A096580 a(n) = smallest m >= 2 such that Sum_{k=2..m} 1/(k*log(k)) >= n.

2, 3, 28, 8718, 51426757439

Offset: 0

Author

N. J. A. Sloane, Aug 13 2004

Keywords

• nonn
• more
• hard

Comments

The sum diverges (see link), so a(n) is well-defined.

Examples

			For m = 27 the sum is 1.992912323604..., for m = 28 it is 2.0036302389..., so a(2) = 28.
For m = 8717 the sum is 2.999991290360..., for m = 8718 it is 3.0000039326..., so a(3) = 8718.
		

Links

• M. Goar, Olivier and Abel on series convergence: An episode from early 19th century analysis, Math. Mag., 72 (No. 5, 1999), 347-355.

Crossrefs

Cf. A016088.

Programs

Mathematica

n = 0;  m = 2; sum = 1/(m*Log[m]); lst = {};
While[n <= 3,
  While[ sum < n, m++; sum += 1/(m*Log[m])];
AppendTo[lst, m];  n++]; lst (* Robert Price, Apr 09 2019 *)

Formula

Since Integral 1/(x*log(x)) dx = log log x, a(n) is close to e^(e^n) (cf. A096232, A096404, A016066).

a(n) is roughly exp(exp(n-k)), where k = 0.7946786454... - Charles R Greathouse IV, Jul 23 2007

Extensions

a(3) from Robert G. Wilson v, Aug 17 2004

a(4) from Charles R Greathouse IV, Jul 23 2007