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A269804
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Numbers having harmonic fractility 1, cf. A270000.
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7
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2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 17, 18, 21, 24, 27, 28, 31, 32, 34, 36, 42, 48, 49, 51, 54, 56, 62, 63, 64, 68, 72, 81, 84, 93, 96, 98, 102, 108, 112, 113, 124, 126, 128, 136, 144, 147, 151, 153, 162, 168, 186, 189, 192, 196, 204, 216, 224, 226, 241, 243
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OFFSET
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1,1
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COMMENTS
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In order to define (harmonic) fractility of an integer m > 1, we first define nested interval sequences. Suppose that r = (r(n)) is a sequence satisfying (i) 1 = r(1) > r(2) > r(3) > ... and (ii) r(n) -> 0. For x in (0,1], let n(1) be the index n such that r(n+1) < x <= r(n), and let L(1) = r(n(1)) - r(n(1)+1). Let n(2) be the largest index n such that x <= r(n(1)+1) + L(1)*r(n), and let L(2) = (r(n(2))-r(n(2)+1))*L(1). Continue inductively to obtain the sequence (n(1), n(2), n(3), ... ) =: NI(x), the r-nested interval sequence of x.
For fixed r, call x and y equivalent if NI(x) and NI(y) are eventually equal (up to an offset). For m > 1, the r-fractility of m is the number of equivalence classes of sequences NI(k/m) for 0 < k < m. Taking r = (1/1, 1/2, 1/3, 1/4, ...) gives harmonic fractility.
For harmonic fractility, r(n) = 1/n, n(j+1) = floor(L(j)/(x - Sum_{i=1..j} L(i-1)/(n(i)+1))) for all j >= 0, L(0) = 1. - M. F. Hasler, Nov 05 2018
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LINKS
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EXAMPLE
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Nested-interval sequences NI(k/m) for m = 6:
NI(1/6) = (6, 1, 1, 1, 1, 1,...)
NI(2/6) = (3, 1, 1, 1, 1, 1,...)
NI(3/6) = (2, 1, 1, 1, 1, 1,...)
NI(4/6) = (1, 3, 1, 1, 1, 1,...)
NI(5/6) = (1, 1, 3, 1, 1, 1,...):
There is only one equivalence class, so that the fractility of 6 is 1.
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MATHEMATICA
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PROG
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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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STATUS
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approved
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