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%I #20 Sep 26 2018 15:47:07
%S 1,2,1,2,3,1,2,4,1,2,3,5,1,2,3,6,1,2,3,4,7,1,2,3,4,8,1,2,3,5,9,1,2,3,
%T 4,5,10,1,2,3,4,6,11,1,2,3,4,6,12,1,2,3,4,5,7,13,1,2,3,4,5,7,14,1,2,3,
%U 4,5,8,15,1,2,3,4,6,8,16,1,2,3,4,5,6,9,17
%N Irregular triangle read by rows where row n lists the least members of the equivalence classes of the equivalence relation xRy defined by floor(n/x) = floor(n/y) with n, x and y in N*.
%C Row n has 1 + floor(sqrt(n)) + floor(sqrt(n+1/4)- 1/2) terms (see Remark 2.5 in Cardinal link). It appears that this is A000267.
%H Jean-Paul Cardinal, <a href="http://arxiv.org/abs/0807.4145">Une suite de matrices symétriques en rapport avec la fonction de Mertens</a>, arXiv:0807.4145 [math.NT], 2008 (in French).
%F T(n, 1) = 1; and last term of row, possibly T(n, A000267(n)) = n+1.
%e The first few rows of the irregular triangle are:
%e 1, 2;
%e 1, 2, 3;
%e 1, 2, 4;
%e 1, 2, 3, 5;
%e 1, 2, 3, 6;
%e 1, 2, 3, 4, 7;
%e 1, 2, 3, 4, 8;
%e 1, 2, 3, 5, 9;
%e ...
%e For n=3, floor(3/k) is 3, 1, 1, 0, 0, 0. So the 3 equivalence classes are {1}, {2, 3} and {4, 5, 6, ...} and the least members 1, 2 and 4.
%e On the other hand, the greatest members are 1, 3 and infinity (this is the option taken in the Cardinal link).
%t row[n_] := (First /@ Split[Table[{k, Floor[n/k]}, {k, 1, n+1}], #1[[2]] == #2[[2]]&])[[All, 1]];
%t Array[row, 16] // Flatten (* _Jean-François Alcover_, Sep 26 2018 *)
%o (PARI) leastr(n) = {my(vn = []); my(vr = []); for (k=1, n+1, r = n\k; if (! vecsearch(vr, r), vr = vecsort(concat(vr, r),, 8); vn = concat(vn, k));); vn;}
%o tabf(nn) = for (n = 1, nn, vn = leastr(n); for (k=1, #vn, print1(vn[k], ", ")); print(););
%K nonn,tabf
%O 1,2
%A _Michel Marcus_, Nov 26 2015
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