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A014015
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Denominators of sign-alternating Egyptian fraction expansion of e - 2.
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3
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OFFSET
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0,2
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COMMENTS
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The greedy alternating Egyptian fraction for e-2 must begin with 1/1 as the (-1)^0 term. - Greg Huber, May 17 2018
a(n) >= a(n-1)^2 + a(n-1) for n >= 1, so the ratio log(a(n))/2^n is strictly increasing. But does it approach a limit? Conjecture: lim_{n->infinity} log(a(n))/2^n = 0.9748... - Jon E. Schoenfield, Jun 22 2018
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LINKS
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Initial 2 removed, offset corrected and Name clarified by Greg Huber, May 17 2018
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STATUS
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approved
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