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A227965
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a(1) = least k such that 1 + 1/2 < H(k) - H(2); a(2) = least k such that H(a(1)) - 1/2 < 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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5
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11, 53, 249, 1164, 5435, 25371, 118428, 552798, 2580343, 12044484, 56221045, 262427666, 1224955522, 5717827134, 26689578960, 124581175389, 581517950673, 2714399875409, 12670230858892, 59141894115145, 276061555506087, 1288595564424512, 6014885070144844
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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) = (1,2); H(a(n)) - H(a(n-1)) approaches a limit 1.540684... given by A227966, and a(n)/a(n-1) approaches a limit 4.6677834... given by A227967. It is unknown whether the sequence a(n) is linearly recurrent.
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LINKS
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EXAMPLE
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The first two values (a(1),a(2)) = (11,53) match the beginning of the following inequality chain (and partition of the harmonic numbers): 1/1 + 1/2 < 1/3 + ... + 1/11 < 1/12 + ... + 1/53 < ...
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MATHEMATICA
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z = 300; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2;
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}];
t = N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 60]
Last[RealDigits[t, 10]] (* A227966 *)
t = N[Table[a[t]/a[t - 1], {t, 2, z, 50}], 60]
Last[RealDigits[t, 10]] (* A227967 *)
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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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