login
A383183
Square spiral numbers of the n-th grid point visited by a king always moving to the unvisited point labeled with the smallest possible prime or else composite number.
3
0, 2, 3, 5, 7, 23, 47, 79, 48, 24, 8, 1, 11, 13, 31, 29, 53, 27, 9, 10, 26, 25, 49, 83, 50, 51, 52, 28, 12, 30, 54, 55, 89, 131, 179, 129, 87, 127, 85, 84, 124, 173, 229, 293, 227, 169, 223, 167, 119, 80, 81, 82, 122, 120, 121, 168, 170, 171, 123, 172, 228, 292, 226, 224, 225, 287, 359, 439
OFFSET
0,2
COMMENTS
The infinite 2D grid is labeled along a square spiral as shown in A316328, starting with 0 at the origin (0,0), where the n-th shell contains the 8n points with sup norm n, as follows:
.
16--15--14--13--12 :
| | :
17 4---3---2 11 28
| | | | |
18 5 0---1 10 27
| | | |
19 6---7---8---9 26
| |
20--21--22--23--24--25
.
The cursor is moving like a chess king to the von Neumann neighbor not visited earlier and labeled with the smallest prime number if possible, otherwise with the smallest possible composite number.
After the 171th move, the cursor is trapped in the point (3,0) labeled a(171) = 33. All eight neighbors were then already visited earlier, so the king has no more any possible move.
LINKS
M. F. Hasler, Table of n, a(n) for n = 0..171 (full sequence).
M. F. Hasler, Path plot of A383183(0..171). (The 8 dark blue points around the red ending point represent the 8 already visited neighbors.) May 13, 2025.
EXAMPLE
From the starting point (0,0) labeled a(0) = 0, the king can reach the point (1,1) labeled 2, which is the smallest possible prime number, so a(1) = 2.
Then the king can reach (1,0) labeled 3 which is the next smaller prime number, so a(2) = 3. From there it can go to (-1,0) labeled 5 = a(3), and then to (0,-1) labeled a(4) 7 = a(4). From there, the only available prime number is 23 = a(5) at (1,-2).
The king continues in that south-east direction, before walking in north-east direction and then back and further south-east up to and beyond the point (10,-10). Then it goes back in the opposite north-west direction up to (-8,7) and (-5, 8), before heading to the point (3,0) where it gets stuck.
PROG
(Python)
from sympy import isprime # = A010051
def square_number(z): return int(4*y**2-y-x if (y := z.imag) >= abs(x := z.real)
else 4*x**2-x-y if -x>=abs(y) else (4*y-3)*y+x if -y>=abs(x) else (4*x-3)*x+y)
def A383183(n, moves=(1, 1+1j, 1j, 1j-1, -1, -1-1j, -1j, 1-1j)):
if not hasattr(A:=A383183, 'terms'): A.terms=[0]; A.pos=0; A.path=[0]
while len(A.terms) <= n:
try: _, s, z = min((1-isprime(s), s, z) for d in moves if
(s := square_number(z := A.pos+d)) not in A.terms)
except ValueError:
raise IndexError(f"Sequence has only {len(A.terms)} terms")
A.terms.append(s); A.pos = z; A.path.append(z)
return A.terms[n]
A383183(999) # gives IndexError: Sequence has only 172 terms
A383183.terms # shows the full sequence; append [:N] to show only N terms
import matplotlib.pyplot as plt # this and following to plot the path:
plt.plot([z.real for z in A383183.path], [z.imag for z in A383183.path])
plt.show()
CROSSREFS
Cf. A383184 (the same with "diamond spiral" numbering).
Cf. A383185 (similar with |a(n)-a(n+1)| > 2 instead of the prime number condition).
Cf. A335856 (same with the square spiral and indices starting at 1).
Cf. A316328 (knight path on square spiral numbered board).
Sequence in context: A068710 A120805 A177119 * A096265 A356271 A056041
KEYWORD
nonn,walk,fini,full
AUTHOR
M. F. Hasler, May 13 2025
STATUS
approved