A116020 -id:A116020 - cp's OEIS frontend

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

Giovanni Resta, Feb 13 2006

(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

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

Cf. A116017, A116018, A116020.

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

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

3 more terms from Farideh Firoozbakht, Aug 17 2006

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

cp's OEIS Frontend

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

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