

A116029


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


0



1, 2, 3, 11, 12, 14, 16, 32, 37, 216, 325, 1851, 2962, 6836, 18125, 178569, 3652175, 7404814, 10599021, 196259690, 370355439, 2962962962, 7355242534, 7404880354, 31328532351
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OFFSET

1,2


COMMENTS

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^A0040231)/9 is a subsequence of this sequence. 37 is the only multidigit 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
(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*10^2911)/27 (a 29digit 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


LINKS



EXAMPLE

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


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



