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a(n) = A134205(n)/n.
3

%I #24 Oct 17 2017 09:28:19

%S 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,

%T 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,

%U 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

%N a(n) = A134205(n)/n.

%C 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.

%C 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).

%H M. F. Hasler and Michael De Vlieger, <a href="/A134206/b134206.txt">Table of n, a(n) for n = 1..10000</a> (First 1000 terms from M. F. Hasler).

%t 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 *)

%Y Cf. A134204, A134205.

%K nonn

%O 1,1

%A _Leroy Quet_, Oct 14 2007

%E More terms from _Robert Israel_, Oct 14 2007