1,1

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 (mymontain(AT)yahoo.com), Mar 24 2006

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

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

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 (mymontain(AT)yahoo.com), Mar 24 2006

Cf. A001229.

Sequence in context: A081957 A122758 A122756 this_sequence A115506 A057852 A140539

Adjacent sequences: A116080 A116081 A116082 this_sequence A116084 A116085 A116086

nonn

Farideh Firoozbakht (mymontain(AT)yahoo.com), Mar 12 2006