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Minimal sum of integers filling an n X n square, as explained in the Comments and Example sections.
1

%I #13 May 21 2019 23:56:30

%S 0,9,59,198,510,1095

%N Minimal sum of integers filling an n X n square, as explained in the Comments and Example sections.

%C Take an infinite orthogonal grid of empty cells. Fill one cell with 0. This 0 is now the upper-left corner of an n X n square that must be filled by different integers - one per cell. You are forced to sew your integers one by one on the grid, starting with 1, then 2, then 3, etc. - but in a certain way: start on the 0 and go Up, Right, Down or Left, jumping over 0 cell [this is, put a 1 in a cell that shares a side with the 0-cell (as no diagonal jumps are admitted)]. From there, jump over exactly 1 cell and put a 2 where you land. From there jump over exactly 2 cells and write a 3 where you land (the cell must be free: the cells of the grid are either empty or occupied by a single integer). The general moving rule is thus: from a cell, jump orthogonally over k cells in any direction and write the number k + 1 in that cell. The challenge is to fill the n X n initial square with the smallest possible integers. When the n X n square is filled, the sum of the integers inside the square is a(n).

%e Here is how the 3 X 3 square was filled: start on 0, go to 1, jump to 2, jump to 3, etc. until 16 is reached (a cell is represented by 2 dots).

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. 11 .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .0 .1 16 .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. 10 .8 .6 .. .. .. .. .. .. .7 .9

%e 13 .. .. .. .. .. .. .. .. .. .. .3 .. 12 .2 .4 .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .5 .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

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%e 14 .. .. .. .. .. .. .. .. .. .. .. .. .. .. 15 .. .. .. .. .. .. .. ..

%e .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

%e From 0, the successive moves are RDLRDURLRLUDLDRU.

%e Hence a(3) = 0+1+16+10+8+6+12+2+4 = 59.

%K nonn,more

%O 1,2

%A _Eric Angelini_ and Tom Karzes, May 21 2019