# [seqfan] Towards a proof of the conjectured formula for A289233?

jonscho at hiwaay.net jonscho at hiwaay.net
Sun Dec 24 19:42:38 CET 2017

```All,

It's been proved that A289233(n) >= floor(n*(n+1)/6) - 1 in all cases
in which n mod 12 is 3, 5, 6, or 8, and I strongly suspect that that
lower bound is, in fact, the actual value for all such cases, but I
haven't yet been able to prove (or disprove) that suspicion.

An observation: for the cases I've checked in which n mod 12 is 0, 2,
9, or 11 -- in each of those cases, all the points can be assembled
into triples, so a(n) = n*(n+1)/6) -- the numbers of upward-pointing
triangles and downward-pointing triangles have followed a fairly
simple pattern:  the ups (U) have exceeded the downs (D) by exactly
U-D = n/3 when n is divisible by 3, and exactly U-D = (n+1)/3 otherwise.

Is there some straightforward counting argument that proves that, when
n mod 12 is 0, 2, 9, or 11, these relationships give the value of U-D
for every solution that assembles all the points into triples?

For the cases in which n mod 12 is 3, 5, 6, or 8, such a result for
U-D can never hold, because it would result in noninteger values of U
and D.  Is there some straightforward counting argument that proves
that, when n mod 12 is 3, 5, 6, or 8, it's impossible to assemble all
the points into triples?  If so, then the exact value of A289233(n)
would be known for all n >= 0.

Thanks!

-- Jon

Jon E. Schoenfield

```