[seqfan] Re: Numbers x such that the base 10 representation of x^2 forms an arithmetic sequence when split into equal-sized chunks

[Charles Greathouse]

Hmm. Given a 4n-digit number the first n digits can be arbitrary (except
that the first can't be 0) and the last has only the modular constraint
that the difference between it and the first block is divisible by 3, but
then the other two blocks are entirely constrained. So the 'probability'
that a given 4n-digit number is of this form is about 1/(3 * 100^n).
Between 10^(4n-1) and 10^(4n) there are about 100^n * (1 - sqrt(1/10)) =
0.68... * 100^n squares, and so out of these the expected number of
examples is . So the expected number of cases up to x is about 0.228 *
log_10(x) or 0.0990 * log(x).

(insert standard disclaimer about heuristics here)

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Wed, Jul 2, 2014 at 2:11 PM, Alex M <timeroot.alex at gmail.com> wrote:

> In particular, if this is true (and the distribution is random) -- i'm very
> curious what that constant rate for the 4-term case would be, I.e. what is
> the average number of solutions for a fixed length?
> On Jul 2, 2014 11:04 AM, "Alex M" <timeroot.alex at gmail.com> wrote:
>
> > Speaking very roughly, if these are something like random distribution --
> > one would expect to find arithmetic sequences of 3 terms fairly easily,
> but
> > 4 will be tricky, and for 5 or more would only occur finitely often
> (where
> > that finite number might be 0!). For the three-term sequences, if we
> write
> > them as (a, a+b, a+2b), then roughly the order of each term must be the
> > same -- they all have the same number of digits. If it's k digits, then
> we
> > have -- up to constant factor -- about 10^k possibilities for a and 10^k
> > possibilities for b. The resultant concatenation is of order 10^3k, which
> > is chosen random has a roughly 1 / 10^(3k/2) chance of being a perfect
> > square. Thus for a k we have -- again, up to a constant factor -- about
> > 10^2k degrees of freedom, each with a 10^(-3k/2) chance of succeeding,
> for
> > a total of 10^(k/2) expected instances for each k. That is, the more
> > digits, the more solutions there are (finding them still becomes harder
> as
> > we go up, though.) A similar heuristic tells us that for 4 terms we'd
> find
> > a constant number for each k, for 5+ we'd find 10^(-nk/2) which when
> summed
> > over k converges.
> >
> > Of course this relies on the notion that it will actually behave
> randomly,
> > but I get the feeling that "base" sequences usually do, accurately (but
> are
> > extremely difficult to prove so).
> >
> > -Alex Meiburg
> >
> > ~6 out of 5 statisticians say that the
> > number of statistics that either make
> > no sense or use ridiculous timescales
> > at all has dropped over 164% in the
> > last 5.62474396842 years.
> > Looks interesting enough to be added to the OEIS - so please go ahead!
> >
> > On Wed, Jul 2, 2014 at 11:21 AM, Christian Perfect
> > <christianperfect at gmail.com> wrote:
> > > The twitter feed @onthisdayinmath tweeted the fact that 183^2 = 183184.
> > > This leads immediately to A030467. I came up with the following
> > questions:
> > >
> > > - are there any x such that x^3 = (a)(a+1)(a+2)?
> > > - are there any x such that x^3 = (a)(a-1)(a-2)?
> > > - are there any x such that x^3 = (a)(a)(a)?
> > >
> > > The answer to all of those, as far as I can see, is "no, for for x <
> > > 10000000". A disclaimer: I'm a middling mathematician and I haven't
> come
> > up
> > > with any reasons why these sequences might not exist.
> > >
> > > So, I decided to widen my search: are there any (x,n), with n>2, such
> > that
> > > the base 10 representation of x^n forms an arithmetic sequence when
> split
> > > into three or more equal-sized chunks? The answer to that also appears
> to
> > > be "no, for x < a fairly large number". I wonder if I'm just asking for
> > > something so specific that I need to look at orders of magnitude more
> > > candidates.
> > >
> > > Anyway, in defeat, I decided to see if I could get numbers whose
> squares
> > > form arithmetic sequences when you split them into three or more
> > > equal-sized chunks. I got the following:
> > >
> > >
> >
> 11142,11553,14088,16713,18801,22284,23097,23718,26787,28818,323589,327939,328992,416103,438357,459069,
> > > ...
> > >
> > > For example, 11142^2 = 124144164, and 124, 144, 164 is an arithmetic
> > > sequence.
> > >
> > > This doesn't seem to be in the OEIS, but my route to it was so
> convoluted
> > > that I'm not sure whether it's worth adding. By the way, these all
> split
> > > into three chunks - I haven't found a number yet which gives an
> > arithmetic
> > > sequence of 4 chunks.
> > >
> > > So, should I add the above sequence?
> > >
> > > _______________________________________________
> > >
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> >
> >
> >
> > --
> > Dear Friends, I have now retired from AT&T. New coordinates:
> >
> > Neil J. A. Sloane, President, OEIS Foundation
> > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> > Phone: 732 828 6098; home page: http://NeilSloane.com
> > Email: njasloane at gmail.com
> >
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