A132660 : wrong terms(?!)

[zak seidov]

According to Mmca,
the terms <10^6 are;

339542,355877,372554,385263,388943,415683,423413,435267,487386,513728,514377,538271,578979,625958,629721,657971,661977,731618,738792,739278,753921,797454,853892,856529,963458

Zak
--- zak seidov <zakseidov at yahoo.com> wrote:

> Smth's wrong with A132660:
> 
> Only 7 terms are OK while 8 not(?!):
> 
> {True, False, True, False, False, False, True, True,
> True, False, True, False, False, True, False}
> 
> Zak
> 
> %I A132660
> %S A132660
>
355877,377395,423413,459764,465392,489154,538271,629721,731618,736989,
> %T A132660 753921,823134,835149,856529,961635
> %N A132660 Numbers which, together with their
> squares
> in base ten, contain all digits from 1 to 9 exactly
> twice.
> %C A132660 Also of note is the unique number 314,
> which together with its sixth power 958468597212736
> has the property that each digit from 1 to 9 occurs
> exactly once. Aside from trivial cases that use
> zeroth
> powers, these are all the cases where two numbers
> that
> are powers of the same base have this property.
> %C A132660 Found by exhaustive search.
> %e A132660 355877^2 = 126648439129.
> %K A132660 base,fini,full,nonn,new
> %O A132660 1,1
> %A A132660 Wei-Hwa Huang (onigame(AT)gmail.com), Nov
> 15 2007
> 
> 
> 
>      
>
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Jonathan,  Perhaps a more natural sequence would be to 
write the numbers one, two, three, ... in Latin,
and then sum the values of the letters.  

Here are the first few numbers, just to
get started - can you work out the start of
the sequence and send it in?  Thanks!

unus duo tres quattuor quinque sex septem octo novem decem ...

http://www.jessiegietl.com/Latinnum.html  has more!

Neil