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A227653
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a(1) = least k such that 1/2 + 1/3 < H(k) - H(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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3
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8, 21, 54, 138, 352, 897, 2285, 5820, 14823, 37752, 96148, 244872, 623645, 1588311, 4045140, 10302237, 26237926, 66823230, 170186624, 433434405, 1103878665, 2811378360, 7160069791, 18235396608, 46442241368, 118279949136, 301237536249, 767197263003
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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 A227965, (x,y) = (2,3); H(a(n)) - H(a(n-1)) approaches a limit 0.9348448455..., and a(n)/a(n-1) approaches a limit 2.546818276...
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LINKS
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FORMULA
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a(n) = 3*a(n-1) - a(n-2) - a(n-4) (conjectured).G.f.: (8 - 3 x - x^2 - 3 x^3)/(1 - 3 x + x^2 + x^4) (conjectured).
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EXAMPLE
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The first two values (a(1),a(2)) = (8,21) match the beginning of the following inequality chain: 1/2 + 1/3 < 1/4 + ... + 1/8 < 1/9 + ... + 1/21 < ...
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MATHEMATICA
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z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 2; 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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