

A120116


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


1



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
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OFFSET

1,2


COMMENTS

Theorem: If 2^p1 is prime (a Mersenne prime) and 0<=k<=5 then m=2^p*(2^2^k1) is in the sequence. Proof: If k=0 then m=2^p and sigma(uphi(m))=sigma(uphi(2^p))=sigma(2^p1)=2^p=m. Now if k>0 then m=2^p*(2^2^0+1)*(2^2^1+1)*...*(2^2^(k1)+1) since 2^2^i+1 for i=0,1,...,4 are primes(Fermat primes) we conclude that uphi(m)=(2^p1)*(2^2^0)*(2^2^1)*...*(2^2^(k1))=(2^p1)*2^ (2^0+2^1+...+2^(k1))=(2^p1)*2^(2^k1) hence sigma(uphi(m))= sigma(2^p1)*sigma(2^(2^k1))=2^p*(2^2^k1)=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^k1).


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((21)*(31)*(71)*(191)*(1271)*(10931) = sigma(2*6*18*126*1092) = sigma(2^6*3^6*7^2*13) = (2^71)*((3^71)/2)*((7^31)/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



