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A227816
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a(1) = greatest k such that H(k) - H(6) < H(6) - H(3); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(6), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.
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3
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16, 41, 103, 257, 640, 1592, 3958, 9839, 24457, 60792, 151107, 375596, 933591, 2320556, 5768028, 14337143, 35636731, 88579473, 220175161, 547272407, 1360312788, 3381224518, 8404448844, 20890289891, 51925381404, 129066913288, 320811665802, 797416799492
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OFFSET
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1,1
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COMMENTS
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Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. It appears that for many choices of (x,y), the sequence c(n) is linearly recurrent with signature of length less than 10; (3,6) is not one of them. H(c(n)) - H(c(n-1)) and c(n)/c(n-1) approach limits given by A227817 and A227818.
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LINKS
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EXAMPLE
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The first two values (a(1),a(2)) = (16,41) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 3 ):
1/3 + 1/4 + 1/5 + 1/6 > 1/7 + ... + 1/16 < 1/17 + ... + 1/41 < ...
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MATHEMATICA
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z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 6; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* (A227804) Peter J. C. Moses, Jul 23 2013 *)
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CROSSREFS
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KEYWORD
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nonn,frac,easy
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AUTHOR
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STATUS
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approved
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