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Search: id:A071864
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| A071864 |
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Composite n such that the number of elements in the continued fraction for Sum( d|n, 1/d ) equals tau(n), the number of divisors of n. |
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+0
1
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| 1, 4, 9, 14, 15, 21, 25, 49, 51, 55, 57, 63, 95, 98, 99, 115, 116, 121, 147, 161, 169, 172, 175, 188, 195, 203, 236, 244, 245, 247, 265, 284, 287, 289, 297, 299, 322, 328, 329, 351, 356, 361, 363, 370, 371, 374, 387, 406, 412, 413, 418, 423, 425, 437, 465, 488
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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If p is prime p^2 is in the sequence since the continued fraction for Sum( d|p^2, 1/d ) is [1, p-1, p+1] and there are 3 divisors for p^2.
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PROGRAM
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(PARI) for(n=1, 1000, if(length(contfrac(sumdiv(n, d, 1/d)))==numdiv(n)*(1-isprime(n)), print1(n, ", ")))
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CROSSREFS
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Sequence in context: A140343 A091880 A141560 this_sequence A050986 A034259 A010452
Adjacent sequences: A071861 A071862 A071863 this_sequence A071865 A071866 A071867
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KEYWORD
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easy,nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 09 2002
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