A104899 Numbers n such that sum of the proper divisors of n is equal to sigma(phi(n)).

2, 4, 8, 16, 28, 32, 64, 128, 256, 512, 1024, 2048, 3100, 4096, 8192, 16384, 32768, 65536, 118458, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296

Offset

1,1

Comments

This sequence is infinite because for all natural numbers n, 2^n is in the sequence. 28, 3100 & 118458 are in the sequence but they aren't of the form 2^n. Are there numbers n such that sum of the proper divisors of n is equal to phi(sigma(n))?

Links

Table of n, a(n) for n=1..35.

Formula

a(5)=28; a(13)=3100; a(19)=118458 & If n<5 then a(n)=2^n, if 5<n<13 then a(n)=2^(n-1), if 13<n<19 then a(n)=2^(n-2) and if 19<n<30 then a(n)=2^(n-3).

Example

m=2^n(n>0), phi(m)=2^(n-1) so sigma(phi(m))=2^n-1=(2^(n+1)-1)-2^n=
sigma(m)-m hence m is in the sequence.

Mathematica

Do[If[m + DivisorSigma[1, EulerPhi[m]] == DivisorSigma[1, m], Print [m]], {m, 70000000}]

Crossrefs

Cf. A037172, A100416.

Sequence in context: A258624 A236292 A280783 * A057975 A260881 A089055

Adjacent sequences: A104896 A104897 A104898 * A104900 A104901 A104902

Keyword

nonn

Author

Farideh Firoozbakht, Mar 31 2005

Extensions

a(30)-a(33) from Donovan Johnson, Jul 29 2009

a(34)-a(35) from Donovan Johnson, Jan 25 2013

Status

approved