A335261 Irregular triangle S(n,k) = denominators of k*A002110(n)/A005867(n) for 1 <= k <= A005867(n).
1, 1, 1, 1, 4, 2, 4, 1, 4, 2, 4, 1, 8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8, 1, 16, 8, 16, 4, 16, 8, 16, 2, 16, 8, 16, 4, 16, 8, 16, 1, 16, 8, 16, 4, 16, 8, 16
Offset: 1
Examples
Table begins:
1;
1;
1, 1;
4, 2, 4, 1, 4, 2, 4, 1;
8, 4, 8, 2, 8, 4, 8, 1, ..., 8, 1;
...
Row n = 4 contains the denominators of (35/8)*k for 1 <= k <= A005867(4): 35/8, 35/4, 105/8, 35/2, 175/8, 105/4, 245/8, 35, 315/8, 175/4, 385/8, 105/2, 455/8, 245/4, 525/8, 70, 595/8, 315/4, 665/8, 175/2, 735/8, 385/4, 805/8, 105, 875/8, 455/4, 945/8, 245/2, 1015/8, 525/4, 1085/8, 140, 1155/8, 595/4, 1225/8, 315/2, 1295/8, 665/4, 1365/8, 175, 1435/8, 735/4, 1505/8, 385/2, 1575/8, 805/4, 1645/8, 210.
The mean of row 4: (A060753(4+1)/(A038110(4+1)*2))*(A005867(4)+1) = (35/(8*2))*(48+1) = (35/16)*49 = 1715/16.
Programs
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Mathematica
Table[Denominator[P Range[EulerPhi[P]]/EulerPhi[P]], {P, FoldList[Times, Prime@ Range@ 5]}] (* or, more efficiently for larger datasets: *) Flatten@ Block[{nn = 7, s, t}, s = Array[Numerator@ Product[1 - 1/Prime[k], {k, # - 1}] &, nn]; t = Nest[Append[#, #[[-1]] (Prime[Length@ #] - 1)]&, {1}, nn]; MapIndexed[Function[{m, D, i}, PadRight[{}, t[[i]], ReplacePart[ConstantArray[0, m], Flatten@ Map[Function[d, Map[# -> d &, m/d Select[Range[d], GCD[#, d] == 1 &]]], D]]]] @@ {#1, Divisors@ #1, First[#2]} &, s]] (* or, to generate a single denominator of T(n,k) *) f[n_, k_] := #/GCD[#, Mod[k, #]] &@ Numerator@ Product[1 - 1/Prime[i], {i, n - 1}]
Comments