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

Fri Jul 9 16:48:33 CEST 2010

```> 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."

RG> The conjecture is true!
[...]
RG> Next: for every n>0 integer the 2^n*3 is a Zumkeller number.

Thus A007283 is (for n>0) a subsequence of A083207.
Perhaps this should be added as a comment to A007283.
And perhaps also to A000396: perfect numbers are Zumkeller.
6     [6]   [1, 2, 3]
28    [28]  [1, 2, 4, 7, 14]
496   [496] [1, 2, 4, 8, 16, 31, 62, 124, 248]

RG> Proof: sigma(2^n*3)=4*(2^(n+1)-1)
RG> 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
RG> so 2,2^2,..,2^(n-1),3*2^n is in one partition. This is good.

Yes. The tower of Zumkeller partitions that I used is similar.
....
24    [6, 24]           [1, 2, 3, 4, 8, 12]
48    [2, 12, 48]       [1, 3, 4, 6, 8, 16, 24]
96    [6, 24, 96]       [1, 2, 3, 4, 8, 12, 16, 32, 48]
192   [2, 12, 48, 192]  [1, 3, 4, 6, 8, 16, 24, 32, 64, 96]
....

RG> And last: if you choose 12 consecutive numbers then you will see 2 numbers
RG> divisible by 6, and at least one of them isn't divisible by 9 so has got
RG> form 2^N*3*M, where N>0 and gcd(2^N*3,M)=1, from the previous two lemmas
RG> 2^N*3*M is a Zumkeller number. Proving the conjecture.

Cool. Cheers, Peter.

P.S. Can we find a 'grid' of Zumkeller numbers such that every
other Zumkeller number is reachable within 12 steps and which has