A057555 Lexicographic ordering of N x N, where N = {1,2,3...}.
1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 3, 1, 1, 4, 2, 3, 3, 2, 4, 1, 1, 5, 2, 4, 3, 3, 4, 2, 5, 1, 1, 6, 2, 5, 3, 4, 4, 3, 5, 2, 6, 1, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 6, 2, 7, 1, 1, 8, 2, 7, 3, 6, 4, 5, 5, 4, 6, 3, 7, 2, 8, 1, 1, 9, 2, 8, 3, 7, 4, 6, 5, 5, 6, 4, 7, 3, 8, 2, 9, 1, 1, 10, 2, 9, 3, 8, 4, 7, 5, 6, 6, 5, 7, 4, 8, 3, 9, 2, 10, 1, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6, 6, 7, 5, 8, 4, 9, 3, 10, 2, 11, 1, 1, 12, 2, 11, 3, 10, 4, 9, 5, 8, 6, 7, 7, 6, 8, 5, 9, 4, 10, 3, 11, 2, 12, 1
Offset: 1
Keywords
Examples
Flatten the ordered lattice points (1,1) < (1,2) < (2,1) < (1,3) < (2,2) < ... as 1,1, 1,2, 2,1, 1,3, 2,2, ...
Programs
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Mathematica
lexicographicLattice[{dim_,maxHeight_}]:= Flatten[Array[Sort@Flatten[(Permutations[#1]&)/@IntegerPartitions[#1+dim-1,{dim}],1]&,maxHeight],1]; Flatten@lexicographicLattice[{2,12}] (* Peter J. C. Moses, Feb 10 2011 *) u[x_] := Floor[3/2 + Sqrt[2*x]]; v[x_] := Floor[1/2 + Sqrt[2*x]]; n[x_] := x - v[x]*(v[x] - 1)/2; k[x_] := 1 - x + u[x]*(u[x] - 1)/2; Flatten[Table[{n[m], k[m]}, {m, 45}]] (* L. Edson Jeffery, Jun 20 2015 *) -
PARI
a(n)= if(n<1, 0, 1+(-1)^(n%2) * (binomial((n+1)%2+(sqrtint(4*n)+1)\2, 2)-n\2)) /* Michael Somos, Mar 06 2004 */
Formula
Let p(i,j) be the position of (i,j) in the ordering. Then p(i,j) = ((i+j)^2-i-3j+2)/2. Inversely, the pair (i,j) in a given position p is given by i=p-q(q-1)/2 and j=q+1-i, where q=floor((1+sqrt(8k-7))/2).
Extensions
Extended by Clark Kimberling, Feb 10 2011