cp's OEIS Frontend

For every prime p sigma(p)-phi(p) is 2, so that case is trivial.

(I). If both numbers p=4*10^n+1 & q=(4*10^n-13)/9 are primes then m=p*q is in the sequence because sigma(m)-phi(m)=8*(10^(n+1)-1)/9 is a repdigit number. Conjecture: 123, 17243 & 1772443 are all such terms. - Farideh Firoozbakht, Aug 24 2006

(II). If p=(10^n-7)/3 is prime then m=2p is in the sequence because sigma(m)-phi(m)=2p+4=6*(10^n-1)/9 is a repdigit number. 62 is the smallest such terms. - Farideh Firoozbakht, Aug 24 2006

(III). If p=(4*10^n-31)/9 is prime then m=3p is in the sequence because sigma(m)-phi(m)=2p+6=8*(10^n-1)/9 is a repdigit number. 123 is the smallest such terms. - Farideh Firoozbakht, Aug 24 2006

(IV). If p=(8*10^n-17)/9 is a prime then both numbers 4p & 46p are in the sequence because sigma(4p)-phi(4p)=5p+9=4*(10^(n+1)-1)/9 & sigma(46p)-phi(46p)=50p+94=4*(10^(n+2)-1)/9 are repdigit numbers. 28 & 322 are the smallest such terms. - Farideh Firoozbakht, Aug 24 2006

(V). If p=(4*10^n-13)/9 is a prime greater than 3 then m=6p is in the sequence because sigma(m)-phi(m)=10p+14=4*(10^(n+1)-1)/9 is a repdigit number. 258 is the smallest such terms. - Farideh Firoozbakht, Aug 24 2006

(VI). If p=(8*10^(2n+1)-179)/99 is prime then m=8p is in the sequence because sigma(m)-phi(m)=11p+19=8*(10^(2n+1)-1)/9 is a repdigit number. 632 is the smallest such terms. - Farideh Firoozbakht, Aug 24 2006

(VII). If p=(10^(3n+1)-37)/27 is prime then m=12p is in the sequence because sigma(m)-phi(m)=24p+32=8*(10^(3n+1)-1)/9 is a repdigit number. 4444444428 is the smallest such terms. - Farideh Firoozbakht, Aug 24 2006

All primes of the form 11...1 are in the sequence because if p=11...1 is a prime then sigma(p)+phi(p)+p=3p=33...3 is a repdigit number, so (10^A004023-1)/9 is a subsequence of this sequence. 37 is the only multi-digit prime term of the sequence which is not of the form 11...1 - the proof is easy. Next term is greater than 2.3*10^10. - Farideh Firoozbakht, Aug 24 2006

Also we have the following two assertions. - Farideh Firoozbakht, Aug 24 2006

(a). If p=(2*10^(3n+2)-11)/27 is prime then m=2p is in the sequence because sigma(m)+phi(m)+m=6p+2=4*(10^(3n+2)-1)/9 is a repdigit number. 2*(2*10^29-11)/27 (a 29-digit number)is the smallest such terms of the sequence and the next such term(if it exists) has more than 20000 digits. - Farideh Firoozbakht, Aug 24 2006

(b). If p=(4*10^(3n+1)-13)/27 is prime then m=2p is in the sequence because sigma(m)+phi(m)+m=8*(10^(3n+1)-1)/9 is a repdigit number. 2962 is the smallest such terms of the sequence. - Farideh Firoozbakht, Aug 24 2006

a(26) > 10^11. - Donovan Johnson, Feb 19 2013