A chess knight dropping figures

(Dall-e creation)

We place a chess knight on a square of an infinite grid plane. This square is marked by zero. The knight jumps, lands on a first square and marks this square with a 1. He jumps again and marks the landing square with a 2. Another jump and we have a square marked 3, etc.
.............
......3......
....2........
......1......
........0....
.............
.............
At this stage we see 4 numbers on the grid – which sum up to 6.
But as we want to maximise the sum, we try another path for the knight:
.............
.............
........2....
......1......
.......30....
.............
.............
The sum is now 33. This is, I guess, impossible to beat.
What about 4 jumps? What is the maximum sum we can reach? Is it 73?
.............
.............
........2....
.....41......
.......30....
.............
.............
With 5 jumps, is the maximum 123?
.............
.............
.......52....
.....41......
.......30....
.............
.............
Since this knight can only place one digit per square, he will have to place the number 10 in two jumps – the first jump for the digit 1 and the second jump for the digit 0.
(The grid does not accept numbers with leading zeros at any stage of the construction.)
What is the maximum sum we can reach with the 11 jumps needed to place the 11 digits of the integers 0 to 10?
Here are a few tries:
---------------
...............
...............
......852......
....7411.......
.....96300..... sum = 852 + 7411 + 96300 = 104563
...............
...............
---------------
...............
...............
.....3811......
.......6....... 
....472900..... sum = 3811 + 6 + 472900 + 5 = 476722
......5........
...............
---------------
...............
...............
........81.....
........5......
......490720... sum = 81 + 5 + 490720 + 631 = 491437
.......631.....
...............
---------------
We could submit to the OEIS the largest possible sums reachable in n jumps.
The sequence K starts perhaps like this:
K = 1, 3, 33, 73, 123, ...
____________________
Next morning updates
Giorgos Kalogeropoulos and Jean-Marc Falcoz were quick to submit the hereunder results and extend K to 10 terms:
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We thus have so far for K (the blue terms are confirmed maximums):
K = 1, 3, 33, 73, 546, 723, 7669, 11259, 98892, 947305, 9472116, 103192539, ...
Merci and bravo to both Giorgos and Jean-Marc !
____________________
April 16th update by Nicolas Graner

(Dall-e creation)


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