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A225922
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.
1
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
OFFSET
1,2
COMMENTS
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
FORMULA
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
EXAMPLE
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... .
MATHEMATICA
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]]
CROSSREFS
Cf. A225918.
Sequence in context: A253428 A005905 A177890 * A235510 A220974 A063521
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 21 2013
EXTENSIONS
a(10)-a(17) from Robert G. Wilson v, May 22 2013
a(18) from Robert G. Wilson v, Jun 13 2013
a(19) from Jinyuan Wang, Jun 14 2020
Terms a(20) on from Max Alekseyev, Jan 27 2022
STATUS
approved