[seqfan] Re: Very nice new sequence A329126 [1, 6, 42, 60, 139810, 126, ...]

[Ali Sada]

 Hi Everyone, 

I think thatanother way to look at this sequence if we construct a 3D array with thisdefinition: 


A(i,j,k)=(i^j)mod k. 


i=2,3,4,5,..(the numeral system.)

j=1,2,3,4,…(j is the power, and we can’t start with j=0 because we would never get numbersdivisible by k.)

k=2,3,4,5,…(we can start with k=1 but we will only get zeros.)

  


The firstsheet is (i^j) mod 2. It is obvious that the shortest way to get mod 2=0 is byadding the terms of the first two columns (i.e. i^1+i^2).


 

The secondsheet is (i^j) mod 3. The shortest way to get mod 3=0 is by adding the terms inthe second, fourth, and sixth columns (i.e. i^1+i^3+i^5).

 
 
The thirdsheet is for (i^j) mod 4, and so on.
 
 
In eachsheet, we don’t need to go beyond i=k+1 because it will periodic.
 

Here areimages of the first 5 sheets of the array.

 

https://justpaste.it/1ycwb

 

I think wemight see a general rule using this array.

 
 

 

Best,

 

 

 

Ali

 

 


    On Wednesday, November 13, 2019, 10:26:30 AM EST, Allan Wechsler <acwacw at gmail.com> wrote:  
 
 Yes, my "explanation" was laughably wrong -- I even made an off-by-one
error counting zeroes in the early entries.

The true structure is much richer. Readers are encouraged to read the entry
carefully to see the cool things that Neil has discovered, and there is
clearly more work to be done.

On Wed, Nov 13, 2019 at 4:36 AM Eric Schmidt <eric41293 at comcast.net> wrote:

> I can't reproduce the given terms from this formula. Calling it b(n), I get
>
> b(1) = 1
> b(2) = 10
> b(3) = 1001001
> b(4) = 100
> b(5) = 100001000010000100001
> b(6) = 10010010
> b(7) = 1000000100000010000001000000100000010000001
> b(8) = 1000
>
> Also, looking at, for instance, the given value of a(15), it doesn't
> appear that a formula as simple as this can work in all cases.
>
> On 11/12/2019 7:56 PM, Allan Wechsler wrote:
> > There's a clear pattern here, and I wonder if it persists. Express n in
> the
> > form 2^k * d, where d is odd. Then (conjecture) the solution is 1
> (0^(d-1)
> > 1)^(d-1) 0^k. This fits all the examples given, and seems plausible.
> >
> >
> > On Tue, Nov 12, 2019 at 10:32 PM Neil Sloane <njasloane at gmail.com>
> wrote:
> >
> >> A new sequence sent in by Alon Ran, A329126,
> >> which starts 1, 110, 101010, 111100, 100010001000100010, 1111110,
> >>
> >> a(n) is defined to be the lex. earliest string of numbers such that when
> >> read in every base b >= 2, a(n) is divisible by n.
> >>
> >> It follows from the definition that all the digits are 0 and 1.  The
> author
> >> gives a number of other remarks, but it was not completely clear to me
> >> which were empirical observations and which were theorems. I do not have
> >> time to work on it, but it looks like a lovely problem. He gives 21
> terms.
> >>
> >> The strings are huge. If they are rewritten in base 10 you get A329000
> >> = 1,6,42,60,139810,126,..., which makes it a lot easier to remember the
> >> first few terms.
> >>
> >> I hope someone will look into this and figure out what is going on.
> >>
> >> It has keyword "base", but it does not depend on any one base - it
> depends
> >> on all bases (so maybe that keyword is not justified).
> >>
> >> Neil
> >>
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
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