[seqfan] Re: Nice new sequence based on the "sticks problem"

[Don Reble]

> As the author says, none of the rows or columns seem to be in the
> OEIS, and no formulas are known!

    The second row is floor(3K/2).

    Part A) With K-1 ones and floor(K/2) twos, one can't make a K-sided
            polygon. (Total sticks: floor(3K/2)-1.)
    There aren't enough ones for a length-1 K-polygon.
    The total length is <= (K-1)+K, not enough for a length-2 K-polygon.

    Part B) With A ones and floor(3K/2)-A twos, one can make a K-sided
            polygon.
    If there are K or more ones, make a length-1 K-polygon.
    If not, then A<K. In any case, one can make length-2 polygon with
    Q=[floor(3K/2)-A + floor(A/2)] sides.

        A<K implies A-floor(A/2) = ceiling(A/2) <= floor(K/2)

    Q = floor(3K/2)-A + floor(A/2)
      = floor(3K/2) - (A - floor(A/2))
     >= floor(3K/2) - floor(K/2)
      = K + floor(K/2) - floor(K/2) = K

    So the length-2 polygon has at least K sides.

    ---

    The third row is 2K-1. The 2K-2 counterexamples are [0,K-1,K-1]
    (as hinted by Bertagnolli's 4.2).

    Part B) With A ones, B twos, and 2K-1-(A+B) threes, one can make a
            K-sided polygon.
    If there are K or more ones, or K or more twos, make a 1-or-2-length.
    If not, then A<K and B<K. In any case, one can make a length-3
    polygon with Q=[2K-1-(A+B) + min(A,B)] sides.

    A+B = max(A,B) + min(A,B), so

    Q = 2K-1 - (A+B) + min(A,B) = 2K-1 - max(A,B) > 2K-1 - K = K-1
        The length-3 polygon has more than K-1 sides, so at least K.



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    P.S. Hey! While I composed this message, my program finished
         calculating A059958(18). But I have to check some things.
         Film at 11...

-- 
Don Reble  djr at nk.ca