A225920 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+5) and a(1) = 1.

1, 13, 48, 150, 447, 1312, 3831, 11167, 32531, 94748, 275938, 803605, 2340292, 6815476, 19848236, 57802615, 168334451, 490228448, 1427657419, 4157665074, 12108072013, 35261476137, 102689486632, 299055281267, 870917405325, 2536310757258, 7386317253546, 21510645891422

Offset: 1

Author

Clark Kimberling, May 21 2013

Keywords

• nonn

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

Examples

			a(1) = 1 by decree; a(2) = 13 because 1/7 + ... + 1/17 < 1 < 1/7 + ... + 1/(13+5), so that a(3) = 48 because 1/19 + ... + 1/52 < 1/7 + ... + 1/18 < 1/19 + ... + 1/(48+5).
Successive values of a(n) yield a chain: 1 < 1/(1+6) + ... + 1/(13+5) < 1/(13+6) + ... + 1/(48+5) < 1/(48+6) + ... + 1/(150+5) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.912229..., it appears that lim_{n->infinity} b(n) = log(R) = 1.068918... .
		

Crossrefs

Cf. A225918.

Programs

Mathematica

nn = 11; f[n_] := 1/(n + 5); 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]]

Formula

For n>=3, a(n) = ceiling( (a(n-1)+5.5)^2 / (a(n-2)+5.5) - 5.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022

Extensions

a(12)-a(16) from Robert G. Wilson v, May 22 2013

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

a(18) corrected by and a(19) from Jinyuan Wang, Jun 14 2020

Terms a(20) on from Max Alekseyev, Jan 27 2022