A116029 - OEIS

%I #11 Feb 19 2013 04:25:43

%S 1,2,3,11,12,14,16,32,37,216,325,1851,2962,6836,18125,178569,3652175,

%T 7404814,10599021,196259690,370355439,2962962962,7355242534,

%U 7404880354,31328532351

%N Numbers n such that n + sigma(n) + phi(n) is a repdigit.

%C 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

%C Also we have the following two assertions. - _Farideh Firoozbakht_, Aug 24 2006

%C (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

%C (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

%C a(26) > 10^11. - _Donovan Johnson_, Feb 19 2013

%e 3652175 + sigma(3652175) + phi(3652175) = 11111111.

%Y Cf. A116019, A004023.

%K nonn,base

%O 1,2

%A _Giovanni Resta_, Feb 13 2006

%E More terms from _Farideh Firoozbakht_, Aug 24 2006

%E a(22)-a(25) from _Donovan Johnson_, Feb 19 2013