A134206 a(n) = A134205(n)/n.

5, 4, 4, 5, 6, 6, 6, 8, 8, 6, 6, 4, 6, 9, 8, 9, 8, 7, 8, 9, 10, 9, 14, 14, 6, 11, 14, 9, 14, 13, 8, 9, 8, 5, 6, 9, 18, 20, 16, 12, 14, 15, 8, 12, 12, 11, 16, 15, 12, 15, 14, 9, 8, 11, 20, 19, 12, 13, 8, 5, 12, 11, 6, 12, 12, 8, 12, 13, 12, 10, 18, 19, 16, 17, 14, 12, 12, 14, 22, 21, 10, 9, 6, 8

Offset: 1

Author

Leroy Quet, Oct 14 2007

Keywords

• nonn

Comments

A134205(n) is divisible by n for every n, by definition of A134204. But it is unknown whether A134204(n), and therefore also A134205 and A134206, are defined for all n>0.

Conjecture: Apart from a(2)=a(3)=a(12)=4 and a(1)=a(4)=a(34)=a(60)=5, all a(n) exceed 5. - M. F. Hasler, Feb 12 2013. Reply from David Applegate, Dec 11 2013: The conjecture is false: A134206(73397) = A134206(213138) = A134206(790306) = 2.(those are the only 2's for n <= 10^6).

Links

• M. F. Hasler and Michael De Vlieger, Table of n, a(n) for n = 1..10000 (First 1000 terms from M. F. Hasler).

Crossrefs

Cf. A134204, A134205.

Programs

Mathematica

With[{nn = 84}, MapIndexed[Total[#1]/First@ #2 &, Partition[#, 2, 1]] &@ Fold[Append[#1, SelectFirst[Prime@ Range[2, Ceiling@ Log2[nn] nn], Function[p, And[FreeQ[#1, p], Divisible[Last@ #1 + p, #2]]]]] &, {2}, Range@ nn]] (* Michael De Vlieger, Oct 16 2017 *)

Extensions

More terms from Robert Israel, Oct 14 2007