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A319385 Assuming the truth of the Collatz conjecture, let {m, f(m), f(f(m)), ..., 1} be the set where f is the Collatz function. The sequence lists the numbers m such that m/phi(m) + f(m)/phi(f(m)) + f(f(m))/phi(f(f(m))) + ... + 1/phi(1) is an integer, where phi is the Euler totient function A000010. 0

%I #28 Jan 10 2024 16:19:52

%S 1,2,4,8,16,26,32,64,128,256,512,1024,1664,2048,3392,4096,8192,16384,

%T 32768,65536,106496,131072,262144,524288,1048576,2097152,4194304,

%U 6815744,8388608,16777216,27918336,33554432,67108864,134217728,268435456,436207616,536870912

%N Assuming the truth of the Collatz conjecture, let {m, f(m), f(f(m)), ..., 1} be the set where f is the Collatz function. The sequence lists the numbers m such that m/phi(m) + f(m)/phi(f(m)) + f(f(m))/phi(f(f(m))) + ... + 1/phi(1) is an integer, where phi is the Euler totient function A000010.

%C The corresponding integers are 1, 3, 5, 7, 9, 21, 11, 13, 15, 17, 19, 21, 34, 23, 36, 25, 27, 29, 31, 33, 47, 35, 37, 39, 41, 43, 45, 60, 47, 49, 81, 51, 53, 55, 57, 73, 59, 61, 63, ... Conjecturally, it seems that all odd numbers are present, and the even numbers are rare: 34, 36, 60, ...

%C We observe that the non-powers of 2 of the sequence: 26, 1664, 3392, 106496, 6815744, 27918336, 436207616, ... are of the form q*2^k with q in the set {13, 53, 213, ...}.

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

%e 26 is in the sequence because the Collatz trajectory is 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1, and 26/phi(26) + 13/phi(13) + 40/phi(40) + 20/phi(20) + 10/phi(10) + 5/phi(5) + 16/phi(16) + 8/phi(8) + 4/phi(4) + 2/phi(2) + 1/phi(1) = 26/12 + 13/12 + 40/16 + 20/8 + 10/4 + 5/4 + 16/8 + 8/4 + 4/2 + 2/1 + 1/1 = 21 is an integer.

%p with(numtheory):nn:=10^6:

%p for n from 1 to 100000 do:

%p T:=array(1..1000,[0$1000]):

%p it:=0:m:=n:k:=0:

%p for i from 1 to nn while(m<>1) do:

%p if irem(m, 2)=0

%p then

%p k:=k+1:T[k]:=m:m:=m/2:

%p else

%p k:=k+1:T[k]:=m:m:=3*m+1:

%p fi:

%p od:

%p k:=k+1:T[k]:=1:

%p s:=sum(ā€˜T[i]/phi(T[i])ā€™, ā€˜iā€™=1..k):

%p if s=floor(s)

%p then

%p printf (`%d %d \n`,n,s):

%p else fi:

%p od:

%Y Cf. A000010, A000079, A006370, A070165.

%K nonn

%O 1,2

%A _Michel Lagneau_, Sep 18 2018

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