1,1

If 5*2^n-1 is prime (that is, n is in A001770) then m = 2^n*(5*2^n-1) is in the sequence. Because 6*phi(m)-sigma(m) = 6*2^(n-1)*(5*2^n-2) -(2^(n+1)-1)*5*2^n = 30*2^(2n-1)-6*2^n-5*2^(2n+1)+5*2^n = 5*2^(2n)-2^n = 2^n(5*2^n-1) = m. The first six terms of the sequence are of such form. Is this true for all terms of the sequence?

6*phi(76)-sigma(76)=6*36-140=76 so 76 is in the sequence.

Do[If[n==6*EulerPhi[n]-DivisorSigma[1, n], Print[n]], {n, 85000000}]

Cf. A001770, A136540.

Adjacent sequences: A136536 A136537 A136538 this_sequence A136540 A136541 A136542

Sequence in context: A136964 A136962 A137146 this_sequence A061618 A060190 A017792

more,nonn

Farideh Firoozbakht (mymontain(AT)yahoo.com), Jan 05 2008, Feb 01 2008