[seqfan] Re: like A004080 but for 1/(2i) and more

[M. F. Hasler]

On Tue, Jun 23, 2020 at 10:33 PM David Rabahy <DavidRabahy at comcast.net>
wrote:

> Least k such that H2(k) >= n, where H2(k) is the even harmonic number
> sum_{i=1..k} 1/(2i);
> n
> 1 - 5
> 2 - 31
> 3 - 227
> 4 - 1674
>

Unless I'm wrong it should be 4, not 5, for n=1 above,
and more generally, A4080(2n), for obvious reasons :

Clearly this idea could be generalized, i.e. where Hq(k) is the qth
> harmonic number sum_{i=1..k} 1/(qi).
>

which is the same as H(k)/q, whence your sequence should be A4080(qn).

- Maximilian