[seqfan] Re: Digit-counters updating themselves
M. F. Hasler
oeis at hasler.fr
Sat Oct 11 21:28:43 CEST 2014
Eric,
my program seems to confirm your data (congrats !),
I submitted a proposal as https://oeis.org/draft/A248034
-- Maximilian
(PARI)
c=vector(10);print1(a=0);for(n=1,99,apply(d->c[d+1]++,if(a,digits(a)));print1(","a=c[1+a%10]))
On Sat, Oct 11, 2014 at 1:59 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>
> D=0,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,1,10,2,2,3,2,4,2,5,2,6,2,7,2,8,2,9,2,10,3,3,4,3,5,3,6,3,7,3,8,3,9,3,10,4,4,5,4,6,4,7,4,8,4,9,4,10,5,5,6,5,7,5,8,5,9,5,10,6,6,7,6,8,6,9,6,10,7,7,8,7,9,7,10,8,8,9,8,10,9,9,10,10,11,20,12,...
>
> Hello SeqFans,
> pick any comma in D.
> Immediately to the left of the comma
> there is a digit 'd'.
> Immediately to the right of the comma
> there is an integer d(n).
> D is such that there are d(n) digit 'd'
> so far in D [from the start of D up to the comma].
>
> In other words, the rightmost digit of d(n) is present d(n+1) times in D, counting from d(1) to d(n).
>
> I'm wondering: do all integers appear
> at least once in D?
>
> P.-S.
> It is possible to compute similar sequences
> for every base. I guess the binary-one is:
>
> B = 0,1,1,10,10,11,110,100,110,111,1110,...
>
> Best,
> É.
>
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