A136010 ?

[Eric]

20, 9, 243, 1782, 14661, 118665, 962604, 7806213, 63306927, 513404406

at the 20 position i need a 9
so as you say each number could be divide by 9
so to have 9 at position 20....
i do the number - the sum of each term of the number...

like this
1 sum of numbers 1 :o) 1-1=0
etc until 10
after i have 9 until 20

so 243 at 280
1990 for 1782
16300 for 14661...

(i think it will be another one stupid mail from me but it's not a problem
lol :o) )


-----Message d'origine-----
De : Alexander Povolotsky [mailto:apovolot at gmail.com]
Envoyé : mercredi 26 mars 2008 14:37
À : seqfan at ext.jussieu.fr
Objet : A136010 ?


 Sorry for bothering everyone ...

But did anyone looked at recently submitted A136010 ?
It looks like a puzzle to me since no verbal definition of the pattern
or mathematical rule of generation is given ...

As far as I could only see is that:

a) The terms starting from the second one are dividable by 9 ...
b) The terms starting from the third one are dividable by 81 ....

Alex
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A136010  From an online IQ test (Adaptive IQ ). Solution not known to
the submitter.

 20, 9, 243, 1782, 14661, 118665, 962604, 7806213, 63306927, 513404406
(list; graph; listen)

 OFFSET  1,1

 FORMULA  nonn,unkn

 EXAMPLE  20=4*5, 4+5=9.

But 9 = 3+6 = 3*3, while 243 = 3^5?

1782=162*11=81*22?

14661=81*181?

 KEYWORD  nonn,new

 AUTHOR  Jordan Giedd (jordyg365(AT)gmail.com), Mar 20 2008