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Search: id:A111303
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| A111303 |
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Numbers n such that 2^tau(n) = n + 1 (where tau(n) = number of divisors of n). |
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+0
1
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OFFSET
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1,2
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COMMENT
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It is clear that n+1 must be a power of 2. Hence n=2^k-1 for some k. Found k=1, 2, 4, 6, 8, 16, 32. No other k<150. - T. D. Noe (noe(AT)sspectra.com), Nov 04 2005
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FORMULA
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Note that this is different from the sequence A019434(n)-2.
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MATHEMATICA
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Select[Range[10^6], 2^DivisorSigma[0, # ] == # + 1 &]
2^Select[Range[150], DivisorSigma[0, 2^#-1]==#&] - 1 (Noe)
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CROSSREFS
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Cf. A046801 (number of divisors of 2^n-1).
Sequence in context: A062211 A024036 A103454 this_sequence A118339 A083858 A151241
Adjacent sequences: A111300 A111301 A111302 this_sequence A111304 A111305 A111306
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KEYWORD
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nonn
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AUTHOR
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Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Nov 02 2005
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EXTENSIONS
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One more term from T. D. Noe (noe(AT)sspectra.com), Nov 04 2005
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