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A100366
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a(n) is the least prime number q such that q,q+1,q+2,q+3,...,q+n-1 have 2,4,8,...,2^n divisors respectively.
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2
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OFFSET
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1,1
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COMMENTS
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Any run of 8 or more consecutive integers must include at least one number k of the form 8j+4; in the prime factorization of k, the prime factor 2 appears with multiplicity exactly 2, so the number of divisors of k is divisible by 3 (which is not a power of 2). Thus, there is no term a(8): the sequence is complete, ending with a(7). - Jon E. Schoenfield, Nov 12 2017
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LINKS
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EXAMPLE
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a(4)=613: q=613 (a prime, hence two divisors), q+1 = 614 = 2*307 (4 divisors), q+2 = 615 = 3*5*41 (8 divisors), and q+3 = 616 = 2^3 * 7 * 11 (16 divisors).
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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EXTENSIONS
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Keywords fini and full added and Example section edited by Jon E. Schoenfield, Nov 12 2017
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STATUS
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approved
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