[seqfan] Re: like A004080 but for 1/(2i) and more
rgwv at rgwv.com
rgwv at rgwv.com
Wed Jun 24 21:22:59 CEST 2020
This is just the bisection of A002387.
-----Original Message-----
From: SeqFan <seqfan-bounces at list.seqfan.eu> On Behalf Of M. F. Hasler
Sent: Tuesday, June 23, 2020 11:48 PM
To: DavidRabahy at comcast.net
Cc: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: like A004080 but for 1/(2i) and more
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
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