[seqfan] Re: A conjecture on Hamming distance

Vladimir Shevelev shevelev at bgu.ac.il
Wed Apr 18 17:52:17 CEST 2012

I proved the conjecture in cases k=2,3. Moreover, if to define the property S_k also for even n, then I shown that n possesses S_2  iff n has not one of the forms 8*t and 8*t+1 (see comment in A182187) and that n possesses S_3 iff n has not one of the forms 32*t and 32*t+1. 


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Thursday, April 5, 2012 22:00
Subject: [seqfan] A conjecture on Hamming distance
To: seqfan at list.seqfan.eu

> Dear SeqFans,
> A conjecture which I formulate below arose after publication 
> sequences A209544 and A209554. In the first one, R. Mathar asks: 
> are these related to A141174,A045390 or A007519?; in the second 
> one,  he again asks: how relate these to A133870?  Using the 
> definition of operation <+> (which was introduced by me in 
> comment in A206853), I now formulate a general conjecture.
> Let n>=3 be odd and k>=2. We say that n possesses a property 
> S_k, if for every integer m from interval [0,n) with the Hamming 
> distance D(m,n) in [2,k], there exists an integer h from (m,n) 
> with D(m,h)=D(m,n).
> Conjecture. Odd n>3 possesses the property S_k iff n has the 
> form n=2^(2*k-1)*t+1.
> Example. Let k=2, t=1. Then n=9=(1001)_2. All numbers m from 
> [0,9) with D(m,9)=2 are 0,3,5.
> For m=0, we can take h=3, since 3 from (0,9) and D(0,3)=2; for 
> m=3, we can take h=5, since 5 from (3,9) and D(3,5)=2; for m=5, 
>  we can take h=6, since 6  from (5,9) and D(5,6)=2.
> You are welcome to prove (disprove) this conjecture!
> Regards,
> Vladimir
>  Shevelev Vladimir‎
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 Shevelev Vladimir‎

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