partitions of 6n+3 element set
Joshua Zucker
joshua.zucker at gmail.com
Fri Mar 11 01:44:27 CET 2005
How many ways are there to partition a 6n+3 element set into 2n+1 sets
of 3 elements each, such that each 3-element set has the same sum?
I have no clue about an analytic solution.
My fairly brute-force non-clever program cranked out a few terms:
1, 2, 11, 84, 1296, 24293,
and I submitted these few to OEIS. The n = 6 term is going to take my
program an hour or so to produce, and n = 7 would take a day or so ...
obviously my program could be improved a lot.
It's pretty easy to prove this impossible for 6n (since 6n(6n+1)/2 is
not divisible by 2n). So those terms are all 0. I didn't bother
entering them into OEIS: I just entered the terms for 6n+3. Is that
the normal behavior, or should my sequence go 1, 0, 2, 0, 11, 0, ...
instead? Or even 0, 0, 1, 0, 0, 0, 0, 0, 2, ...?
Also, my general program can count the number of ways to partition an
n-element set into n/k sets of k elements each so that all the
k-element sets have the same sum. Are there any other interesting
values? Should I construct the whole triangle of (n,k) -- that feels
silly, with so many 0s in it!
Thanks,
--Joshua Zucker
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