A262057 Array based on the Stanley sequence S(0), A005836, by antidiagonals.

0, 2, 1, 7, 5, 3, 21, 8, 6, 4, 23, 22, 16, 11, 9, 64, 26, 24, 17, 14, 10, 69, 65, 50, 25, 19, 15, 12, 71, 70, 67, 53, 48, 20, 18, 13, 193, 80, 78, 68, 59, 49, 34, 29, 27, 207, 194, 152, 79, 73, 62, 51, 35, 32, 28, 209, 208, 196, 161, 150, 74, 63, 52, 43, 33, 30

Offset: 1

Author

Max Barrentine, Nov 29 2015

Keywords

• nonn
• tabl

Comments

This array is similar to a dispersion in that the first column is the minimal nonnegative sequence that contains no 3-term arithmetic progression, and each next column is the minimal sequence consisting of the numbers rejected from the previous column that contains no 3-term arithmetic progression.

A100480(n) describes which column n is sorted into.

The columns of the array form the greedy partition of the nonnegative integers into sequences that contain no 3-term arithmetic progression. - Robert Israel, Feb 03 2016

Examples

			From the top-left corner, this array starts:
   0   2   7  21  23  64
   1   5   8  22  26  65
   3   6  16  24  50  67
   4  11  17  25  53  68
   9  14  19  48  59  73
  10  15  20  49  62  74
		

Links

• Max Barrentine and Robert Israel, Table of n, a(n) for n = 1..10011 (first 141 antidiagonals, flattened; n=1..77 from Max Barrentine)

Crossrefs

First column is A005836.

First row is A265316.

Cf. A006997, A074940, A100480.

Programs

MATLAB

function  A = A262057( M, N )
% to get first M antidiagonals using x up to N
B = cell(1,M);
F = zeros(M,N+1);
countdowns = [M:-1:1];
for x=0:N
    if max(countdowns) == 0
        break
    end
    for i=1:M
        if F(i,x+1) == 0
            newforb = 2*x - B{i};
            newforb = newforb(newforb <= N & newforb >= 1);
            F(i,newforb+1) = 1;
            B{i}(end+1) = x;
            countdowns(i) = countdowns(i)-1;
            break
        end
    end
end
if max(countdowns) > 0
    [~,jmax] = max(countdowns);
    jmax = jmax(1);
    error ('Need larger N: B{%d} has only %d elements',jmax,numel(B{jmax}));
end
A = zeros(1,M*(M+1)/2);
k = 0;
for n=1:M
    for i=1:n
        k=k+1;
        A(k) = B{n+1-i}(i);
    end
end
end % Robert Israel, Feb 03 2016

Maple

M:= 20: # to get the first M antidiagonals
for i from 1 to M do B[i]:= {}: F[i]:= {}: od:
countdowns:= Vector(M,j->M+1-j):
for x from 0 while max(countdowns) > 0 do
  for i from 1 do
     if not member(x, F[i]) then
       F[i]:= F[i] union map(y -> 2*x-y, B[i]);
       B[i]:= B[i] union {x};
       countdowns[i]:= countdowns[i] - 1;
     break
    fi
  od;
od:
seq(seq(B[n+1-i][i], i=1..n),n=1..M); # Robert Israel, Feb 03 2016

Formula

A006997(A(n, k)) = k - 1. - Rémy Sigrist, Jan 06 2024