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A361919
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The number of primes > A000040(n) and <= (A000040(n)^c + 1)^(1/c), where c = 0.567148130202... is defined in A038458.
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1
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1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 4, 4, 3, 2, 1, 1, 3, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 2, 1, 3, 5, 4, 3, 3, 3, 2, 3, 3, 4, 4, 3, 3, 2, 1, 3, 3, 3, 2, 2, 4, 4, 4, 3, 3, 3, 4, 3, 4, 3, 3, 3, 3, 4, 5, 4, 4, 4, 5, 5
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OFFSET
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1,7
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COMMENTS
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Let c = 0.567148130202... (see A038458), the solution to 127^x - 113^x = 1. c is conjectured by Smarandache to be the smallest real number x such that A000040(n+1)^x - A000040(n)^x = 1 has a solution. This conjecture is equivalent to saying that the terms of the present sequence are always positive, but that if c were replaced by a larger real number, there would be zeros in the sequence. However, note that a(30) is not the last occurrence of 1: a(46) = a(61) = 1 as well.
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LINKS
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EXAMPLE
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a(30) is the number of primes > A000040(30), which is 113, and <= (113^c + 1)^(1/c) = 127. This relatively large interval contains only the prime 127.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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