

A269526


Square array T(n,k) (n>=1, k>=1) read by antidiagonals upwards in which each term is the least positive integer satisfying the condition that no row, column, diagonal, or antidiagonal contains a repeated term.


51



1, 2, 3, 3, 4, 2, 4, 1, 5, 6, 5, 2, 6, 1, 4, 6, 7, 3, 2, 8, 5, 7, 8, 1, 5, 9, 3, 10, 8, 5, 9, 4, 1, 7, 6, 11, 9, 6, 4, 7, 2, 8, 5, 12, 13, 10, 11, 7, 3, 5, 6, 9, 4, 14, 8, 11, 12, 8, 9, 6, 10, 3, 7, 15, 16, 14, 12, 9, 13, 10, 11, 14, 4, 15, 16, 17, 7, 18, 13, 10, 14, 11, 3, 4, 8, 16, 9, 6, 12, 15, 7
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OFFSET

1,2


COMMENTS

An infinite Sudokutype array.
In the definition, "diagonal" means a diagonal line of slope 1, and "antidiagonal" means a diagonal line of slope +1.
Theorem C (Bob Selcoe, Jul 01 2016): Every column is a permutation of the natural numbers.
Proof: Fix k, and suppose j is the smallest number missing from that column. For this to happen, every entry T(n,k) for sufficiently large n in that column must see a j in the NW diagonal through that cell or in the row to the W of that cell. But there are at most k1 copies of j in the columns to the left of the kth column, and if n is very large the entry T(n,k) will be unaffected by those j's, and so T(n,k) would then be set to j, a contradiction. QED
Proof: Fix n, and suppose j is the smallest number missing from that row. For this to happen, every entry T(n,k) for sufficiently large k in that row must see a j in the column to the N, or in the NW diagonal through that cell or in the SW diagonal through that cell.
Rows 1 through n1 contain at most n1 copies of j, and their influence on the entries in the nth row only extend out to the entry T(n,k_0), say. We take k to be much larger than k_0 and consider the entry T(n,k). We will show that for large enough k it can (and therefore must) be equal to j, which is a contradiction.
Consider the triangle bounded by row n, column 1, and the SW antidiagonal through cell (n,k). Replace every copy of j in this triangle by a queen and think of these cells as a triangular chessboard. These are nonattacking queens, by definition of the sequence, and by the result in A274616 there can be at most 2*k/3 + 1 such queens. However, there are kk_0 cells in row n that have to be attacked, and for large k this is impossible since kk_0 > 2*k/3+1. If a cell (n,k) is not attacked by a queen, then T(n,k) can take the value j. QED
Presumably every diagonal is also a permutation of the natural numbers, but the proof does not seem so straightforward. Of course the antidiagonals are not permutations of the natural numbers, since they are finite in length.  N. J. A. Sloane, Jul 02 2016
For an interpretation of this array in terms of SpragueGrundy values, see A274528.
Let b(n) be the position in column n where 1 appears, i.e., such that T(b(n),n) = 1. Then b(n) is A065188, which is Antti Karttunen's "Greedy Queens" permutation.
Let b'(n) be the position in row n where 1 appears, i.e., such that T(n,b'(n)) = 1. Then b'(n) is A065189, the inverse "Greedy Queens" permutation. (End)
The same sequence arises if we construct a triangle, by reading from left to right in each row, always choosing the smallest positive number which does not produce a duplicate number in any row or diagonal.  N. J. A. Sloane, Jul 02 2016
It appears that the numbers generally appear for the first time in or near the first few rows.  Omar E. Pol, Jul 03 2016
The last comment in the FORMULA section seems wrong: It seems that columns 4, 5, 6, 7, 8, 9, ...(?) all have first differences which become 16periodic from, respectively, term 8, 17, 52, 91, 92, 131, ... on, rather than having period 4^(k1) from term k on.  M. F. Hasler, Sep 26 2022


LINKS



FORMULA

Theorem 1: T(n,1) = n.
Proof by induction. T(1,1)=1 by definition. When calculating T(n,1), the only constraint is that it be different from all earlier entries in the first column, which are 1,2,3,...,n1. So T(n,1)=n. QED
Theorem 2 (Based on a message from Bob Selcoe, Jun 29 2016): Write n = 4t+i with t >= 0, i=1,2,3, or 4. Then T(n,2) = 4t+3 if i=1, 4t+4 if i=2, 4t+1 if i=3, 4t+2 if i=4. This implies that the second column is the permutation A256008.
Proof: We check that the first 4 entries in column 2 are 2,5,6,3. From then on, to calculate the entry T(n,2), we need only look to the N, NW, W, and SW (we need never look to the East). After we have found the first 4t entries in the column, the column contains all the numbers from 1 to 4t. The four smallest free numbers are 4t+1, 4t+2, 4t+3, 4t+4. Entry T(4t+1,2) cannot be 4t+1 or 4t+2, but it can (and therefore must) be 4t+3. Similarly T(4t+2,2)=4t+4, T(4t+3,2)=4t+1, and T(4t+4,2)=4t+2. The column now contains all the numbers from 1 to 4t+4. Repeating this argument established the theorem. QED
From Theorem 2, column 2 (i.e., terms a((j^2+j+4)/2), j>=1) is a permutation. After a(3)=3, the differences of successive terms follow the pattern a(n) = 3 [+1, 3, +1, +5], so a(5)=4, a(8)=1, a(12)=2, a(17)=7, a(23)=8, a(30)=5...
Similarly, column 3 (i.e., terms a((j^2+j+6)/2), j>=2) appears to be a permutation, but with the pattern after a(6)=2 and a(9)=5 being 5 [+1, 3, 2, +8, 5, +3, +1, +5, +1, 3, +1, 2, +8, 3, +1, +5]. (See A274614 and A274615.)
I conjecture that other similar cyclical difference patterns should hold for any column k (i.e., terms a(j^2+j+2k)/2), j>=k1), so that each column is a permutation.
Also, the differences in column 1 are a 1cycle ([+1]), in column 2 a 4cycle after the first term, and in column 3 a 16cycle after the second term. Perhaps the cycle lengths are 4^(k1) starting after j=k1. (End) WARNING: These comments may be wrong  see COMMENTS section.  N. J. A. Sloane, Sep 26 2022


EXAMPLE

The array is constructed along its antidiagonals, in the following way:
.
a(1) a(3) a(6) a(10)
a(2) a(5) a(9)
a(4) a(8)
a(7)
.
See the link from Peter Kagey for an animated example.
The beginning of the square array is:
1, 3, 2, 6, 4, 5, 10, 11, 13, 8, 14, 18, 7, 20, 19, 9, 12, ...
2, 4, 5, 1, 8, 3, 6, 12, 14, 16, 7, 15, 17, 9, 22, 21, 11, ...
3, 1, 6, 2, 9, 7, 5, 4, 15, 17, 12, 19, 18, 21, 8, 10, 23, ...
4, 2, 3, 5, 1, 8, 9, 7, 16, 6, 18, 17, 11, 10, 23, 22, 14, ...
5, 7, 1, 4, 2, 6, 3, 15, 9, 10, 13, 8, 20, 14, 12, 11, 17, ...
6, 8, 9, 7, 5, 10, 4, 16, 2, 1, 3, 11, 22, 15, 24, 13, 27, ...
7, 5, 4, 3, 6, 14, 8, 9, 11, 18, 2, 21, 1, 16, 10, 12, 20, ...
8, 6, 7, 9, 11, 4, 13, 3, 12, 15, 1, 10, 2, 5, 26, 14, 18, ...
9, 11, 8, 10, 3, 1, 14, 6, 7, 13, 4, 12, 24, 18, 2, 5, 19, ...
10, 12, 13, 11, 16, 2, 17, 5, 20, 9, 8, 14, 4, 6, 1, 7, 3, ...
11, 9, 14, 12, 10, 15, 1, 8, 21, 7, 16, 20, 5, 3, 18, 17, 32, ...
12, 10, 11, 8, 7, 9, 2, 13, 5, 23, 25, 26, 14, 17, 16, 15, 33, ...
...


MAPLE

# The following Maple program was provided at my request by Alois P. Heinz, who said that he had not posted it himself because it stores the data in an inefficient way.  N. J. A. Sloane, Jul 01 2016
A:= proc(n, k) option remember; local m, s;
if n=1 and k=1 then 1
else s:= {seq(A(i, k), i=1..n1),
seq(A(n, j), j=1..k1),
seq(A(nt, kt), t=1..min(n, k)1),
seq(A(n+j, kj), j=1..k1)};
for m while m in s do od; m
fi
end:
[seq(seq(A(1+dk, k), k=1..d), d=1..15)];


MATHEMATICA

A[n_, k_] := A[n, k] = If[n == 1 && k == 1, 1, s = {Table[A[i, k], {i, 1, n1}], Table[A[n, j], {j, 1, k1}], Table[A[nt, kt], {t, 1, Min[n, k]  1}], Table[A[n+j, kj], {j, 1, k1}]} // Flatten; For[m = 1, True, m++, If[FreeQ[s, m], Return[m]]]];
Table[Table[A[1+dk, k], {k, 1, d}], {d, 1, 15}] // Flatten (* JeanFrançois Alcover, Jul 21 2016, translated from Maple *)


PROG

(Haskell)
import Data.List ((\\))
a269526 n = head $ [1..] \\ map a269526 (a274080_row n)
(PARI) {M269526=Map(); A269526=T(r, c)=c>1 && !mapisdefined(M269526, [r, c], &r) && mapput(M269526, [r, c], r=sum(k=1, #c=Set(concat([[T(r+k, c+k)k<[1min(r, c)..1]], [T(r, k)k<[1..c1]], [T(k, c)k<[1..r1]], [T(r+ck, k)k<[1..c1]]])), c[k]==k)+1); r} \\ M. F. Hasler, Sep 26 2022


CROSSREFS

A065188 and A065189 say where the 1's appear in successive columns and rows.
If all terms are reduced by 1 and the offset is changed to 0 we get A274528.
A274650 and A274651 are triangles in the shape of a right triangle and with a similar definition.
See A274630 for the case where both queens' and knights' moves must avoid duplicates.


KEYWORD



AUTHOR



EXTENSIONS



STATUS

approved



