%I #38 Jul 30 2022 08:18:51

%S 1,2,3,4,6,8,12,15,16,20,24,30,32,40,48,60,64,80,96,120,128,160,192,

%T 240,255,256,272,320,340,384,408,480,510,512,544,640,680,768,816,960,

%U 1020,1024,1088,1280,1360,1536,1632,1920,2040,2048,2176,2560,2720,3072

%N Let d(1), d(2), ..., d(q) be the q divisors of a number m. The sequence lists the numbers m such that {phi(d(1)), phi(d(2)), ..., phi(d(q))} is the set of the divisors of a number k, where phi is the Euler totient function.

%C The sequence is infinite because the powers of 2 are in the sequence.

%C Conjecture: the corresponding numbers k of the sequence is a sequence b(n) of powers of 2.

%C The sequence b(n) begins with 1, 1, 2, 2, 2, 4, 4, 8, 8, 8, 8, 8, 16, 16, 16, 16, 32, 32, 32, 32, 64, 64, 64, 64, 128, 128, 128, 128, 128, 128, 128, 128, 128, 256, 256, 256, 256, 256, 256, 256, 256, 512, 512, 512, 512, 512, 512, 512, 512, 1024, 1024, 1024, 1024, ...

%C The number of terms of the form 2^i is given by the sequence c(i,2^i) = 2, 3, 2, 5, 4, 4, 4, 9, 8, 8, 8, 8, 8, 8, 8, 17, ... for i = 0, 1, 2, ...

%C Property of the prime factors of a(n) for n > 1:

%C We observe periodic sequences of prime factors of the form 2^m + 1.

%C +---------------------------------------+------------------------------+

%C | subsequences of consecutive | corresponding prime factors |

%C | values of a(n) | |

%C +---------------------------------------+------------------------------+

%C |4, 6 |{2},{2, 3} |

%C |8, 12 |{2},{2, 3} |

%C |16, 20, 24, 30 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|

%C |32, 40, 48, 60 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|

%C |64, 80, 96, 120 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|

%C |128, 160, 192, 240 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|

%C |256, 272, 320, 340, 384, 408, 480, 510 |{2}, {2, 17},...,{2, 3, 5, 17}|

%C |512, 544, 640, 680, 768, 816, 960, 1020|{2}, {2, 17},...,{2, 3, 5, 17}|

%C |1024, 1088, 1280, 1360, 1536, ..., 2040|{2}, {2, 17},...,{2, 3, 5, 17}|

%C |2048, 2176, 2560, 2720, 3072, ..., 4080|{2}, {2, 17},...,{2, 3, 5, 17}|

%C |4096, 4352, 5120, 5440, 6144, ..., 8160|{2}, {2, 17},...,{2, 3, 5, 17}|

%C ............................................................

%H Amiram Eldar, <a href="/A309353/b309353.txt">Table of n, a(n) for n = 1..300</a>

%e 272 is in the sequence because the divisors are {1, 2, 4, 8, 16, 17, 34, 68, 136, 272}, and {phi(1), phi(2), phi(4), phi(8), phi(16), phi(17), phi(34), phi(68), phi(136), phi(272)} = {1, 2, 4, 8, 16, 32, 64, 128} is the set of the divisors of 128.

%p with(numtheory):nn:=3000:

%p for n from 1 to nn do:

%p d:=divisors(n):n0:=nops(d):A:={op(d),d[n0]}:lst:={}:

%p for k from 1 to n0 do:

%p lst:=lst union {phi(d[k])}:

%p od:

%p n1:=nops(lst):B:={op(lst),lst[n1]}:

%p d1:=divisors(lst[n1]):n2:=nops(d1):C:={op(d1),d1[n2]}:

%p if B=C

%p then

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

%p else

%p fi:

%p od:

%t Select[Range[3100], (d = Union @ EulerPhi @ Divisors @ #) == Divisors[LCM @@ d] &] (* _Amiram Eldar_, Aug 02 2019 *)

%Y Cf. A000010, A000079 (powers of 2).

%K nonn

%O 1,2

%A _Michel Lagneau_, Aug 02 2019