sum of digits of n^3 is 8: more terms?
Max Alekseyev
maxale at gmail.com
Tue Feb 12 03:41:52 CET 2008
As of the patterns in the digit sums of cubes, I suggest the following sequence:
Positive integers n such that there exist positive integers x and y
such with the digit sum of (x*10^k + y)^3 equal n for all large enough
integers k.
In other words, it will contain those n that cubes of all large enough
numbers x0...0y have the digit sum equal n.
, while from below it follows that n<8 do not.
Basically, to check whether n belongs to this sequence one needs to
solve the following equation in positive integers:
digitsum(x^3) + digitsum(3*x^2*y) + digitsum(3*x*y^2) + digitsum(y^3) = n
or to prove that there are no solutions. Note that two middle terms
are multiples of 3 and, hence, n cannot be smaller than 8. And as we
have seen, n=8 belongs to this sequence - so, the first term is
identified ;)
Here it may be useful to have terms from your previous sequence ready
to quickly solve subequations like digitsum(x^3) = k for k < n.
Variations of this sequence may restrict solutions to x=y, or x being
a reverse of y (in which case the numbers x*10^k + y will be
palindromes).
Regards,
Max
On Feb 11, 2008 6:10 PM, Max Alekseyev <maxale at gmail.com> wrote:
> 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
> >
>
More information about the SeqFan
mailing list