14
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A diamond of numbers is an arrangement of circles in the shape of a trapezoid (see figure) in which the number in any circle above its central (longest) row is the sum of the two numbers in the circles that it lies on, while the number in any circle below the central row is the (absolute) difference of the two numbers in the circles that lie on it. The diamond is perfect if all the numbers used are different positive integers.

What is the least number that can occupy the top circle of a perfect diamond whose central row consists of four circles?

What if the central row consists of 5 or 6 circles?

enter image description here

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    $\begingroup$ You should consider submitting this sequence (of which the puzzle's answers are the 4th, 5th, and 6th terms) to oeis.org ! $\endgroup$ Sep 17 at 19:31
  • $\begingroup$ @GregMartin I shall do so, thanks Greg! $\endgroup$ Sep 17 at 22:02
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    $\begingroup$ I added terms from 1 to 7 to my answer. $\endgroup$
    – RobPratt
    Sep 18 at 14:35
14
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Four circles:

37

     37
   14  23
 10   4  19
9   1   3  16
  8   2  13
    6  11
      5

Five circles:

103

        103
      49   54
    32   17  37
  22  10    7  30
18   4    6   1  29
  14   2    5  28
    12    3  23
       9   20
         11
 

Six circles:

267

           267
        105   162
      68    37   125
    54   14    23   102
  44  10     4    19   83
35   9    1     3    16  67
  26   8     2    13   51
    18    6    11    38
      12     5    27
          7    22
            15

I used mixed integer linear programming to find these optimal solutions, and the first several optimal values are:

1, 4, 13, 37, 103, 267, 645

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    $\begingroup$ If these are proven optimal then it would be good to say so and say how. (Though I think I can already guess how.) $\endgroup$
    – Gareth McCaughan
    Sep 18 at 13:11
  • $\begingroup$ Are you able to post more of these here or somewhere else? I have a partial answer that tries to solve this analytically but I need more data $\endgroup$
    – Pureferret
    Sep 19 at 12:09
  • $\begingroup$ Do you want the actual solutions or just the number in the top circle? $\endgroup$
    – RobPratt
    Sep 19 at 12:17
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Best I've found so far is

       40
     13  27
    9   4  23
  8   1   3  20
    7   2  17
      5  15
       10

but I haven't exhausted alternatives just yet

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2
  • $\begingroup$ Thanks, hexomino. Apparently one of those days...^_^b $\endgroup$ Sep 16 at 20:01
  • $\begingroup$ No problem, good answer. $\endgroup$
    – hexomino
    Sep 16 at 20:02
3
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Could it be:

68?

       68
     17  51
    4  13  38
  1   3  10  28
    2   7  18
      5  11
        6

If so,

I just naively proceeded with the assumption that I should start with the smallest numbers possible in the middle row; I adjusted these up whenever I encountered a conflict downstream. (e.g. The second circle in the middle row couldn't be 2, because the difference between 1 and 2 would be 1, a number already used.)

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