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A116019
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Numbers n such that sigma(n) + phi(n) is a repdigit.
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2
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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
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OFFSET
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1,2
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COMMENTS
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(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
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LINKS
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EXAMPLE
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sigma(314000)+phi(314000)=888888.
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MATHEMATICA
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Do[If[Length[Union[IntegerDigits[EulerPhi[n] + DivisorSigma[1, n]]]]==1, Print[n]], {n, 280000000}] - Farideh Firoozbakht, Aug 17 2006
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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