OFFSET
1,2
COMMENTS
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.
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)
LINKS
Peter Kagey, Table of n, a(n) for n = 1..32896 (first 256 rows, flattened)
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
FORMULA
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
EXAMPLE
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.
MAPLE
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
CROSSREFS
AUTHOR
Peter Kagey, Dec 27 2016
STATUS
approved