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Search: id:A049096
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| A049096 |
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Numbers n such that 2^n + 1 is divisible by a square. |
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4
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| 3, 9, 10, 15, 21, 27, 30, 33, 39, 45, 50, 51, 55, 57, 63, 68, 69, 70, 75, 78, 81, 87, 90, 93, 99, 105, 110, 111, 117, 123, 129, 130, 135, 141, 147, 150, 153, 159, 165, 170, 171, 177, 182, 183, 189, 190, 195, 201, 204, 207, 210, 213, 219, 225, 230, 231, 234, 237, 243
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Conjecture: lim n -> infinity a(n)/n = C exists and 4<C<9/2. There seems to be a sequence of primes p such that p^2 never divides numbers of the form 2^x+1: the first few are 5,7,23,31. - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 20 2002
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FORMULA
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For any a(n+1)-a(n) <= 6 since numbers of form 3^a*(2k+1) a>0, k>=0, are in the sequence (2^(3*(2k+1)+1 is divisible by 9). So are numbers of the form 20k+10 since 2^(20k+10)+1 is divisible by 25, 110k+55 since 2^(110k+55)+1 is divisible by 11^2, 78+156k since 2^(156k+78)+1 is divisible by 13^2 ... - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 20 2002
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EXAMPLE
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9 is here because 2^9+1=513 is divisible by a square, 9. 99 is here because 2^99+1=3^3*19*67*683*5347*20857*242099935645987 is divisible by 3^2, i.e. is not square-free.
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CROSSREFS
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Cf. A049093, A049094, A049095.
Cf. A086982, which is just the same with base b=10 instead of b=2.
Sequence in context: A030608 A118515 A138923 this_sequence A030794 A134073 A088005
Adjacent sequences: A049093 A049094 A049095 this_sequence A049097 A049098 A049099
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KEYWORD
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nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu)
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EXTENSIONS
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More terms from James A. Sellers (sellersj(AT)math.psu.edu), Dec 16 1999
More terms from Vladeta Jovovic (vladeta(AT)Eunet.yu), Apr 12 2002
Missing term 182 added by Rainer Rosenthal (r.rosenthal(AT)web.de), Nov 01 2005
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