OFFSET
1,2
COMMENTS
The sequence is infinite because the powers of 2 are in the sequence.
Conjecture: the corresponding numbers k of the sequence is a sequence b(n) of powers of 2.
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, ...
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, ...
Property of the prime factors of a(n) for n > 1:
We observe periodic sequences of prime factors of the form 2^m + 1.
+---------------------------------------+------------------------------+
| subsequences of consecutive | corresponding prime factors |
| values of a(n) | |
+---------------------------------------+------------------------------+
|4, 6 |{2},{2, 3} |
|8, 12 |{2},{2, 3} |
|16, 20, 24, 30 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|
|32, 40, 48, 60 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|
|64, 80, 96, 120 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|
|128, 160, 192, 240 |{2}, {2, 5}, {2, 3}, {2, 3, 5}|
|256, 272, 320, 340, 384, 408, 480, 510 |{2}, {2, 17},...,{2, 3, 5, 17}|
|512, 544, 640, 680, 768, 816, 960, 1020|{2}, {2, 17},...,{2, 3, 5, 17}|
|1024, 1088, 1280, 1360, 1536, ..., 2040|{2}, {2, 17},...,{2, 3, 5, 17}|
|2048, 2176, 2560, 2720, 3072, ..., 4080|{2}, {2, 17},...,{2, 3, 5, 17}|
|4096, 4352, 5120, 5440, 6144, ..., 8160|{2}, {2, 17},...,{2, 3, 5, 17}|
............................................................
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..300
EXAMPLE
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.
MAPLE
with(numtheory):nn:=3000:
for n from 1 to nn do:
d:=divisors(n):n0:=nops(d):A:={op(d), d[n0]}:lst:={}:
for k from 1 to n0 do:
lst:=lst union {phi(d[k])}:
od:
n1:=nops(lst):B:={op(lst), lst[n1]}:
d1:=divisors(lst[n1]):n2:=nops(d1):C:={op(d1), d1[n2]}:
if B=C
then
printf(`%d, `, n):
else
fi:
od:
MATHEMATICA
Select[Range[3100], (d = Union @ EulerPhi @ Divisors @ #) == Divisors[LCM @@ d] &] (* Amiram Eldar, Aug 02 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Aug 02 2019
STATUS
approved