

A116083


Numbers n such that phi(sigma(n))sigma(phi(n))=1.


0




OFFSET

1,1


COMMENTS

If k is a natural number less than 6 then 2^(2^k1) is in the sequence because phi(sigma(2^(2^k1)) = phi(2^(2^k)1) = phi((2^(2^0)+1)*(2^(2^1)+1)*...*(2^(2^(k1))+1)) = 2^(2^0)*2^(2^1)*...* 2^(2^(k1)) = 2^(2^0+2^1+...+2^(k1)) = 2^(2^k1) and sigma(phi(2^(2^k1))) = sigma(2^(2^k2)) = 2^(2^k1)1 so phi(sigma(2^(2^k1))) sigma(phi(2^(2^k1))) = 1(note that for i = 0,1,2,3 & 4 the Fermat number 2^2^i+1 is prime). Next term is greater than 7*10^8.
Also if n is a natural number less than 6 then 3*2^(2^n1) is in the sequence, the proof is similar to the case m=2^(2^n1). Note that all known terms of the sequence are the ten numbers m*2^(2^n1) m=1 & 3 and n=1,2,3,4 & 5. Conjecture: There is no other term.  Farideh Firoozbakht, Mar 24 2006
There are no more terms up to 11*10^9. [From Farideh Firoozbakht, Apr 13 2010]


LINKS

Table of n, a(n) for n=1..10.


FORMULA

For n<11, a(n)=((1)^n+2)*2^(2^Floor[(n+1)/2]1).  Farideh Firoozbakht, Mar 24 2006


EXAMPLE

phi(sigma(2147483648)) = 2147483648 and sigma(phi(2147483648)) = 2147483647 so phi(sigma(2147483648))sigma(phi(2147483648)) = 1. Hence 2147483648 is in the sequence.


MATHEMATICA

Do[If[EulerPhi[DivisorSigma[1, n]]DivisorSigma[1, EulerPhi[n]]==1, Print[n]], {n, 700000000}]
Table[((1)^n + 2)*2^(2^Floor[(n + 1)/2]  1), {n, 10}]  Farideh Firoozbakht, Mar 24 2006


PROG

(PARI) is(n)=eulerphi(sigma(n))sigma(eulerphi(n))==1 \\ Charles R Greathouse IV, May 15 2013


CROSSREFS

Cf. A001229.
Sequence in context: A193946 A189850 A189358 * A115506 A243552 A057852
Adjacent sequences: A116080 A116081 A116082 * A116084 A116085 A116086


KEYWORD

nonn


AUTHOR

Farideh Firoozbakht, Mar 12 2006


STATUS

approved



