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Rand al'Thor
  • 118k
  • 29
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  • 637

Here's an example that beats the OP's:

  |11  41   7|  59
  |47   3  17|  67
  |13  83   5| 101
  ------------
23 71 127  29   19

I got this by putting the smallest possible number (3) in the most central position (which is involved in 3four of the 8eight sums) and then the next smallest numbers (5, 7, 11, 13) in the positions involved in 2three of the 8eight sums, so that $S_1+\dots+S_8$ is minimised. I believe this may be the best possible.

Here's an example that beats the OP's:

  |11  41   7|  59
  |47   3  17|  67
  |13  83   5| 101
  ------------
23 71 127  29   19

I got this by putting the smallest possible number (3) in the most central position (which is involved in 3 of the 8 sums) and then the next smallest numbers (5, 7, 11, 13) in the positions involved in 2 of the 8 sums, so that $S_1+\dots+S_8$ is minimised. I believe this may be the best possible.

Here's an example that beats the OP's:

  |11  41   7|  59
  |47   3  17|  67
  |13  83   5| 101
  ------------
23 71 127  29   19

I got this by putting the smallest possible number (3) in the most central position (which is involved in four of the eight sums) and then the next smallest numbers (5, 7, 11, 13) in the positions involved in three of the eight sums, so that $S_1+\dots+S_8$ is minimised.

added 77 characters in body
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Rand al'Thor
  • 118k
  • 29
  • 325
  • 637

Here's an example that beats the OP's:

  |13|11  541 11| 29 7|  59
  |31|47   3 19| 5317|  67
  |29|13 23 83 7| 59 5| 101
  ------------
4323 7371 31127 37 29 23  19

I got this by putting the smallest possible number (3) in the most central position (which is involved in 3 of the 8 sums) and proceeding from therethen the next smallest numbers (5, leaving7, 11, 13) in the positions only featuredinvolved in two2 of the 8 sums, so that (i.e$S_1+\dots+S_8$ is minimised. the squares orthogonally adjacent to the central square) untilI believe this may be the endbest possible.

Here's an example that beats the OP's:

  |13  5 11| 29
  |31  3 19| 53
  |29 23  7| 59
  ----------
43 73 31 37  23

I got this by putting the smallest number (3) in the most central position and proceeding from there, leaving the positions only featured in two of the sums (i.e. the squares orthogonally adjacent to the central square) until the end.

Here's an example that beats the OP's:

  |11  41   7|  59
  |47   3  17|  67
  |13  83   5| 101
  ------------
23 71 127  29   19

I got this by putting the smallest possible number (3) in the most central position (which is involved in 3 of the 8 sums) and then the next smallest numbers (5, 7, 11, 13) in the positions involved in 2 of the 8 sums, so that $S_1+\dots+S_8$ is minimised. I believe this may be the best possible.

Source Link
Rand al'Thor
  • 118k
  • 29
  • 325
  • 637

Here's an example that beats the OP's:

  |13  5 11| 29
  |31  3 19| 53
  |29 23  7| 59
  ----------
43 73 31 37  23

I got this by putting the smallest number (3) in the most central position and proceeding from there, leaving the positions only featured in two of the sums (i.e. the squares orthogonally adjacent to the central square) until the end.