A120116 Numbers n such that sigma(uphi(n)) = n where uphi is the unitary totient (or unitary phi) function (see A047994).

1, 3, 4, 8, 12, 15, 24, 32, 60, 96, 120, 128, 255, 384, 480, 1020, 1920, 2040, 2418, 8160, 8192, 24576, 32640, 65535, 122880, 131072, 262140, 370986, 393216, 524280, 524288, 1572864, 1966080, 2088960, 2097120, 7864320, 8388480, 33423360, 110771178, 133693440

Offset

1,2

Comments

Theorem: If 2^p-1 is prime (a Mersenne prime) and 0<=k<=5 then m=2^p*(2^2^k-1) is in the sequence. Proof: If k=0 then m=2^p and sigma(uphi(m))=sigma(uphi(2^p))=sigma(2^p-1)=2^p=m. Now if k>0 then m=2^p*(2^2^0+1)*(2^2^1+1)*...*(2^2^(k-1)+1) since 2^2^i+1 for i=0,1,...,4 are primes(Fermat primes) we conclude that uphi(m)=(2^p-1)*(2^2^0)*(2^2^1)*...*(2^2^(k-1))=(2^p-1)*2^ (2^0+2^1+...+2^(k-1))=(2^p-1)*2^(2^k-1) hence sigma(uphi(m))= sigma(2^p-1)*sigma(2^(2^k-1))=2^p*(2^2^k-1)=m and the proof is complete. 2418, 370986 & 110771178 are the only terms up to 15*10^8 which aren't of the form 2^p*(2^2^k-1).

Links

Donovan Johnson, Table of n, a(n) for n = 1..50 (terms < 2^35)

Example

110771178 = 2*3*7*19*127*1093 is in the sequence because sigma(uphi(2*3*7*19*127*1093)) = sigma((2-1)*(3-1)*(7-1)*(19-1)*(127-1)*(1093-1) = sigma(2*6*18*126*1092) = sigma(2^6*3^6*7^2*13) = (2^7-1)*((3^7-1)/2)*((7^3-1)/6)*14 = 110771178.

Mathematica

uphi[n_] := (A = FactorInteger[n]; l = Length[A]; Product[A[[k]][[1]] ^A[[k]][[2]] - 1, {k, l}]); Do[If[DivisorSigma[1, uphi[n]] == n, Print[n]], {n, 150000000}]

Crossrefs

Cf. A047994, A019434.

Sequence in context: A050102 A022432 A271474 * A199880 A063227 A293462

Adjacent sequences: A120113 A120114 A120115 * A120117 A120118 A120119

Keyword

nonn

Author

Farideh Firoozbakht, Jul 11 2006

Extensions

Name corrected by Donovan Johnson, May 01 2013

Status

approved