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

T. D. Noe noe at sspectra.com
Fri Jul 9 00:42:57 CEST 2010

Nice result!


At 10:12 PM +0200 7/8/10, Robert Gerbicz wrote:
>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]
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>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.
>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.
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