[seqfan] k-chunks sum and division by k
Eric Angelini
Eric.Angelini at kntv.be
Fri Mar 22 13:16:45 CET 2013
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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