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Lexicographic ordering of N x N x N x N x N, where N={1,2,3,...}.
3

%I #20 Feb 09 2023 22:23:29

%S 1,1,1,1,1,1,1,1,1,2,1,1,1,2,1,1,1,2,1,1,1,2,1,1,1,2,1,1,1,1,1,1,1,1,

%T 3,1,1,1,2,2,1,1,1,3,1,1,1,2,1,2,1,1,2,2,1,1,1,3,1,1,1,2,1,1,2,1,2,1,

%U 2,1,1,2,2,1,1,1,3,1,1,1,2,1,1,1,2,2,1,1,2,1,2,1,2,1,1,2,2,1,1,1,3,1,1,1,1,1,1,1,1,4,1,1,1,2,3,1,1,1,3,2,1,1,1,4,1,1,1,2,1,3,1,1,2,2,2,1,1,2,3,1,1,1,3,1,2,1,1,3,2,1,1,1,4,1,1,1,2,1,1,3

%N Lexicographic ordering of N x N x N x N x N, where N={1,2,3,...}.

%C By changing a single number, the Mathematica code suffices for other dimensions: N x N (A057555), N x N x N (A057557), N x N x N x N (A057559), and higher.

%H G. C. Greubel, <a href="/A186006/b186006.txt">Table of n, a(n) for n = 1..1000</a>

%e First, list the 5-tuples in lexicographic order: (1,1,1,1,1) < (1,1,1,1,2) < (1,1,1,2,1) < (1,1,2,1,1) < ... < (1,2,2,1,1) < (1,1,3,1,1) < ... Then flatten the list, leaving 1,1,1,1,1, 1,1,1,1,2, 1,1,1,2,1, 1,1,2,1,1, ...

%t lexicographicLattice[{dim_,maxHeight_}]:= Flatten[Array[Sort@Flatten[(Permutations[#1]&)/@IntegerPartitions[#1+dim-1,{dim}],1]&,maxHeight],1];

%t lexicographicLatticeHeightArray[{dim_,maxHeight_,axis_}]:= Array[Flatten@Position[Map[#[[axis]]&, lexicographicLattice[{dim,maxHeight}]],#]&,maxHeight]

%t Take[Flatten@lexicographicLattice[{5,12}],160]

%t (* _Peter J. C. Moses_, Feb 10 2011 *)

%Y Cf. A057555, A057557, A057559.

%K nonn

%O 1,10

%A _Clark Kimberling_, Feb 10 2011