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A225605
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a(1) = least k such that 1/3 < H(k) - 1/3; a(2) = least k such that H(a(1)) - H(3) < H(k) - H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.
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1
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5, 9, 16, 29, 53, 97, 178, 327, 601, 1105, 2032, 3737, 6873, 12641, 23250, 42763, 78653, 144665, 266080, 489397, 900141, 1655617, 3045154, 5600911, 10301681, 18947745, 34850336, 64099761, 117897841, 216847937, 398845538, 733591315, 1349284789, 2481721641
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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 a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?
For A225605, (x,y) = (3,3); it appears that H(a(n)) - H(a(n-1)) approaches 0.60937786343... ; it is conjectured a(n)/a(n-1) approaches the constant given at A058265..
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LINKS
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FORMULA
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a(n) = 2*a(n-1) - a(n-4) (conjectured).
G.f.: (5 - x - 2 x^2 - 3 x^3)/(1 - 2 x + x^4) (conjectured)
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EXAMPLE
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The first two values (a(1),a(2)) = (5,9) match the beginning of the following inequality chain:
1/3 < 1/4 + 1/5 < 1/6 + 1/7 + 1/8 + 1/9 < ...
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MATHEMATICA
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z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 3;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
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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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