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A351041
Minimal number of steps for a Racetrack car (using Moore neighborhood) to go around a circle of radius n.
8
7, 9, 12, 13, 15, 16, 18, 18, 19, 21, 22, 22, 24, 24, 25, 26, 27, 27, 28, 28, 30, 31, 31, 31, 32, 33, 33, 34, 35, 35, 36, 36, 37, 37, 37
OFFSET
1,1
COMMENTS
The car moves according to the rules of the game of Racetrack, i.e., if P, Q, and R are three successive positions of the car, both coordinates of the second difference (acceleration vector) P - 2Q + R must be 1, 0, or -1. The car starts with zero velocity at a point (x,0) for some integer x >= n, and finishes when it passes, or lands on, the positive x-axis after a complete counterclockwise lap around the origin. The line segments between successive positions must be outside or on the circle with center in (0,0) and radius n.
FORMULA
a(n) = min {k >= 6; A351349(k)/A351350(k) >= n^2}.
a(n) <= A351042(n).
a(n) >= A027434(n) + A027434(2*n) + A002024(n). This can be seen by looking at the y-coordinate only: First, the car must go up to at least y = n and reduce the speed in the y-direction to zero in order to turn downwards; this requires at least A027434(n) steps. Then down to y = -n or below with speed reduced to zero; this requires at least A027434(2*n) steps. Finally, up to at least y = 0 (not necessarily reducing the speed); this requires at least A002024(n) steps.
It appears that a(n) = A027434(n) + A027434(2*n) + A002024(n) + 1 if n is a triangular number (A000217), otherwise a(n) = A027434(n) + A027434(2*n) + A002024(n).
EXAMPLE
The following diagrams show examples of optimal trajectories for n = 1, 2, 3. The origin is marked with an asterisk.
.
a(1) = 7:
. 2 . 1 . .
3 . * . 0 7
. 5 . 6 . .
(The car stands still on the fourth step.)
.
a(2) = 9:
. 3 . 2 . .
4 . . . 1 .
. . * . 0 9
5 . . . 8 .
. 6 . 7 . .
.
a(3) = 12:
. . . 4 3 . . . .
. 5 . . . . 2 . .
. . . . . . . . .
6 . . . . . . 1 .
7 . . . * . . 0 12
. . . . . . . . .
. 8 . . . . . 11 .
. . . 9 . 10 . . .
CROSSREFS
KEYWORD
nonn,more
AUTHOR
STATUS
approved