New Sequence from wrong Comment in A083088

[Rainer Rosenthal]

David W. Cantrell wrote:

> For your sequence giving the greatest number of consecutive integer
> reciprocals, beginning at 1/n, which may be added without exceeding 1,
> I conjecture that
> 
> a(n) = floor( (e - 1)n - (e + 1)/2 + (e + 1/e)/(12(2n - 1)) )
> 
> Perhaps one could give a heuristic argument (similar to that mentioned
> in http://www.research.att.com/~njas/sequences/A002387) that, with
> high probability, my formula should always work.

Using Robert Israel's easy Maple code I suspected a(n)/n -> e - 1
already and checked it with a(2^n)/2^n and a(3^n)/3^n for not too
large n. My PC is still busy with a(3^12)/3^12 though. I know:

                   a(3^11)/3^11  =  1.7182735...

It's wonderful that you managed to guess an exact formula.
Yes, it's really wonderful.

> But even with that correction, A054414(n) = A083088(n) + n is
> incorrect, first failing for n = 24.
> 
> Thus, his comment for A054414 should be removed.

It's a pity he did such floppy and misleading remarks. But on the other
hand he didn't mean it and he surely believed he did something nice.
I already thought it might be worthwile to consider b(n) := a(n) + n.
Who knows ... maybe there is a similar but even simpler expression
for b(n)? On second thought this may be another bad remark as floor
is floor and so a(n)+n is as easy or as difficult as a(n).

While my PC was busy I tried to find a proof for my conjecture

         a(n+1) - a(n) < 3     for all n       (Conj_1_2008/RR)

Now, with your lovely formula, this will be easy. I had the impression
I could derive a proof directly from the way that sequence a(n) is built.

Thanks for your best wishes for 2008 - they already have become true :-)

Happy Seq2008 to all of you - including Robin Saunders, who may feel a
bit miserable now (cheer up and be more careful!)

Rainer Rosenthal
r.rosenthal at web.de