

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
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*10^2911)/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
a(26) > 10^11.  Donovan Johnson, Feb 19 2013


LINKS

Table of n, a(n) for n=1..25.


EXAMPLE

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


CROSSREFS

Cf. A116019, A004023.
Sequence in context: A085305 A189818 A116032 * A060812 A212129 A172409
Adjacent sequences: A116026 A116027 A116028 * A116030 A116031 A116032


KEYWORD

nonn,base


AUTHOR

Giovanni Resta, Feb 13 2006


EXTENSIONS

More terms from Farideh Firoozbakht, Aug 24 2006
a(22)a(25) from Donovan Johnson, Feb 19 2013


STATUS

approved



