[seqfan] Re: A083207 On an observation of Frank Buss.

[Robert Gerbicz]

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 [1]).
>
> 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
>
> [1] http://www.luschny.de/math/seq/ZumkellerNumbers.html
> [2] http://www.research.att.com/~njas/sequences/?q=id%3AA083207<http://www.research.att.com/%7Enjas/sequences/?q=id%3AA083207>
>
>
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The conjecture is true!

Proof:
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.
Proof: sigma(2^n*3)=4*(2^(n+1)-1)
and sum(i=1,n-1,2^i)+2^n*3=2^n-2+3*2^n=4*2^n-2=sigma(2^n*3)/2
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.