|
|
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^k-1) is in the sequence because phi(sigma(2^(2^k-1)) = phi(2^(2^k)-1) = phi((2^(2^0)+1)*(2^(2^1)+1)*...*(2^(2^(k-1))+1)) = 2^(2^0)*2^(2^1)*...* 2^(2^(k-1)) = 2^(2^0+2^1+...+2^(k-1)) = 2^(2^k-1) and sigma(phi(2^(2^k-1))) = sigma(2^(2^k-2)) = 2^(2^k-1)-1 so phi(sigma(2^(2^k-1))) -sigma(phi(2^(2^k-1))) = 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^n-1) is in the sequence, the proof is similar to the case m=2^(2^n-1). Note that all known terms of the sequence are the ten numbers m*2^(2^n-1) m=1 & 3 and n=1,2,3,4 & 5. Conjecture: There is no other term. - Farideh Firoozbakht, Mar 24 2006
|
|
LINKS
|
|
|
FORMULA
|
|
|
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}]
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|