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User: Tom Karzes

Tom Karzes has authored 2 sequences.

A360923 Table T(i,j), i >= 0, j >= 0, read by antidiagonals giving the smallest number of moves needed to win Integer Lunar Lander, starting from position (i,j). Game rules in comments.

0, 2, 1, 3, 3, 4, 4, 4, 5, 7, 4, 5, 6, 7, 9, 5, 5, 6, 8, 10, 12, 5, 6, 7, 8, 10, 12, 14, 6, 6, 7, 9, 10, 12, 14, 17, 6, 7, 8, 9, 11, 13, 15, 17, 19, 6, 7, 8, 9, 11, 13, 15, 17, 19, 21, 7, 7, 8, 10, 11, 13, 15, 17, 19, 22, 24, 7, 8, 9, 10, 12, 13, 15, 17, 20, 22, 24, 26

Offset: 0

Allan C. Wechsler, Feb 25 2023, using information about the original game from Eric Angelini, Hans Havermann, and M. F. Hasler, and data from Tom Karzes

A position in the game of Integer Lunar Lander consists of an ordered pair (i,j) of integers. There are always 3 legal moves, to (i+1,j+i+1), (i,j+i), and (i-1,j+i-1). The object of the game is to reach (0,0) in the minimum possible number of moves, T(i,j). The listed data was provided by Tom Karzes.
The ordered pair may be interpreted as the upward velocity and altitude of a vehicle; the object is to land the vehicle at zero velocity. In this version negative altitudes are permitted (that is, there is no crash detection, and the optimal trajectory is allowed to go underground without ill effects).
Are any of the numbers different if crashes are forbidden? It seems not, slowing to a soft landing always seems to produce a better solution than overshooting the surface and climbing back up. Of course from some initial positions like (-5,1) it is impossible not to crash. But it appears that if a soft landing is possible, then the optimal solution doesn't crash.
The game was invented in the early 1970's as a trivial version of a two-dimensional pencil-and-paper game that has been played at least since the first half of the 20th century.
Solution lengths for the corresponding game on triples would be interesting for (0,0,n) or (n,0,0).
Conjecture 1: For all i, j, T(i,j) differs from its eight neighbors by at most 3. The array of differences {T(i+1,j) - T(i,j)} might be worth studying, as well as the array {T(i,j) mod 2}. - N. J. A. Sloane, Feb 25 2023
Conjecture 2 (Start)
The zeroth row (0,j) is T(0,j) = 1+ floor(sqrt(4j-3)). j>0.
Let t_k = k(k+1)/2 be a triangular number.
Then the i-th row, {(i,j), j>=0}, is given by
  T(i,j) = T(0, j+t_{i-1}) + i for i>0.
In other words, to get the i-th row, shift the zeroth row to the left by i*(i-1)/2 places and add i to each term.
This would imply that the leading column, T(i,0), is equal to i+1+floor(sqrt(2*i^2-2*i-3)) for i >= 2.
(End) - N. J. A. Sloane, Feb 25 2023
Conjecture 1 seems to hold for all (i,j) except (1,0) and (2,1). - M. F. Hasler, Feb 27 2023

			For the starting position (2,3), a 6-move solution is (1,4), (0,4), (-1,3), (-2,1), (-1,0), (0,0). No shorter solution is possible, so T(2,3) = 6.
The upper left corner of the table i:
 0  2  3  4  4  5
 1  3  4  5  5
 4  5  6  6
 7  7  8
 9 10
12
The first few antidiagonals are
 0
 2 1
 3 3 4
 4 4 5 7
 4 5 6 7 9

Martin Gardner, "Sim, Chomp, and Race Track", "Mathematical Games", Scientific American, January, 1973. Reprinted in "Knotted Doughnuts and Other Mathematical Entertainments", W. H. Freeman, 1986, pp. 109-122.

Rémy Sigrist, Table of n, a(n) for n = 0..10010
Hans Havermann, Transcription of Gardner's account of the 2-D game.
Rémy Sigrist, C++ program
Rémy Sigrist, Scatterplot of (i, j) at an even number of moves from (0, 0) with abs(i), abs(j) <= 400
Rémy Sigrist, Scatterplot of (i, j) at up to 400 moves from (0, 0) (the color is function of the number of moves)
Wikipedia, Racetrack (game) (a 2D version), as of June 2022.

Cf. A360924, A360925, A360926.

M360923=Map(); T360923=0; S360923=[[0, 0]]; A360923(v, x)={iferr(mapget( M360923, [v,x]), E, while(!setsearch(S360923, [v,x]), foreach(S360923, vx, vx[1]>=0 && mapput(M360923, vx, T360923)); S360923 = Set(concat([[[u,vx[2]-vx[1]] | u<-[vx[1]-1..vx[1]+1], !mapisdefined(M360923, [u,vx[2]-vx[1]])] | vx <- S360923, vx[2]>=vx[1]])); T360923 += 1); T360923)} \\ M. F. Hasler, Feb 27 2023
T(i,j)=if(i,T(0,j+i*(i-1)\2)+i,j,sqrtint(4*j-3)+1) \\ Implementing Conjecture 2. - M. F. Hasler, Feb 27 2023

# From M. F. Hasler, Feb 27 2023: (Start)
def A360923(v, x, A={0:0, 1:{(0,0)}}):
    if (v,x) in A: return A[v,x]
    while (v,x) not in A[1]:
        A.update((vx,A[0]) for vx in A[1] if vx[0]>=0); A[0] += 1
        A[1] = {(u,x-v) for v,x in A[1] if x >= v
                for u in range(v-1,v+2) if (u,x-v) not in A}
    return A[0] # (End)

More terms from Rémy Sigrist, Feb 26 2023

A308353 Minimal sum of integers filling an n X n square, as explained in the Comments and Example sections.

Original entry on oeis.org

0, 9, 59, 198, 510, 1095

Offset: 1

Eric Angelini and Tom Karzes, May 21 2019

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).

			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).
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. 11 .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .0 .1 16 .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. 10 .8 .6 .. .. .. .. .. .. .7 .9
13 .. .. .. .. .. .. .. .. .. .. .3 .. 12 .2 .4 .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .5 .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
14 .. .. .. .. .. .. .. .. .. .. .. .. .. .. 15 .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
From 0, the successive moves are RDLRDURLRLUDLDRU.
Hence a(3) = 0+1+16+10+8+6+12+2+4 = 59.

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User: Tom Karzes

A360923 Table T(i,j), i >= 0, j >= 0, read by antidiagonals giving the smallest number of moves needed to win Integer Lunar Lander, starting from position (i,j). Game rules in comments.

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