[seqfan] Re: Sum divides concatenation

[John W. Layman]

Actually, I reinterpreted Eric's statement
    "The sum of the first n integers divides the concatenation of the 
first n integers"
to mean
    "The sum of the first n terms divides the concatenation of the first 
n terms".
Otherwise, I felt that he should have used a(n) rather than n in his 
definition.

John



zak seidov wrote:
> May be you mean a(n)>a(n-1)?
> Otherwise,  why not a(4)=0:
> 1+2+6+0=9 and 9|1260?
> Zak
>
> --- On Tue, 9/29/09, John W. Layman <layman at math.vt.edu> wrote:
>
>   
>> From: John W. Layman <layman at math.vt.edu>
>> Subject: [seqfan] Re: Sum divides concatenation
>> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
>> Cc: Eric.Angelini at kntv.be
>> Date: Tuesday, September 29, 2009, 12:28 PM
>> Eric Angelini wrote:
>>     
>>> Hello SeqFans,
>>> I don't know how to find more terms -- thus I don't
>>>       
>> know
>>     
>>> if this is old hat in Neil's database:
>>>
>>> 1, 2, 6, ...
>>>
>>> The sum of the first n integers divides the
>>>       
>> concatanion of
>>     
>>> the first n integers; always prolong the seq with the
>>>       
>> smal-
>>     
>>> lest _natural_ number not yet used:
>>>
>>> 1 divides 1  --> HIT
>>> 1+2   (=3) divides 12 --> HIT
>>> 1+2+3 (=6) does not divide 123
>>> 1+2+4 (=7) does not divide 124
>>> 1+2+5 (=8) does not divide 125
>>> 1+2+6 (=9) divides 126 --> HIT
>>> etc.
>>>
>>> Best,
>>> É.
>>>
>>>
>>>    
>>>       
>> It takes Mathematica a couple of minutes to get
>>     {1, 2, 6, 250488, 279786, 1060566, 1414088,
>> 1767610, 2447460},
>> which is not in the OEIS.
>>
>> Regards,
>> John
>>
>>
>>
>>
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