A256008 -id:A256008 - cp's OEIS frontend

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.

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

Offset: 1

Alec Jones, Apr 07 2016

An infinite Sudoku-type 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 k-1 copies of j in the columns to the left of the k-th 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
Theorem R (Rob Pratt, Bob Selcoe, N. J. A. Sloane, Jul 02 2016): Every row is a permutation of the natural numbers.
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 n-1 contain at most n-1 copies of j, and their influence on the entries in the n-th 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 non-attacking 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 k-k_0 cells in row n that have to be attacked, and for large k this is impossible since k-k_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 Sprague-Grundy values, see A274528.
From Don Reble, Jun 30 2016: (Start)
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 16-periodic from, respectively, term 8, 17, 52, 91, 92, 131, ... on, rather than having period 4^(k-1) from term k on. - M. F. Hasler, Sep 26 2022

			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, ...
...
  - _N. J. A. Sloane_, Jun 29 2016

Peter Kagey, Table of n, a(n) for n = 1..10000
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Peter Kagey, Animated example illustrating the first fifteen terms.

First 4 rows are A274315, A274316, A274317, A274791.
Main diagonal is A274318.
Column 1 is A000027, column 2 is A256008(n) = A004443(n-1)+1 = 1 + (nimsum of n-1 and 2), column 3 is A274614 (or equally, A274615 + 1), and column 4 is A274617 (or equally, A274619 + 1).
Antidiagonal sums give A274530. Other properties of antidiagonals: A274529, A275883.
Cf. A274080 (used in Haskell program), A274616.
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.

import Data.List ((\\))
a269526 n = head $ [1..] \\ map a269526 (a274080_row n)
-- Peter Kagey, Jun 10 2016

# 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..n-1),
                 seq(A(n,j), j=1..k-1),
                 seq(A(n-t,k-t), t=1..min(n,k)-1),
                 seq(A(n+j,k-j), j=1..k-1)};
            for m while m in s do od; m
         fi
     end:
[seq(seq(A(1+d-k, k), k=1..d), d=1..15)];

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

{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<-[1-min(r, c)..-1]], [T(r,k)|k<-[1..c-1]], [T(k,c)|k<-[1..r-1]], [T(r+c-k,k)|k<-[1..c-1]]])), c[k]==k)+1); r} \\ M. F. Hasler, Sep 26 2022

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,...,n-1. 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
Comments from Bob Selcoe, Jun 29 2016: (Start)
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+2*k)/2), j>=k-1), so that each column is a permutation.
Also, the differences in column 1 are a 1-cycle ([+1]), in column 2 a 4-cycle after the first term, and in column 3 a 16-cycle after the second term. Perhaps the cycle lengths are 4^(k-1) starting after j=k-1. (End)  WARNING: These comments may be wrong - see COMMENTS section. - N. J. A. Sloane, Sep 26 2022

Definition clarified by Omar E. Pol, Jun 29 2016

A004443 Nimsum n + 2.

Original entry on oeis.org

2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13, 18, 19, 16, 17, 22, 23, 20, 21, 26, 27, 24, 25, 30, 31, 28, 29, 34, 35, 32, 33, 38, 39, 36, 37, 42, 43, 40, 41, 46, 47, 44, 45, 50, 51, 48, 49, 54, 55, 52, 53, 58, 59, 56, 57, 62, 63, 60, 61, 66, 67, 64, 65

Offset: 0

N. J. A. Sloane

A self-inverse permutation of the natural numbers. - Philippe Deléham, Nov 22 2016

E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 60.
J. H. Conway, On Numbers and Games. Academic Press, NY, 1976, pp. 51-53.

N. J. A. Sloane, Table of n, a(n) for n = 0..20000
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Index entries for sequences related to Nim-sums
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Cf. A003987, A004442, A269526.
Essentially the same as A256008 - 1.
Also the second column of A274528.
Cf. A002162.

nimsum := proc(a,b) local t1,t2,t3,t4,l; t1 := convert(a+2^200,base,2); t2 := convert(b+2^200,base,2); t3 := evalm(t1+t2); map(x->x mod 2, t3); t4 := convert(evalm(%),list); l := convert(t4,base,2,10); sum(l[k]*10^(k-1), k=1..nops(l)); end;
f := n -> n + 2*(-1)^floor(n/2); # N. J. A. Sloane, Jul 06 2019

Table[BitXor[n, 2], {n, 0, 100}] (* T. D. Noe, Feb 09 2013 *)

a(n)=bitxor(n,2) \\ Charles R Greathouse IV, Oct 07 2015

for n in range(20): print(2^n) # Oliver Knill, Feb 16 2020

a(n) = n XOR 2. - Joerg Arndt, Feb 07 2013
G.f.: (2-x-2x^2+3x^3)/((1-x)^2(1+x^2)). - Ralf Stephan, Apr 24 2004
The sequences 'Nimsum n + m' seem to have the general o.g.f. p(x)/q(x) with p, q polynomials and q(x) = (1-x)^2*Product_{k>=0} (1+x^(2^e(k))), with Sum_{k>=0} 2^e(k) = m. - Ralf Stephan, Apr 24 2004
a(n) = n + 2(-1)^floor(n/2). - Mitchell Harris, Jan 10 2005
a(n) = OR(n,2) - AND(n,2). - Gary Detlefs, Feb 06 2013
E.g.f.: 2*(sin(x) + cos(x)) + x*exp(x). - Ilya Gutkovskiy, Jul 01 2016
Sum_{n>=0,n<>2} (-1)^n/a(n) = -log(2) = -A002162. - Peter McNair, Aug 07 2023

A280172 Lexicographically earliest table of positive integers read by antidiagonals such that no row or column contains a repeated term.

Original entry on oeis.org

1, 2, 2, 3, 1, 3, 4, 4, 4, 4, 5, 3, 1, 3, 5, 6, 6, 2, 2, 6, 6, 7, 5, 7, 1, 7, 5, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 7, 5, 7, 1, 7, 5, 7, 9, 10, 10, 6, 6, 2, 2, 6, 6, 10, 10, 11, 9, 11, 5, 3, 1, 3, 5, 11, 9, 11, 12, 12, 12, 12, 4, 4, 4, 4, 12, 12, 12, 12, 13, 11, 9, 11, 13, 3, 1, 3, 13, 11, 9, 11, 13

Offset: 1

Peter Kagey, Dec 27 2016

The table is symmetrical about the main diagonal.
The first row/column is A000027.
The second row/column is A103889.
The third row/column is A256008.
The fourth row/column is A113778.
Conjecture: The (2^k)-th antidiagonal consists entirely of 2^k.
Similar in spirit to A269526, A274528. - N. J. A. Sloane, Dec 27 2016
From Daniel Forgues, Sep 14 2019: (Start)
Plot of a(n) looks like a transform of a Sierpinski equilateral triangle.
Considering t(a(n)) = a(n)*(a(n)+1)/2: top edge of plot would be linear, but left & right sides of [concave curved] triangles would grow/decrease quadratically. a(n), a univalued sequence, tries to plot a Sierpinski triangle, which requires a multivalued sequence: a(n) uses t(2^k) terms to draw a Sierpinski triangle of width & height 2^k.
Conjecture: T(2n, k) = 2 * T(n, ceiling(k/2)), n >= 1, 1 <= k <= 2n. E.g.
  row  5:      5,     3,    1,    3,     5
  row 10:   10, 10, 6, 6, 2, 2, 6, 6, 10, 10 (End)
From Daniel Forgues, Sep 15 2019: (Start)
Conjectured algorithm for equilateral triangle (1-indexed rows and row terms), whose concatenated rows give this sequence: T(1, 1) = 1;
For each k >= 0, the height of the Sierpinski triangle is doubled:
* Left and right triangles: for 1 <= i <= 2^k, 1 <= j <= i:
  T(2^k + i, j) = T(2^k + i, 2^k + i + 1 - j) = T(i, j) + 2^k;
* Central triangle: for 1 <= i <= 2^k - 1, 1 <= j <= i:
  T(2^(k+1) - i, 2^k - i + j) = T(i, j).
Left and right triangles copies rows 1 to 2^k, terms augmented by 2^k.
Central triangle is mirrored through row 2^k.
When n is t(2^k), k >= 0, i.e., a triangular number with index a power of 2, a phase of the Sierpinski triangle plot is neatly completed. (End)

			As table (upper anti-triangular matrix) (concat. antidiagonals):
  1 2 3 4 5 6 7 8
  2 1 4 3 6 5 8
  3 4 1 2 7 8
  4 3 2 1 8
  5 6 7 8
  6 5 8
  7 8
  8
As equilateral triangle (concat. rows): (see formula section)
         1
        2 2
       3 1 3
      4 4 4 4
     5 3 1 3 5
    6 6 2 2 6 6
   7 5 7 1 7 5 7
  8 8 8 8 8 8 8 8
Lexicographically earliest equilateral triangle of positive integers read by rows such that no diagonal or antidiagonal contains a repeated term.

Peter Kagey, Table of n, a(n) for n = 1..32896 (first 256 rows, flattened)
Eric Gottlieb, Matjaž Krnc, and Peter Muršič, Nim on Integer Partitions and Hyperrectangles, arXiv:2506.04991 [math.CO], 2025. See p. 16.
Peter Kagey, Bitmap of first 2^10 = 1024 rows and columns. (Black pixels correspond to numbers divisible by 3; white pixels to all other numbers.)
Rémy Sigrist, Scatterplot of (n, a(n)*(a(n)+1)/2) for n = 1..2100225

Cf. A003987, A269526, A274528.

Rows (or columns) 1 to 4: A000027, A103889, A256008, A113778.

A280172 := (n, k) -> 1 + Bits:-Xor(k-1, n-k):
seq(print(seq(A280172(n, k), k=1..n)), n=1..14); # Peter Luschny, Sep 21 2019

T(n, k) = ( (n-1) XOR (k-1) ) + 1 = A003987(n-1, k-1) + 1. - Rémy Sigrist, Sep 18 2019

a(n) = T(row, n - t(row - 1)), n >= 1, where row = ceiling((-1 + sqrt(1 + 8*n))/2) and t(i) = i*(i+1)/2. - Daniel Forgues, Sep 20 2019

A292576 Permutation of the natural numbers partitioned into quadruples [4k-1, 4k-3, 4k-2, 4k], k > 0.

Original entry on oeis.org

3, 1, 2, 4, 7, 5, 6, 8, 11, 9, 10, 12, 15, 13, 14, 16, 19, 17, 18, 20, 23, 21, 22, 24, 27, 25, 26, 28, 31, 29, 30, 32, 35, 33, 34, 36, 39, 37, 38, 40, 43, 41, 42, 44, 47, 45, 46, 48, 51, 49, 50, 52, 55, 53, 54, 56, 59, 57, 58, 60, 63, 61, 62

Offset: 1

Guenther Schrack, Sep 19 2017

nonn

Partition the natural number sequence into quadruples starting with (1,2,3,4); swap the second and third elements, then swap the first and the second element; repeat for all quadruples.

Guenther Schrack, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).
Index entries for sequences that are permutations of the natural numbers

Inverse: A056699(n+1) - 1 for n > 0.
Sequence of fixed points: A008586(n) for n > 0.
Subsequences:
   elements with odd index: A042964(A103889(n)) for n > 0.
   elements with even index: A042948(n) for n > 0.
   odd elements: A166519(n) for n>0.
   indices of odd elements: A042963(n) for n > 0.
   even elements: A005843(n) for n>0.
   indices of even elements: A014601(n) for n > 0.
Sum of pairs of elements:
   a(n+2) + a(n) = A163980(n+1) = A168277(n+2) for n > 0.
Difference between pairs of elements:
   a(n+2) - a(n) = (-1)^A011765(n+3)*A091084(n+1) for n > 0.
Compound relations:
   a(n) = A284307(n+1) - 1 for n > 0.
   a(n+2) - 2*a(n+1) + a(n) = (-1)^A011765(n)*A132400(n+1) for n > 0.
Compositions:
   a(n) = A116966(A080412(n)) for n > 0.
   a(A284307(n)) = A256008(n) for n > 0.
   a(A042963(n)) = A166519(n-1) for n > 0.
   A256008(a(n)) = A056699(n) for n > 0.

a = [3 1 2 4]; % Generate b-file
max = 10000;
for n := 5:max
   a(n) = a(n-4) + 4;
end;

for(n=1, 10000, print1(n + ((-1)^(n*(n-1)/2)*(2 - (-1)^n) - (-1)^n)/2, ", "))

a(1)=3, a(2)=1, a(3)=2, a(4)=4, a(n) = a(n-4) + 4 for n > 4.
O.g.f.: (2*x^3 + x^2 - 2*x + 3)/(x^5 - x^4 - x + 1).
a(n) = n + ((-1)^(n*(n-1)/2)*(2-(-1)^n) - (-1)^n)/2.
a(n) = n + (cos(n*Pi/2) - cos(n*Pi) + 3*sin(n*Pi/2))/2.
a(n) = n + n mod 2 + (ceiling(n/2)) mod 2 - 2*(floor(n/2) mod 2).
Linear recurrence: a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
First Differences, periodic: (-2, 1, 2, 3), repeat; also (-1)^A130569(n)*A068073(n+2) for n > 0.

A298364 Permutation of the natural numbers partitioned into quadruples [4k-2, 4k-1, 4k-3, 4k] for k > 0.

Original entry on oeis.org

2, 3, 1, 4, 6, 7, 5, 8, 10, 11, 9, 12, 14, 15, 13, 16, 18, 19, 17, 20, 22, 23, 21, 24, 26, 27, 25, 28, 30, 31, 29, 32, 34, 35, 33, 36, 38, 39, 37, 40, 42, 43, 41, 44, 46, 47, 45, 48, 50, 51, 49, 52, 54, 55, 53, 56, 58, 59, 57, 60, 62, 63, 61, 64, 66, 67, 65

Offset: 1

Guenther Schrack, Jan 18 2018

Partition the natural number sequence into quadruples starting with (1,2,3,4); swap the first and second elements, then swap the second and third elements; repeat for all quadruples.

Guenther Schrack, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Inverse: A292576.
Sequence of fixed points: A008586(n) for n > 0.
First differences: (-1)^floor(n^2/4)*A068073(n-1) for n > 0.
Subsequences:
  elements with odd index: A042963(A103889(n)) for n > 0.
  elements with even index A014601(n) for n > 0.
  odd elements: A166519(n-1) for n > 0.
  indices of odd elements: A042964(n) for n > 0.
  even elements: A005843(n) for n > 0.
  indices of even elements: A042948(n) for n > 0.
Other similar permutations: A116966, A284307, A292576.

a = [2 3 1 4];
max = 10000;    % Generation of b-file.
for n := 5:max
   a(n) = a(n-4) + 4;
end;

Nest[Append[#, #[[-4]] + 4] &, {2, 3, 1, 4}, 63] (* or *)
Array[# + ((-1)^# + ((-1)^(# (# - 1)/2)) (1 - 2 (-1)^#))/2 &, 67] (* Michael De Vlieger, Jan 23 2018 *)
LinearRecurrence[{1,0,0,1,-1},{2,3,1,4,6},70] (* Harvey P. Dale, Dec 12 2018 *)

for(n=1, 100, print1(n + ((-1)^n + ((-1)^(n*(n-1)/2))*(1 - 2*(-1)^n))/2, ", "))

O.g.f.: (3*x^3 - 2*x^2 + x + 2)/(x^5 - x^4 - x - 1).
a(1) = 2, a(2) = 3, a(3) = 1, a(4) = 4, a(n) = a(n-4) + 4 for n > 4.
a(n) = n + ((-1)^n + ((-1)^(n*(n-1)/2))*(1 - 2*(-1)^n))/2.
a(n) = n + (cos(n*Pi) - cos(n*Pi/2) + 3*sin(n*Pi/2))/2.
a(n) = 2*floor((n+1)/2) - 4*floor((n+1)/4) + floor(n/2) + 2*floor(n/4).
a(n) = n + (-1)^floor((n-1)^2/4)*A140081(n) for n > 0.
a(n) = A056699(n+1) - 1, n > 0.
a(n+2) = A168269(n+1) - a(n), n > 0.
a(n+2) = a(n) + (-1)^floor((n+1)^2/4)*A132400(n+2) for n > 0.
Linear recurrence: a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.
First differences: periodic, (1, -2, 3, 2) repeat.
Compositions:
  a(n) = A080412(A116966(n-1)) for n > 0.
  a(n) = A284307(A256008(n)) for n > 0.
  a(A067060(n)) = A133256(n) for n > 0.
  A116966(a(n+1)-1) = A092486(n) for n >= 0.
  A056699(a(n)) = A256008(n) for n > 0.

cp's OEIS Frontend

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.

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A004443 Nimsum n + 2.

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A280172 Lexicographically earliest table of positive integers read by antidiagonals such that no row or column contains a repeated term.

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A292576 Permutation of the natural numbers partitioned into quadruples [4k-1, 4k-3, 4k-2, 4k], k > 0.

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A298364 Permutation of the natural numbers partitioned into quadruples [4k-2, 4k-1, 4k-3, 4k] for k > 0.

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