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A225922
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a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+7) and a(1) = 1.
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1
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1, 16, 58, 176, 507, 1436, 4043, 11359, 31890, 89506, 251193, 704933, 1978258, 5551574, 15579326, 43720081, 122691130, 344306598, 966223316, 2711500429, 7609249779, 21353742568, 59924740904, 168166051476, 471922288540, 1324349620283, 3716505787761, 10429583743512
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OFFSET
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1,2
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COMMENTS
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Conjecture: a(n) is linearly recurrent. See A225918 for details.
The sequence does not satisfy any linear recurrence of order below 50, which suggests it's unlikely to exist. - Max Alekseyev, Jan 27 2022
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LINKS
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FORMULA
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For n>=3, a(n) = ceiling( (a(n-1)+7.5)^2 / (a(n-2)+7.5) - 7.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022
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EXAMPLE
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a(1) = 1 by decree; a(2) = 15 because 1/9 + ... + 1/21 < 1 < 1/9 + ... + 1/(15+7), so that a(3) = 58 because 1/23 + ... + 1/57 < 1/9 + ... + 1/22 < 1/23 + ... + 1/(58+7).
Successive values of a(n) yield a chain: 1 < 1/(1+8) + ... + 1/(15+7) < 1/(15+8) + ... + 1/(58+7) < 1/(58+8) + ... + 1/(176+7) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.80628..., it appears that lim_{n->infinity} b(n) = log(R) = 1.03186... .
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MATHEMATICA
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nn = 11; f[n_] := 1/(n + 7); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &]; s = 0; a[3] = NestWhile[# + 1 &, a[2] + 1, ! (s += f[#]) >= g[a[2]] - f[1] &]; Do[s = 0; a[z] = NestWhile[# + 1 &, a[z - 1] + 1, ! (s += f[#]) >= g[a[z - 1]] - g[a[z - 2]] &], {z, 4, nn}]; m = Map[a, Range[nn]]
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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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