A116019 Numbers k such that sigma(k) + phi(k) is a repdigit.

1, 2, 3, 4, 10, 11, 21, 49, 186, 207, 221, 342, 406, 3324, 4443, 33324, 43375, 222221, 314000, 344032, 389924, 414806, 987652, 2222221, 190476186, 222087442, 222222221, 422720878, 2222222221, 4444444443

Offset: 1

Author

Giovanni Resta, Feb 13 2006

Keywords

• nonn
• base

Comments

(1). If m=(2*10^n-11)/9 is product of two distinct primes then m is in the sequence because phi(m)+sigma(m)=phi(p*q)+sigma(p*q)=2(p*q+1)=2m+2=4*(10^n-1)/9, so phi(m)+sigma(m) is a repdigit number. 21, 221, 222221, 2222221, 222222221,... are such terms. - Farideh Firoozbakht, Aug 17 2006

(2). If m=(4*10^n-13)/9 is product of two distinct primes then m is in the sequence because phi(m)+sigma(m)=phi(p*q)+sigma(p*q)=2(p*q+1)=2m+2=8*(10^n-1)/9, so phi(m)+sigma(m) is a repdigit number. 4443, 4444444443, 44444444443,... are such terms. - Farideh Firoozbakht, Aug 17 2006

(3). If p=(25*10^(n-1)-7)/9 is an odd prime then m=12*p is in the sequence because phi(m)+sigma(m)=32p+24=8*(10^(n+1)-1)/9 so phi(m)+sigma(m) is a repdigit number. 3324, 33324, 33333333324,... are such terms. - Farideh Firoozbakht, Aug 17 2006

(4). If n is a nonnegative integer and p=(8*10^(3n+2)-17)/27 is prime then m=14*p is in the sequence because phi(m)+sigma(m)=30p+18=8*(10^(3n+3)-1)/9 is a repdigit number. 406, 414806, 414814814814814814814806, ... are such terms of the sequence. - Farideh Firoozbakht, Aug 01 2014

Examples

			sigma(314000)+phi(314000)=888888.
		

Crossrefs

Cf. A116017, A116018, A116020.

Programs

Mathematica

Do[If[Length[Union[IntegerDigits[EulerPhi[n] + DivisorSigma[1, n]]]]==1, Print[n]], {n, 280000000}] (* Farideh Firoozbakht, Aug 17 2006 *)

PARI

for(n=1,10^7,d=digits(sigma(n)+eulerphi(n));c=0;for(i=1,#d-1,if(d[i]!=d[i+1],c++;break));if(c==0,print1(n,", "))) \\ Derek Orr, Aug 01 2014

Extensions

3 more terms from Farideh Firoozbakht, Aug 17 2006

a(28)-a(30) from Donovan Johnson, Jan 16 2012