666 and godly numbers
Maximilian Hasler
maximilian.hasler at gmail.com
Wed Mar 5 00:58:51 CET 2008
How about:
%N a(n) = Number of positive integers which do not have 'n' (written
in decimal) as substring in their expansion in any base b>1.
%C To ensure the sequence is well defined, we let a(n)=-1 if there is
an infinite number of numbers not having 'n' as subsrting in their
expansion in any base.
(I conjecture that this case never happens.)
%e a(0)=1 since 1 is the only number which does not have "0" in its
expansion in any base
%e a(1)=0 since any positive integer m has the substring "1" in its
expansion in some base (namely, for m>1, as leading digit in base m).
%e a(2)=1 since 1 is the only positive integer which does not have the
substring "1" in any base
%e a(3)=2 since 1 and 2 are the only positive integers which do not
have the substring "3" in their expansion in any base.
Then the current question is that of a(666) (or a(616), or a(888))
Maximilian
PS: I admit that the "n (written in decimal)" is 'base' and as such
very unnatural...
> I would ordinarily apologize to Neil for such a silly sequence, but I
> think that the concept (are there infinitely many godly numbers?)
> generalizes nicely, and so 666 is just an example of a string of
> base-b digits to avoid. If there are infinitely many godly numbers
> then presumably the same will be true for any string in base b.
>
> Does anyone have an answer to that original question, of how many
> numbers there are that avoid the digits 666 in every base? (Is it
> infinitely many, and if so what can we say about their asymptotic
> density?)
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