According to Catalan, 2NCN = ∑I NCI 2. Since N C 0 = N C N = 1, there’s where our 2 comes from. It turns out that N C x where 1 < X < N. Therefore N C I = 2 + ∑ 1
One issue with this: I have yet to find a counterexample that isn’t in the form pq where p and q are both prime. Interesting?
Cool problem huh? Anyone want to conjecture on that final form of the answer?
1 comment:
Excellent puzzle. At the right level for my rusty brain.
Thoroughly enjoyed solving it, though it made me open my old probability book for definitions and properties of combinations. ( Its been years since I studied math. Too much software makes your brain soft. :-) )
No luck with the converse though. I'm going to try the first puzzle now.
Reg. mathml why can't you use something like this? http://pear.math.pitt.edu/mathzilla/itex2mmlFrag.html
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