[seqfan] Re: Is A109924 correct?
Alex M
timeroot.alex at gmail.com
Sun Jun 19 08:09:43 CEST 2011
They are equivalent; "60" is "060" with a leading zero. There would be no
reason to disallow trailing zeroes at all... as given a(12) has a trailing
zero, and by appending one to the front, it has a leading zero and becomes
palindromic.
I think Lars is correct in this change.
~6 out of 5 statisticians say that the
number of statistics that either make
no sense or use ridiculous timescales
at all has dropped over 164% in the
last 5.62474396842 years.
On Sat, Jun 18, 2011 at 1:33 PM, Klaus Brockhaus <
klaus-brockhaus at t-online.de> wrote:
> Am 18.06.2011 07:39, schrieb Lars Blomberg:
>
> A109924 is defined as a(n)="Least palindromic multiple of concatenation
>> 123...n"
>> with the comment "Leading zeros are allowed (but are not shown)".
>>
>> An example of suppressed leading zeros is
>> a(10) = 12345678910*10971041 = 135444,949494,445310
>>
>> But by the same rule shouldn't we have
>> a(2) = 12*5 = 60 not 12*21 = 252
>> a(5) = 12345*3034 = 37,454730 not 12345*43483 = 536,797635
>> a(6) = 123456*37315 = 4606,760640 not 123456*50061 = 6180,330816
>> ?
>>
>> /Lars B
>>
>>
> Your examples have trailing zeros, not leading zeros.
>
> KB
>
>
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