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Gerhard Paseman
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Edit 2017.03.06 GRP: After finding a packing for $n=65$ (leaving $ 21,39,44,$ and $55$ to handle the excess), I find that leaving out $65$ makes it impossible to produce an exact packing for $n=63$, as $13$ needs to be left out of the good subset, and so do a multiple of $7$ and a multiple of $11$, which is too much for the excess over $4$ of the smooth denominators that are candidates for a good packing. So after $30$, the next $n$ to admit an exact packing are $65$ through $82$.

Edit 2017.03.06 GRP: After finding a packing for $n=65$ (leaving $ 21,39,44,$ and $55$ to handle the excess), I find that leaving out $65$ makes it impossible to produce an exact packing for $n=63$, as $13$ needs to be left out, and so do a multiple of $7$ and a multiple of $11$, which is too much for the excess over $4$ of the smooth denominators that are candidates for a good packing. So after $30$, the next $n$ to admit an exact packing are $65$ through $82$.

Edit 2017.03.06 GRP: After finding a packing for $n=65$ (leaving $ 21,39,44,$ and $55$ to handle the excess), I find that leaving out $65$ makes it impossible to produce an exact packing for $n=63$, as $13$ needs to be left out of the good subset, and so do a multiple of $7$ and a multiple of $11$, which is too much for the excess over $4$ of the smooth denominators that are candidates for a good packing. So after $30$, the next $n$ to admit an exact packing are $65$ through $82$.

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Gerhard Paseman
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Edit 2017.03.06 GRP: After finding a packing for $n=65$ (leaving $ 21,39,44,$ and $55$ to handle the excess), I find that leaving out $65$ makes it impossible to produce an exact packing for $n=63$, as $13$ needs to be left out, and so do a multiple of $7$ and a multiple of $11$, which is too much for the excess over $4$ of the smooth denominators that are candidates for a good packing. So after $30$, the next $n$ to admit an exact packing are $65$ through $82$.

Based on gut feel and some trial nomogramming, I suspect the next lowest admissible $n$ will be around $170$ or greater. So far the solutions above are found by hand. The exploration continues. End Edit 2017.03.06 GRP.

Gerhard "Who Wants To Go Further?" Paseman, 2017.03.03.

Gerhard "Who Wants To Go Further?" Paseman, 2017.03.03.

Edit 2017.03.06 GRP: After finding a packing for $n=65$ (leaving $ 21,39,44,$ and $55$ to handle the excess), I find that leaving out $65$ makes it impossible to produce an exact packing for $n=63$, as $13$ needs to be left out, and so do a multiple of $7$ and a multiple of $11$, which is too much for the excess over $4$ of the smooth denominators that are candidates for a good packing. So after $30$, the next $n$ to admit an exact packing are $65$ through $82$.

Based on gut feel and some trial nomogramming, I suspect the next lowest admissible $n$ will be around $170$ or greater. So far the solutions above are found by hand. The exploration continues. End Edit 2017.03.06 GRP.

Gerhard "Who Wants To Go Further?" Paseman, 2017.03.03.

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Gerhard Paseman
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2  4*    8*              16*
  3   6    9*   12          18        24     27*
     5       10        15        20        25*        30
         7            14               21              28
2  4*    8*              16*
  3   6    9*   12          18        24     27*
     5       10        15        20        25*        30
         7            14               21              28
2  4*    8*              16*
  3   6    9*   12          18        24     27*
     5       10        15        20        25*        30
         7            14             21             28
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Gerhard Paseman
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