|
| |
|
|
A120116
|
|
Numbers n such that uphi(sigma(n))=n where uphi is Unitary totient (or unitary phi) function(see A047994).
|
|
0
| |
|
|
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
(list; graph; refs; listen; history; internal format)
|
|
|
|
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).
|
|
|
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: A187579 A050102 A022432 * A199880 A063227 A190158
Adjacent sequences: A120113 A120114 A120115 * A120117 A120118 A120119
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Farideh Firoozbakht (mymontain(AT)yahoo.com), Jul 11 2006
|
| |
|
|
