[seqfan] Re: A083207 On an observation of Frank Buss.
robert.gerbicz at gmail.com
Thu Jul 8 22:10:20 CEST 2010
2010/7/8 peter.luschny <peter.luschny at googlemail.com>
> Recently numbers n whose divisors can be partitioned into two
> disjoint sets whose sums are both sigma(n)/2, which were introduced by
> Reinhard Zumkeller in 2003, got attention in the mathematical literature,
> in a discussion on the Usenet de.sci.mathematik and by T. D. Noe on OEIS.
> Frank Buss examined the difference of consecutive Zumkeller
> numbers up to 27188 and observed that greatest distance in this
> region is 12. (A plot can be seen here ).
> This observation was extended by T. D. Noe. He writes: "The 229026
> Zumkeller numbers less than 10^6 have a maximum difference of 12.
> This leads to the conjecture that any 12 consecutive numbers have
> at least one Zumkeller number."
> Any comments on this conjecture are highly welcome.
> Cheers, Peter
>  http://www.luschny.de/math/seq/ZumkellerNumbers.html
>  http://www.research.att.com/~njas/sequences/?q=id%3AA083207<http://www.research.att.com/%7Enjas/sequences/?q=id%3AA083207>
> Seqfan Mailing list - http://list.seqfan.eu/
The conjecture is true!
If n is Zumkeller number and gcd(n,m)=1 then n*m is also Zumkeller number.
To see this: use the partition of n. If d is a divisor of n*m then it has
got a unique form of d=d1*d2, where d1|n and d2|m, it is well known (use the
fact that gcd(n,m)=1 is true).
Put d to that partition where you would put d1 when you make the partition
of n. This will give us a good partition of n*m.
Next: for every n>0 integer the 2^n*3 is a Zumkeller number.
so 2,2^2,..,2^(n-1),3*2^n is in one partition. This is good.
And last: if you choose 12 consecutive numbers then you will see 2 numbers
divisible by 6, and at least one of them isn't divisible by 9 so has got
form 2^N*3*M, where N>0 and gcd(2^N*3,M)=1, from the previous two lemmas
2^N*3*M is a Zumkeller number. Proving the conjecture.
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