Constant C=0.1688... for A081881 seems to be wrong

[David W. Cantrell]

SeqFan:
Max's understanding is correct:
my formula has not been proven for all n.

Earlier today, very soon after Rainer posted his original message, I
sent a reply to SeqFan, but it seemingly vanished, never having been
disseminated to the list. Can anyone explain how that happened?
Anyway, here's the reply that I had sent:

---------------------------- ----------------------------

Rainer:

You seem to have forgotten what I wrote on Jan. 9 in your sci.math
thread "Inverse Harmonic Series". I not only confirmed the few digits
that BC had given for the constant, but I also gave many more.
Furthermore, I gave a conjectured formula for the sequence.
I wrote, in part:

----------------------------

BTW, concerning A081881...:
There is the comment from 2003 by Benoit Cloitre that
"a(n) is asymptotic to C*exp(n) where C=0.1688..."
Of course, since we can now calculate more elements of the sequence
very easily, we can also specify C with much greater accuracy. Indeed,
it seems that we can even give a formula for the sequence

a(n) = 1 + floor(C*exp(n))

where

C = 0.1688563566671442037316797755009010341...

But this is one of those situations in which, to be able to calculate
a(n) by the formula for all n, we would need C _precisely_. Alas, I
don't know what that precise value is.

----------------------------

Of course, I would still like to know what C is precisely!

David

---------------------------- ----------------------------

----- Original Message ----- 
From: "Max Alekseyev" <maxale at gmail.com>
To: "Rainer Rosenthal" <r.rosenthal at web.de>
Cc: <seqfan at ext.jussieu.fr>; "Benoit Cloitre" <abmt at wanadoo.fr>;
"wouter meeussen" <wouter.meeussen at pandora.be>
Sent: Thursday, January 24, 2008 00:52
Subject: Re: Constant C=0.1688... for A081881 seems to be wrong


> Rainer,
>
> If I understand correctly, David's formula is just a heuristics that
> has not been proved to be correct for all n yet. If so, it should be
> used carefully and not stated as an established fact in OEIS.
> If not and I missed the proof of David's formula, I would be very
> grateful to see it.
>
> Regards,
> Max
>
> On Jan 23, 2008 3:34 PM, Rainer Rosenthal <r.rosenthal at web.de>
> wrote:
>> Thanks to Neil's latest update I am proud to
>> announce A136616 and A136617, dealing with
>> harmonic numbers.
>>
>> #
>> # 1. Extending A081881
>> #
>>
>> A related sequence is Wouter Meeussen's A081881.
>> Using the relation with A136617 and the nice
>> formula from David Cantrell there, it is easy to
>> extend this sequence from 12 elements now to many
>> more. The first 40 elements are:
>>
>> 1, 2, 4, 10, 26, 69, 186, 504, 1369, 3720, 10111,
>> 27483, 74705, 203068, 551995, 1500477, 4078718,
>> 11087104, 30137872, 81923228, 222690421, 605335323,
>> 1645472007, 4472856655, 12158484965, 33050188741,
>> 89839727480, 244209698681, 663830786257, 1804479163453,
>> 4905082919846, 13333397768101, 36243932864644,
>> 98521224097850, 267808453182726, 727978851794328,
>> 1978851684335001, 5379076574743407, 14621846107014725,
>> 39746298571222758, 108041641154662534
>>
>> The Maple code is:
>> restart:Digits:=50:e:=exp(1);A136617 := n ->
>> floor( (e - 1)*(n - 1/2) + (e - 1/e)/(24*(n - 1/2)) );
>> apq := n -> n + A136617(n);A081881 := n -> (apq@@n)(1);
>> seq(A081881(n),n=0..40);
>>
>> #
>> # 2. Checking Benoit Cloitre's formula
>> #
>>
>> The FORMULA section in A081881 says:
>> a(n) is asymptotic to C*exp(n) where C=0.1688...
>> - Benoit Cloitre (abmt(AT)wanadoo.fr), Apr 14 2003
>>
>> Numerical evidence suggests that this limit exists:
>>
>>                         a(n)
>>          C  =   lim   -------
>>                n->oo      n
>>                          e
>>
>> but the value is  C = 0.4589991659480974... as it seems
>> and not 0.1688...
>>
>> I would like to add my Maple code to A081881 and give a
>> cross reference to A136617 and David Cantrell's formula.
>> At the same time I would like to correct the formula of
>> Benoit Cloitre. But I may be wrong and so I ask you as
>> SeqFan.
>>
>> Best regards,
>> Rainer Rosenthal
>> r.rosenthal at web.de
>>
>>
>>