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It is well-known that the binomial coefficient has some monotonicity, and that can be used to find the maximum and minimum of binomial coefficients.

Similarly, now let $\left(\delta_{i,j}\right)_{6\times6}$ be a binary matrix with $\delta_{i,i}=1$ and $\delta_{i,j}=\delta_{j,i}=1$ or $0$ when $i\ne j$ and $\delta_{i}=\sum_{j=1}^{6}\delta_{i,j}$. Let's define $$ f(k):=\min_{\delta_{1}+\delta_{2}+\delta_{3}+\delta_{4}+\delta_{5}+\delta_{6}=2k}\delta_{1}\delta_{2}\delta_{3}\delta_{4}\delta_{5}\delta_{6} $$ for integer $k$ , $3\le k\le18$, then what is the closed form or combinatorial expression of $f(k)$ as a function of $k$?

Thanks.

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  • $\begingroup$ I deleted my comment because there's something still unclear in your problem statement. For odd $k$, there is no matrix of the type you describe, because of the constraint $\delta_{i,j} = \delta_{j,i} $, if indeed that's what you want. $\endgroup$ Commented Feb 25, 2020 at 15:42
  • $\begingroup$ @MichaelEngelhardt Thank you for your comment, and it has now been revised. Thanks. $\endgroup$
    – EmmaT
    Commented Feb 25, 2020 at 15:51
  • $\begingroup$ Just so I understand the idea here — $\delta_{i,j}$ is just a symmetric (0,1) matrix with all diagonal entries 1? Or are there any other constraints? $\endgroup$ Commented Feb 25, 2020 at 17:06
  • $\begingroup$ related math.stackexchange.com/q/3559389/87355 $\endgroup$ Commented Feb 26, 2020 at 10:18

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The minimum is achieved by filling as large an upper left square block as possible with entries 1, and partially filling the next row below that block (and, symmetrically, the next column to the right of that block) with the remaining available entries 1. The value of $f(k)$ achieved in this way is $$ f(k)=c(k)^{c(k)} \cdot [(c(k)+1)/c(k)]^{r(k)} \cdot [r(k)+1] $$ where the aforementioned upper left square block has size $c(k) \times c(k)$, and $2r(k)$ is the number of residual entries 1 available to partially fill the next row and column, as described. It remains to give explicit expressions for $c(k)$ and $r(k)$. There are probably prettier and more succinct forms, but quickly one could just specify $$ c(k)=1+\frac{k}{4}+\frac{k}{6}+\frac{k}{9}+\frac{k}{13}-\frac{k}{8}-2\left( \frac{k}{12} \right) -\frac{k}{18} $$ where division is to be interpreted as integer division, and $$ r(k)=k-3+(1-c(k))c(k)/2 $$ I have verified that these expressions give the correct minimal values by explicit computation of all possibilities.

An alternative expression for $c(k)$ is $$ c(k) = \mbox{int} \left[ \frac{1}{2} + \sqrt{\frac{1}{4} + 2k-6} \right] $$

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  • $\begingroup$ Thank you for your answer, but is the minimum "achieved by filling as large an upper left square block as possible with entries 1" for $k=6$? Thanks. $\endgroup$
    – EmmaT
    Commented Feb 28, 2020 at 15:34
  • $\begingroup$ @EmmaT - Why are you asking these questions? My answer is valid for all $k$. For $k=6$, you have a $3\times 3$ block, and you can easily convince yourself that any relocation of entries will increase the value of the product. $\endgroup$ Commented Feb 28, 2020 at 15:42
  • $\begingroup$ Sorry that I confused the minimal configuration with another configuration for k=6, but how about $k=7$ (do you get 108 by "filling as large an upper left square block as possible with entries 1", and partially filling the next row below that block (and, symmetrically, the next column to the right of that block) with the remaining available entries 1")? Thanks. $\endgroup$
    – EmmaT
    Commented Feb 28, 2020 at 17:04
  • $\begingroup$ Could you possibly express $f$ by combinatorial numbers or perhaps using multinomial coefficients? Thanks a lot. $\endgroup$
    – EmmaT
    Commented Feb 28, 2020 at 17:22
  • $\begingroup$ @EmmaT - For $k=7$, the minimal configuration yields 72, not 108. You can do it in your head - the $\delta_{i} $ are 4,3,3,2,1,1. I don't understand what you're aiming at regarding combinatorial numbers. I've given the result as a function of $k$. If you wish, replace $k$ by the binomial coefficient $\left( k \atop 1 \right)$. One can surely obfuscate it some more such as to seem less facetious. $\endgroup$ Commented Feb 28, 2020 at 17:55

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