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%I #26 Jan 26 2024 02:30:02
%S 1,2,3,4,5,6,17,21,63,167,201,389,603,1667,3795,3889,4465,5926,50394,
%T 166667,510042,2000001,3888889,5185194,5798663,5925926,6000003,
%U 32050435,200000001,335447667,365110755,444766346,600000003,1558138862,1565408702,1587424430
%N Numbers n such that n + phi(n) is a repdigit.
%C (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
%C (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
%C (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
%C (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
%C (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
%H Max Alekseyev, <a href="/A116018/b116018.txt">Table of n, a(n) for n = 1..98</a> (terms 1..48 terms from Hiroaki Yamanouchi, terms 49..61 from Giovanni Resta)
%e 5185194 + phi(5185194) = 6666666.
%o (Python)
%o from sympy import totient
%o A116018 = [n for n in range(1,10**6) if len(set(str(n+totient(n)))) == 1] # _Chai Wah Wu_, Aug 11 2014
%o (PARI)
%o for(n=1,10^9,d=digits(n+eulerphi(n));if(vecmin(d)==vecmax(d),print1(n,", "))) \\ _Derek Orr_, Aug 11 2014
%Y Cf. A116017, A096503, A309835.
%K nonn,base
%O 1,2
%A _Giovanni Resta_, Feb 13 2006
%E More terms from _Farideh Firoozbakht_, Aug 24 2006
%E a(35)-a(36) from _Donovan Johnson_, Feb 19 2013
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