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

Thu Jul 3 07:27:11 CEST 2014

Let's do this in base B rather than just base 10. If a and b are distinct 
k-digit base-B positive numbers (perhaps with leading zeros) with a==b mod 
3, then [a, (2*a+b)/3, (a+2*b)/3, b] = [x_1,x_2,x_3,x_4] is a four-term 
arithmetic sequence of k-digit base-B numbers. If t = B^k, then the base-B 
concatenation of x_1, x_2, x_3, x_4 is x_1*t^3 + x_2*t^2 + x_3*t + x_4 = 
(1/3)*t*(3*t^2+2*t+1)*a+(1/3*(t^2+2*t+3))*b where we want 0 <= a,b < t, 
a<>b.

I searched all t from 2 to 300, and found the following solutions:

t = 4 = 2^2, a = 0, b = 3, c = 3.  Thus 3^3 = 0123_4 = 00011011_2 
t = 39, a = 7, b = 31, c = 76.  Thus 76^3 = [7][15][23][31]_39 
t = 134, a = 26, b = 107, c = 399.  Thus 399^3 = [26][53][80][107]_134
t = 264, a = 0, b = 153, c = 153.  Thus 153^3 = [0][51][102][153]_264
t = 290, a = 147, b = 78, c= 1532.  Thus 1532^3 = [147][124][101][78]_290.

Cheers,
Robert Israel

On Jul 2 2014, Alex M 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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