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A116017
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Numbers n such that n + sigma(n) is a repdigit.
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7
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1, 2, 3, 4, 5, 9, 34, 141, 198, 277, 297, 375, 499, 1420, 2651, 2777, 3554, 4999, 19050, 28660, 29128, 49999, 131061, 506311, 3844863, 3852517, 4761903, 4999999, 22222218, 37560831, 133878933, 506767303, 872011214, 1381799253, 1427435733, 2777777777, 3018915632, 3555555554
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OFFSET
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1,2
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COMMENTS
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(1) If p=(10^(3n+2)-19)/27 is a prime greater than 3 then m=6p is in the sequence because m+sigma(m)=6*(10^(3n+2)-1)/9 (the proof is easy), so m+sigma(m) is a repdigit number. The smallest such terms is 22222218, the next such term is 6*(10^(3*430+2)-1)/9=222...218 which has 1292 digits.
(2) If p=5*10^n-1 is prime then p is in the sequence because p+sigma(p)=10^(n+1)-1, so p+sigma(p) is a repdigit number. 499, 49999, 4999999,... are such terms.
(3) If p=(25*10^(n-1)-7)/9 is prime then p is in the sequence because p+sigma(p)=5*(10^n-1)/9, so p+sigma(p) is a repdigit number. 2, 277, 2777, 2777777777, ... are such terms.
(4) If p=(16*10^(n-1)-7)/9 is prime then m=2p is in the sequence because m+sigma(m)=8*(10^n-1) /9, so m+sigma(m) is a repdigit number. 34, 3554, 3555555554, ... are such terms. (End)
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LINKS
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EXAMPLE
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22222218 + sigma(22222218) = 66666666.
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MATHEMATICA
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Do[If[Length[Union[IntegerDigits[n + DivisorSigma[1, n]]]]==1, Print[n]], {n, 60000000}] (* Farideh Firoozbakht, Aug 17 2006 *)
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PROG
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(PARI)
for(n=1, 10^7, d=digits(sigma(n)+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
(Python)
from sympy import divisors
A116017 = [n for n in range(1, 10**5) if len(set(str(n+sum(divisors(n))))) == 1] # Chai Wah Wu, Aug 11 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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