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    (Original post by generalebriety)
    Ah, ok. Might feed it into Maple and see if it factorises at all.
    A slightly nicer polynomial is:

    k(10+(k-1)(38+(k-2)(\frac{187}{6}+(k-3)(\frac{187}{24}+\frac{11}{20}(  k-4))))

    Edit: k(k+1) are factors, and I find one other root numerically.
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    (Original post by ukgea)
    Without resorting to cheating (*ahem* :P), I get it to be

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    \displaystyle \sum_{r=0}^k \left(\binom{r+2}{2} - (r + 1)\right)\binom{r+2}{2}
    \displaystyle + \sum_{r=k+1}^{2k-1} \left(\binom{r+2}{2} - 3\binom{r - k+1}{2} - (2k - r +1)\right)\left(\binom{r+2}{2} - 3\binom{r - k +1}{2}\right)
    \displaystyle + \sum_{r=0}^k\binom{r+2}{2}^2
    Thinking about it, it should be

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    \displaystyle \sum_{r=0}^k \left(\binom{r+2}{2} - (r + 1)\right)\binom{r+2}{2}
    \displaystyle + \sum_{r=k+1}^{2k} \left(\binom{r+2}{2} - 3\binom{r - k+1}{2} - (2k - r +1)\right)\left(\binom{r+2}{2} - 3\binom{r - k +1}{2}\right)
    \displaystyle + \sum_{r=0}^{k-1}\binom{r+2}{2}^2


    That gives the values

    k=1: 10
    k=2: 96
    k=3: 445
    k=4: 1431
    k=5: 3681
    k=6: 8141
    k=7: 16142
    k=8: 29466
    k=9: 50412

    for k = 1 through 5, the values are the same as the values given by DFranklin's formula, but for k >= 6 DFranklin's formula starts growing faster. Which is very weird, because both formulae are polynomials of degree 5, I think, so they should be equal. Did I make a mistake in my calculations somewhere? ???
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    I think your numbers are correct (well done - I have no idea where your expression comes from, but it obviously works).

    I don't know if I've managed to make a transcription error somewhere - this is the exact code I have for the polynomial (cut/pasted from Visual Studio):

    k*(10+(k-1)*(38+(k-2)*(187/6.0+(k-3)*(187/24.0+11*(k-4)/20.0))));

    and this gives the same numbers for k=1 to 9 as those you provide (evaluated by computer, obviously).
 
 
 
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