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A243050
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Composite integers n such that n-1 divided by the binary period of 1/n (=A007733(n)) equals an integral power of 2.
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1
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12801, 348161, 3225601, 104988673, 4294967297, 7816642561, 43796171521, 49413980161, 54745942917121, 51125767490519041, 18314818035992494081, 18446744073709551617
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OFFSET
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1,1
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COMMENTS
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All terms are odd. If even n belongs to this sequence, then n-1 is odd and thus (n-1)/A007733(n) is also odd and thus must be equal to 1. On the other hand, for even n, A007733(n) < n/2 <= n-1, i.e., (n-1)/A007733(n) > 1, a contradiction.
Contains all composite Fermat numbers A000215(k) = 2^(2^k)+1 (which are composite for 5<=k<=32 and conjecturally for any k>=5). In particular, a(5) = A000215(5), a(12) = A000215(6), and a(13) <= A000215(7) = 2^128+1.
Pseudoprimes n such that (n-1)/ord_{n}(2) = 2^k for some k, where ord_{n}(2) = A002326((n-1)/2) is the multiplicative order of 2 mod n. Composite numbers n such that Od(ord_{n}(2)) = Od(n-1), where ord_{n}(2) as above and Od(m) = A000265(m) is the odd part of m. Note that if Od(ord_{n}(2)) = Od(n-1), then ord_{n}(2)|(n-1). - Thomas Ordowski, Mar 13 2019
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LINKS
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EXAMPLE
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n = a(6) = 7816642561 = 2^15 * 238545 + 1 is the first term, which is not Proth number (A080075). The binary period of 1/n is 954180 = (n-1)/2^13.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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