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Search: id:A096502
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| A096502 |
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a(n) = k is the smallest exponent k such that 2^k-(2n+1) is a prime number, or 0 if no such k exists. |
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+0
7
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| 2, 3, 3, 39, 4, 4, 4, 5, 6, 5, 5, 6, 5, 5, 5, 7, 6, 6, 11, 7, 6, 29, 6, 6, 7, 6, 6, 7, 6, 6, 6, 8, 8, 7, 7, 10, 9, 7, 8, 9, 7, 8, 7, 7, 8, 7, 8, 10, 7, 7, 26, 9, 7, 8, 7, 7, 10, 7, 7, 8, 7, 7, 7, 47, 8, 14, 9, 11, 10, 9, 10, 8, 9, 8, 8, 31, 8, 8, 15, 8, 10, 9, 9, 8, 11, 10, 8, 9, 8, 12, 9, 8, 8, 11, 8, 14
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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As D. W. Wilson observes, this is similar to the Riesel/Sierpinski problem and there is e.g. no prime of the form 2^k - 777149, which is divisible by 3,5,7,13,19,37 or 73 if k is in 1+2Z, 2+4Z, 4+12Z, 8+12Z, 12+36Z, 0+36Z resp. 24+36Z. Already for n=935 it is difficult to find a solution. Is this linked to the fact that 2n+1=1871 is member of a prime quadruple (A007530) and quintuple (A022007)? - M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 07 2008
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EXAMPLE
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a(0)=A000043(1)=2, a(1)=A050414(1)=3, a(2)=A059608(1)=3, a(3)=A059609(1)=39;
For n=110 and n=111 even these smallest exponents are rather large: a(110)=714, a(111)=261 which mean that 2^714-221 and 2^261-223 are the least corresponding prime numbers.
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PROGRAM
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(PARI) A096502(n, k)={ k | k=log(n)\log(2)+1; n=2*n+1; while( !ispseudoprime(2^k++-n), ); k } /* will take a long time for n=935... */ - M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 07 2008
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CROSSREFS
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Sequence in context: A135353 A084956 A100650 this_sequence A101462 A088266 A116889
Adjacent sequences: A096499 A096500 A096501 this_sequence A096503 A096504 A096505
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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), Jul 09 2004
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