OFFSET
1,2
COMMENTS
(I). If p=(2*10^(3n+1)+7)/27 is prime then m=2p is in the sequence because m+phi(m)=3p-1=2*(10^(3n+1)-1)/9 is a repdigit number. m=2*(2*10^811+7)/27 (a 811-digit number) is the smallest such terms and the next such terms has 4219 digits. - Farideh Firoozbakht, Aug 24 2006
(II). If p=(8*10^(3n+1)+1)/27 is prime then m=2p is in the sequence because m+phi(m)=8*(10^(3n+1)-1)/9 is a repdigit number. 5926 is the smallest such terms. - Farideh Firoozbakht, Aug 24 2006
(III). If p=(2*10^n+1)/3 then both numbers 3p & 9p are in the sequence because 3p+phi(3p)=5p-2=3*(10^(n+1)-1)/9 & 9p+ phi(9p)=9*(10^(n+1)-1)/9 are repdigit numbers. 21 & 63 are the smallest such terms. - Farideh Firoozbakht, Aug 24 2006
(IV). All primes p of the form (35*10^n+1)/9 are in the sequence because p+phi(p)=7*(10^n-1)/9 is a repdigit number. 389 is the smallest such terms. - Farideh Firoozbakht, Aug 24 2006
(V). All primes p of the form (10^n+2)/6 are in the sequence because p+phi(p)=2p-1=3*(10^n-1)/9 is a repdigit number. 2, 17 & 167 are such terms. - Farideh Firoozbakht, Aug 24 2006, Dec 19 2007
LINKS
Max Alekseyev, Table of n, a(n) for n = 1..98 (terms 1..48 terms from Hiroaki Yamanouchi, terms 49..61 from Giovanni Resta)
EXAMPLE
5185194 + phi(5185194) = 6666666.
PROG
(Python)
from sympy import totient
A116018 = [n for n in range(1, 10**6) if len(set(str(n+totient(n)))) == 1] # Chai Wah Wu, Aug 11 2014
(PARI)
for(n=1, 10^9, d=digits(n+eulerphi(n)); if(vecmin(d)==vecmax(d), print1(n, ", "))) \\ Derek Orr, Aug 11 2014
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Giovanni Resta, Feb 13 2006
EXTENSIONS
More terms from Farideh Firoozbakht, Aug 24 2006
a(35)-a(36) from Donovan Johnson, Feb 19 2013
STATUS
approved