[seqfan] Re: k-chunks sum and division by k

[Maximilian Hasler]

Dear Eric & SeqFans,
to avoid duplicate efforts, just to say that I confirm
the given values & extended them & fixed some details
in the mentioned (and other related) sequences ;
I will submit those not yet there (k=6,8,10) as A222256 ff
(some unexpected events preventing me from having done this earlier)

Have a nice week-end,

Maximilian


On Fri, Mar 22, 2013 at 8:16 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
> Hello SeqFans,
> I've tried yesterday to build a sequence S where the sum of any k successive terms of S is divisible by k. S being the first lexicographically such sequence and S never showing twice the same integer.
>
> For k odd, S is trivial:
> S=1,2,3,4,5,6,7,8,9,10,11,12,13,...
>
> For k even we have a few interesting things.
>
> Let's start with k=2 and a(1)=0:
> S=0,2,4,6,8,10,12,14,16,18,20,22,...
>
> Well, no revolution here. Let's try a(1)=1:
> S=1,3,5,7,9,11,13,15,17,19,21,23,...
>
> Mmmmh.
>
> For k=4 and a(1)=0 we have:
> S=0,1,2,5,4,9,6,13,8,17,10,21,22,...
> ... which is http://oeis.org/A114752
>     (and which has a quite complicated definition).
>
> For k=4 and a(1)=1 we get:
> S=1,2,3,6,5,10,7,14,9,18,11,22,13...
> ... which is http://oeis.org/A043547
>     (a nice interspersion)
>
> For k=6 and a(1)=0 I get (by hand) the new seq:
> S=0,1,2,3,4,8,6,7,14,9,10,20,...
>
> Explanation:
> The 1st chunk of 6 consecutive integers 0->8 has sum 18,
>                                which is divisible by 6
> The 2nd chunk of 6 consecutive integers 1->6 has sum 24,
>                                which is divisible by 6
> The 3rd chunk of 6 consecutive integers 2->7 has sum 30,
>                                which is divisible by 6
> Etc.
>
> For k=6 and a(1)=1 I get (again, by hand) the new seq:
> S=1,2,3,4,5,9,7,8,15,...
>
> I guess there is a possible new family of seq if we try
> k=8,10,12,14,16,... for values a(1)=0 and a(1)=1.
>
> Interesting patterns might be found...
> Best,
> É.
>
>
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