[seqfan] Re: A083207 On an observation of Frank Buss.
peter.luschny
peter.luschny at googlemail.com
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
some additional interesting properties?
For example a sequence of numbers Z such that
(a) Z is Zumkeller
(b) Z = 2^N*S, where N > 0 and S is an odd squarefree number [A056911]
(c) is minimal in the sens that the deletion of a member makes
at least one Zumkeller number unreachable in 12 steps.
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