A225919 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+4) and a(1) = 1.

1, 11, 40, 124, 367, 1070, 3104, 8989, 26016, 75280, 217815, 630210, 1823388, 5275597, 15263836, 44162700, 127775471, 369691398, 1069624136, 3094731965, 8953954568, 25906380024, 74954649447, 216865477466, 627454543012, 1815407450477, 5252498763364, 15196997925652

Offset

1,2

Comments

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

Links

Table of n, a(n) for n=1..28.

Formula

For n>=3, a(n) = ceiling( (a(n-1)+4.5)^2 / (a(n-2)+4.5) - 4.5 ) 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) = 11 because 1/6 + ... + 1/14 < 1 < 1/6 + ... + 1/(11+4), so that a(3) = 40 because 1/16 + ... + 1/43 < 1/6 + ... + 1/15 < 1/16 + ... + 1/(40+4).
Successive values of a(n) yield a chain: 1 < 1/(1+5) + ... + 1/(11+4) < 1/(11+5) + ... + 1/(40+4) < 1/(40+5) + ... + 1/(124+4) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.8931..., it appears that lim_{n->infinity} 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: A064798 A056124 A356043 * A348586 A064768 A135719

Adjacent sequences: A225916 A225917 A225918 * A225920 A225921 A225922

Keyword

nonn

Author

Clark Kimberling, May 21 2013

Extensions

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

a(17)-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