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A186006 Lexicographic ordering of N x N x N x N x N, where N={1,2,3,...}.

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%I A186006 #20 Feb 09 2023 22:23:29
%S A186006 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 A186006 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 A186006 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 A186006 Lexicographic ordering of N x N x N x N x N, where N={1,2,3,...}.
%C A186006 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 A186006 G. C. Greubel, <a href="/A186006/b186006.txt">Table of n, a(n) for n = 1..1000</a>
%e A186006 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 A186006 lexicographicLattice[{dim_,maxHeight_}]:= Flatten[Array[Sort@Flatten[(Permutations[#1]&)/@IntegerPartitions[#1+dim-1,{dim}],1]&,maxHeight],1];
%t A186006 lexicographicLatticeHeightArray[{dim_,maxHeight_,axis_}]:= Array[Flatten@Position[Map[#[[axis]]&, lexicographicLattice[{dim,maxHeight}]],#]&,maxHeight]
%t A186006 Take[Flatten@lexicographicLattice[{5,12}],160]
%t A186006 (* _Peter J. C. Moses_, Feb 10 2011 *)
%Y A186006 Cf. A057555, A057557, A057559.
%K A186006 nonn
%O A186006 1,10
%A A186006 _Clark Kimberling_, Feb 10 2011