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A225919 a(n) = 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+4) and a(1) = 1. 1
1, 11, 40, 124, 367, 1070, 3104, 8989, 26016, 75280, 217815, 630210, 1823388, 5275597, 15263836, 44162700, 127775471, 369691398 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Conjecture: a(n) is linearly recurrent. See A225918 for details.

LINKS

Table of n, a(n) for n=1..18.

EXAMPLE

a(1) = 1 by decree; a(2) = 11 because

1/5 + ... + 1/10 < 1 < 1/5 + ... + 1/11, so that a(3) = 40 because

1/12 + ... + 1/39 < 1/5 + ... + 1/11 < 1/12 + ... + 1/40.

Successive values of a(n) yield a chain:

1 < 1/5 + ... + 1/11 < 1/12 + ... + 1/40 < 1/41 + ... + 1/124 < ...

Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.8931..., it appears that lim b(n) = log(R) = 1.0623... .

MATHEMATICA

nn = 11; f[n_] := 1/(n + 4); 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: A059142 A064798 A056124 * A064768 A135719 A089776

Adjacent sequences:  A225916 A225917 A225918 * A225920 A225921 A225922

KEYWORD

nonn,more

AUTHOR

Clark Kimberling, May 21 2013

EXTENSIONS

a(11)-a(16) from Robert G. Wilson v, May 22 2013

a(17) and a(18) from Robert G. Wilson v, Jun 13 2013

STATUS

approved

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Last modified May 28 15:55 EDT 2020. Contains 334684 sequences. (Running on oeis4.)