sum of digits of n^3 is 8: more terms?

[Max Alekseyev]

Zak,

Please notice that it is hard to give a precise answer to your
question, since it requires to answer the following particular hard
question:

Let x(m) be the smallest solution to the congruence x^3 == 11 (mod
10^m) that can be also defined explicitly as
x(m) = 11^( (8*10^(m-1)+1)/3 ) mod 10^m.
For example, x(10) = 1816482771 with x(10)^3 = 5993684093583030220000000011.
Is it ever possible that sum of digits of x(m)^3 equals 8?

In other words, for any positive integer m we can find such number n
that the sum of last m digits of n^3 is 2 (x(m) gives a particular
construction), and who knows maybe some of these numbers will have the
sum of all digits equal 8. That's unlikely but still possible unless
proved otherwise.

On the other hand, I can confirm that all other terms of your sequence
below 10^20 follow the pattern 10..01.

Regards,
Max

On Feb 11, 2008 12:09 AM, zak seidov <zakseidov at yahoo.com> wrote:
> I'm so sorry to bother
> those of you not concerned...
> zak
>
> %I A000001
> %S A000001 2,5,8,11,101,1001,10001,100001
> %N A000001 Numbers n such that sum of digits of n^3 is
> 8. Multiples of 10 are omitted.
> %C A000001 All numbers of form 10..01 are eligible.
> Are there larger terms of other forms?
> %e A000001 2^3=8, 5^3=125, 8^3=512, 11^3=1331
> 101^3=1010301, 1001^3=1003003001,
> 10001^3=1000300030001.
> %t A000001
> Do[If[Total[IntegerDigits[m^3]]==8&&Mod[m,10]>0,Print[m]],{m,100000}]
> %O A000001 1
> %K A000001 ,nonn,
> %A A000001 Zak Seidov (zakseido(AT)yahoo.com), Feb 11 2008
>
>
>       ____________________________________________________________________________________
> Never miss a thing.  Make Yahoo your home page.
> http://www.yahoo.com/r/hs
>