[seqfan] Re: Digits concatenation
Harvey P. Dale
hpd1 at nyu.edu
Thu Nov 10 15:16:24 CET 2011
I'm not sure why the largest integer, x/y, is not always
obtained when y=1 and x equals the largest number that can be created
with the remaining digits. For example, for a(7), 765432/1=765432? I
know that 1 is a permissible value for y because that's the example for
a(2) that Paolo provided.
Best,
Harvehy
-----Original Message-----
From: seqfan-bounces at list.seqfan.eu
[mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Claudio Meller
Sent: Thursday, November 10, 2011 9:03 AM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Digits concatenation
HI Paolo, seqfans, I found this values:
without 0:
a(6)= 2346/51= 46, a(7)= 3426/571=6, a(8)= 3456/1728 = 2 a(9)=
13458/6729=2
with 0:
a(6)= 304/152=2, a(7)= 1046/523=2, a(8)= 10352/647=16, a(9)=
10476/5328=2
Best,
2011/11/10 Paolo Lava <paoloplava at gmail.com>
> Dear SeqFans,
>
>
>
> I am thinking to a sequence where at any step "n" we concatenate all
> the digits of all the natural numbers from 1 to n in order to build up
> two numbers x and y that minimize the ratio x/y, being x/y integer
> (leading zeros not admitted).
>
>
>
> Starting with a(1)=1 we have a(2)=2 because 2/1=2, a(3)=4 because
> 12/3=4,
> a(4)=33 because 132/4=33, a(5)=38 because 532/14=38, etc.
>
>
>
> If we start from a(0)=0 we have a(1)=0 (0/1), a(2)=5 (10/2), a(3)=34
> (102/3), a(4)=3 (102/34), etc.
>
>
>
> Is there anybody who is willing to compute additional terms?
>
>
>
> Thanks
>
>
>
> Paolo
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
--
Claudio Meller
http://grageasdefarmacia.blogspot.com
http://todoanagramas.blogspot.com/
http://simplementenumeros.blogspot.com/
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