[seqfan] Re: Interesting property of A001151

[Vladimir Shevelev]

Using induction, show that, if to arrange the first 2^k terms of  A001511 in the cyclic order, then the property satisfies. For k=0 and k=1 the property is trivial. Let it satisfies for k. Then the concatenation of two the same blocks {a(1),...,a(2^k)}, evidently, one can arrange in the cyclic order such that the property satisfies for all blocks of  length less than 2^k. Moreover, if to the last term of the second  block of length 2^k to add 1, then we obtain a new (maximal) number for the concatenation. Therefore, the property now satisfies entirely. It is left to note that, by this way, we obtain the block {a(1),...,a(2^k),a(2^k+1),...,a(2^(k+1)}, i.e., the block of the first 2^(k+1) terms of A001511. Thus our statement is proven. Together with it the property of A001511 is proven as well.
 
Regards,
Vladimir

----- Original Message -----
From: David Wilson <davidwwilson at comcast.net>
Date: Saturday, August 20, 2011 2:26
Subject: [seqfan] Re: Interesting property of A001151
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Umm,  A001511
> 
> On 8/19/2011 5:54 PM, Maximilian Hasler wrote:
> > do you really mean A001151 ?
> > if so, I'm not sure what is your definition of block and/or 
> element ?
> >
> > Maximilian
> >
> >
> >
> >
> > On Fri, Aug 19, 2011 at 5:38 PM, David 
> Wilson<davidwwilson at comcast.net>  wrote:
> >> I believe that in A001151, there is no nonempty finite block 
> of elements
> >> followed immediately by the same block, e.g, you never see 
> 1,1 or 1,2,1,2 or
> >> 1,3,1,2,1,3,1,2 etc in the sequence.
> >>
> >> However, if you prepend the sequence with any positive 
> integer n, the first
> >> 2^(n-1) elements are repeated.
> >>
> >>
> >> _______________________________________________
> >>
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> >>
> 
> 
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 Shevelev Vladimir‎