cp's OEIS Frontend

n is necessarily composite.

From the Math. Rev.: if both q=7*2^{r-2}+2^s-1 and p=49*2^{2r-s-4}+7*2^{r-2}-5*2^{r-s-2}-1 are prime for 1 <= s < r-2, then n=2^r*3*p*q is a solution to the equation phi(n)+sigma(n)=3n . [R. K. Guy]

If 7*2^n-1 is prime then m = 2^(n+2)*3*(7*2^n-1) is in the sequence. Because phi(m) = 2^(n+2)*(7*2^n-2); sigma(m) = 7*2^(n+2)*(2^(n+3)-1) so phi(m)+sigma(m) = 2^(n+2)*((7*2^n-2)+(7*2^(n+3)-7)) = 2^(n+2)*(63*2^n-9) = 3*(2^(n+2)*3*(7*2^n-1)) = 3*m. Hence A112729 = 2^(A001771+2)*3*(7*2^A001771-1) is a subsequence of this sequence. - Farideh Firoozbakht, Dec 01 2005

If both numbers p=7*2^n+2^k-1 & q=49*2^(2n-k)+2^(n-k)*(7*2^k-5)-1 are prime and m=2^(n+2)*3*p*q then phi(m)+sigma(m)=3*m. Namely m is in the sequence. - Farideh Firoozbakht, Jan 11 2007

If (sigma(m)-phi(m))/(4*m-sigma(m)-phi(m)) is a prime integer p not dividing m, then p*m is in the sequence. 135230346701100 is in the sequence and not divisible by 24. - Jens Kruse Andersen, Feb 17 2009

If k=80*m is in the sequence and gcd(m,10) = 1 then 200*m is also in the sequence. Proof: phi(200*m) + sigma(200*m) = phi(200)*phi(m) + sigma(200)*sigma(m) = 80*phi(m) + 465*sigma(k) = (5/2)*(32*phi(m) + 186*sigma(m)) = (5/2)*(phi(80)*phi(m) + sigma(80)*sigma(m)) = (5/2)*(phi(80*m) + sigma(80*m)) = (5/2)*(phi(k) + sigma(k)) = (5/2)*(4*k) = 5/2*(4*80*m) = 4*(200*m) so 200*m is in the sequence. - Farideh Firoozbakht, Mar 30 2009

a(47) is greater than 4*10^8.