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 A264846 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*. 0
 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, 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, 4, 5, 8, 15, 1, 2, 3, 4, 6, 8, 16, 1, 2, 3, 4, 5, 6, 9, 17 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. LINKS Jean-Paul Cardinal, Une suite de matrices symétriques en rapport avec la fonction de Mertens, arXiv:0807.4145 [math.NT], 2008 (in French). FORMULA T(n, 1) = 1; and last term of row, possibly T(n, A000267(n)) = n+1. EXAMPLE The first few rows of the irregular triangle are: 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; ... 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. On the other hand, the greatest members are 1, 3 and infinity (this is the option taken in the Cardinal link). MATHEMATICA row[n_] := (First /@ Split[Table[{k, Floor[n/k]}, {k, 1, n+1}], #1[[2]] == #2[[2]]&])[[All, 1]]; Array[row, 16] // Flatten (* Jean-François Alcover, Sep 26 2018 *) PROG (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; } tabf(nn) = for (n = 1, nn, vn = leastr(n); for (k=1, #vn, print1(vn[k], ", ")); print(); ); CROSSREFS Sequence in context: A243884 A334428 A126260 * A265691 A334430 A214614 Adjacent sequences: A264843 A264844 A264845 * A264847 A264848 A264849 KEYWORD nonn,tabf AUTHOR Michel Marcus, Nov 26 2015 STATUS approved

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Last modified March 30 05:09 EDT 2023. Contains 361603 sequences. (Running on oeis4.)