A116029 Numbers n such that n + sigma(n) + phi(n) is a repdigit.

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

Offset: 1

Author

Giovanni Resta, Feb 13 2006

Keywords

• nonn
• base

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^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

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*10^29-11)/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

(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

Examples

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

Crossrefs

Cf. A116019, A004023.

Extensions

More terms from Farideh Firoozbakht, Aug 24 2006

a(22)-a(25) from Donovan Johnson, Feb 19 2013