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A066679
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Numbers n such that sigma(n) is congruent to n mod phi(n).
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2
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1, 2, 6, 10, 12, 44, 90, 184, 440, 528, 588, 672, 752, 3796, 8928, 9888, 12224, 35640, 37680, 49024, 50976, 89152, 94200, 108192, 146412, 159840, 279864, 1734720, 2554368, 2977920, 12580864, 14239872, 16544880, 28321920, 41362200, 56976480, 60610624
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OFFSET
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1,2
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COMMENTS
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If p=3*2^n-1 is an odd prime then m=2^n*p is in the sequence. Proof: sigma(m)-m=(2^(n+1)-1)*(p+1)-2^n*p=2*(2^(n-1)*(p-1))= 2*phi(m), so sigma(m)=m mod(phi(m)). Hence for n>0, 2^A002235(n)* (3*2^A002235(n)-1) is in the sequence and 2^164987*(3*2^164987-1) is the largest known term of the sequence. - Farideh Firoozbakht, Apr 15 2006
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LINKS
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EXAMPLE
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sigma(10) = 18 is congruent to 10 mod phi(10) = 4, so 10 is a term of the sequence.
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MATHEMATICA
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Select[ Range[ 1, 10^5 ], Mod[ DivisorSigma[ 1, # ], EulerPhi[ # ] ] == Mod[ #, EulerPhi[ # ] ] & ]
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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