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

[Eric Schmidt]

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
>>
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> 
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