[seqfan] Re: Digits concatenation
franktaw at netscape.net
franktaw at netscape.net
Fri Nov 11 10:54:17 CET 2011
Consider also those n for which a(n) = 1; i.e., every digit occurs an
even number of times in 1..n. This sequence starts:
122, 220, 440, 660, 880, 10022, 10100, 10198, 10202, 10918, 11122,
11220, 11440, 11660, ....
Franklin T. Adams-Watters
-----Original Message-----
From: Paolo Lava <paoloplava at gmail.com>
Maximilian & Harvey
Yes, you are right. Let us say that I should add to the
definition:"...that
minimize the ratio x/y, being x/y integer and greater than zero (leading
zeros not admitted)".
This sounds good for the sequence starting with a(1)=1 but for the other
I'd add a(1)=0.
Best
Paolo
2011/11/10 Harvey P. Dale <hpd1 at nyu.edu>
> Indeed, if zero is ever permitted in the numerator all terms
will
> end up being zero, won't they?
> Harvey
>
> -----Original Message-----
> From: maximilian.hasler at gmail.com
[mailto:maximilian.hasler at gmail.com] On
> Behalf Of Maximilian Hasler
> Sent: Thursday, November 10, 2011 11:12 AM
> To: paoloplava at gmail.com; Claudio Meller
> Cc: Harvey P. Dale
> Subject: Re: [seqfan] Re: Digits concatenation
>
> > with 0:
> >
> > a(6)= 304/152=2, a(7)= 1046/523=2, a(8)= 10352/647=16, a(9)=
> > 10476/5328=2
>
> But a(n) = 0/123...n is a smaller solution, that seems to be valid
in
> view of your example a(1),
>
> >> If we start from a(0)=0 we have a(1)=0 (0/1), a(2)=5 (10/2),
a(3)=34
>
> Maximilian
>
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