) = A292675(n)
   [1  4  9 16 25\36 38 38 40 ...]  m=6: A(6,n) = A292676(n)
   [1  4  9 16 25 36\49 51 53 ...]  m=7: A(7,n) = A292677(n)
   [1  4  9 16 25 36 49\64 66 ...]  m=8: A(8,n) = A292678(n)
   [1  4  9 16 25 36 49 64\81 ...]  m=9: A(9,n) = A292679(n)
   [...   ...  ...  ...  ...  ...]

Prog

(PARI) A(m, n, g=matrix(n, n))={my(ok(g, k, i, j, m)=if(m, ok(g[i, ], k)&&ok(g[, j], k)&&ok(concat(Vec(g[max(1, i-m+1)..i, max(1, j-m+1)..min(#g, j+m-1)])), k), !setsearch(Set(g), k))); for(i=1, n, for(j=1, n, for(k=1, n^2, ok(g, k, i, j, m)&&(g[i, j]=k)&&break))); vecmax(g)} \\ without "vecmax" the program returns the full n X n board.
(Python)
def A(m, n): # change m for A292672, ..., A292679
    mx, S, N, b = 0, {1}, range(1, n+1), m # b is block size
    g = [[0 for j in range(n+b)] for i in range(n+b)]
    row, col = {i:set() for i in N}, {j:set() for j in N}
    offsets = [(i, j) for i in range(-b+1, 1) for j in range(-b+1, 1)]
    offsets += [(i, j) for i in range(-b+1, 0) for j in range(1, b)]
    for i in N:
        for j in N:
            rect = set(g[i+o[0]][j+o[1]] for o in offsets)
            e = min(S - row[i] - col[j] - rect)
            g[i][j] = e
            if e > mx:
                mx = e
                S.add(mx+1)
            row[i].add(e)
            col[j].add(e)
    return mx
print([A(m, n) for n in range(1, 12) for m in range(1, n+1)]) # Michael S. Branicky, Apr 13 2023

Crossrefs

Cf. A292670, A292672, A292673, ..., A292679.

Sequence in context: A298043 A325245 A241064 * A210948 A359678 A008133

Adjacent sequences: A292668 A292669 A292670 * A292672 A292673 A292674

Keyword

nonn,tabl

Author

M. F. Hasler, Sep 20 2017

Status

approved