[seqfan] Re: Unexpected help by A052515
rlshepherd2 at gmail.com
Thu Feb 11 18:12:35 CET 2010
Here is a very straightforward (using Ralf Stephan's claim) proof.
On Thu, Feb 11, 2010 at 8:12 AM, Rainer Rosenthal <r.rosenthal at web.de>wrote:
> Now I know (without proof) that integer solutions [a,b]
> for (a+2b)^2 = 2^b are given by
> a = 2*A052515(n)
> b = 2*n+2
> for n > 2.
Let a = 2^(n+1)-4n-4 and b = 2n+2 for n > 2.
Then (a+2b)^2 = (2^(n+1)-4n-4+4n+4)^2 = (2^(n+1))^2 = 2^(2n+2) = 2^b.
Ralf Stephan stated that, for n>2, A052515(n) = 2^n-2n-2. Assuming
that is true, then also a = 2*A052515(n).
> Examples: n=3: a=2*A052515(3)= 0, b=2*3+2 = 8
> n=4: a=2*A052515(4)= 12, b=2*4+2 = 10
> n=5: a=2*A052515(5)= 40, b=2*5+2 = 12
> I checked all available members of A052515 an they all worked
With PARI, earlier I verified this for n up to 100000. (I also confirmed
that the shown values of A052515, a(3) through a(30), match Ralf Stephan's
> I would like to put my experimental findings as another
> comment into the OEIS.
> The most serious comment would be that a=2*A052515(n) and
> b=2*n+2 provide integer solutions for (a+2b)^2 = 2^b.
> A more playful remark could refer to the original problem:
> Finding ternary numbers with S and P as their sum resp. product
> of digits and satisfying S^2 = P. The number of digits 1 and 2
> is then a and b respectively: a=2*A052515(n), b=2*n+2.
Regarding the "playful remark", it could be generalized -- this is obvious
but it highlights the point rather than making a casual glance at
the comment suggest that it's ternary-specific:
Finding (non-trivial (multidigit)) numbers <with the stated properties> in
any base b > 2 while confining the numbers to those containing only the
digits 1 and/or 2, then <conclusion>.
(I tend to err on the side of verbosity so you might want to make this more
> I am quite happy about my findings and like to share them
> with SeqFan. I will wait with my submission of the comment
> since I am hoping for some helpful advice, how to present
> the findings appropriately.
> Thanks for your post.
(Apologies if the line-formatting has become ugly)
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