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Search: id:A079665
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| A079665 |
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Numbers of the form (2^s+1)/(2^r+1) for s>1, 1<=r<=s-1. |
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+0
4
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| 3, 11, 13, 43, 171, 57, 205, 683, 241, 2731, 3277, 10923, 3641, 993, 43691, 52429, 4033, 174763, 61681, 699051, 233017, 16257, 838861, 2796203, 65281, 11184811, 1016801, 13421773, 44739243, 14913081, 261633, 15790321, 178956971, 214748365
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Conjecture: (b^s+1)/(b^r+1) is an integer if and only if: 1) r<s/2, 2) if s==1 (mod 2) then r is divisor of s, 3) if s=k*2^t with gcd(k,2^t)=1 then r is 2^t*u with u dividing k
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EXAMPLE
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a(2)=11 because (2^6+1)/(2^1+1) = 33/3 = 11 and this is the second such number (ordering by s and r, not by values of a(n)).
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PROGRAM
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(PARI) for(x=2, 30, for(y=1, x-1, if(Mod(2^x+1, 2^y+1), 0, print1((2^x+1)\(2^y+1)", "))))
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CROSSREFS
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Cf. A079672, A079581, A079673.
Sequence in context: A018450 A119145 A113049 this_sequence A083992 A105290 A158790
Adjacent sequences: A079662 A079663 A079664 this_sequence A079666 A079667 A079668
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KEYWORD
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nonn
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AUTHOR
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Jose R. Brox (tautocrona(AT)terra.es), Jan 25 2003
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