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, 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

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) = 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... .
		

Crossrefs

Cf. A225918.

Programs

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]]

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

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