This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
A316667(n)=A316328(n-1)+1 \\ M. F. Hasler, Nov 06 2019
The integer 1 occupies the initial position, so its coordinates are {0,0}; therefore a(1)=0 and a(2)=0.
The integer 2 occupies the position immediately to the right of 1, so its coordinates are {1,0}.
The integer 3 occupies the position immediately above 2, so its coordinates are {1,1}; etc.
f[n_] := Block[{k = Ceiling[(Sqrt[n] - 1)/2], m, t}, t = 2k +1; m = t^2; t--; If[n >= m - t, {k -(m - n), -k}, m -= t; If[n >= m - t, {-k, -k +(m - n)}, m -= t; If[n >= m - t, {-k +(m - n), k}, {k, k -(m - n - t)}]]]]; Array[f, 40] // Flatten (* Robert G. Wilson v, Dec 04 2017 *)
f[n_] := Block[{k = Mod[ Floor[ Sqrt[4 If[OddQ@ n, (n + 1)/2 - 2, (n/2 - 2)] + 1]], 4]}, f[n - 2] + If[OddQ@ n, Sin[k*Pi/2], -Cos[k*Pi/2]]]; f[1] = f[2] = 0; Array[f, 90] (* Robert G. Wilson v, Dec 14 2017 *)
f[n_] := With[{t = Round@ Sqrt@ n}, 1/2*(-1)^t*({1, -1}(Abs[t^2 - n] - t) + t^2 - n - Mod[t, 2])]; Table[f@ n, {n, 0, 95}] // Flatten (* Mikk Heidemaa May 23 2020, after Stephen Wolfram *)
apply( {coords(n)=my(m=sqrtint(n), k=m\/2); if(m <= n -= 4*k^2, [n-3*k,-k], n >= 0, [-k,k-n], n >= -m, [-k-n,k], [k,3*k+n])}, [0..99]) \\ Use concat(%) to remove brackets '[', ']'. This function gives the coordinates of n on the spiral starting with 0 at (0,0), as shown in Examples for A174344, A274923, ..., so (a(2n-1),a(2n)) = coords(n-1). To start with 1 at (0,0), change n to n-=1 in sqrtint(). The inverse function is pos(x,y) given e.g. in A316328. - M. F. Hasler, Oct 20 2019
from math import ceil, sqrt
def get_coordinate(n):
k=ceil((sqrt(n)-1)/2)
t=2*k+1
m=t**2
t=t-1
if n >= m - t:
return k - (m-n), -k
else:
m -= t
if n >= m - t:
return -k, -k+(m-n)
else:
m -= t
if n >= m-t:
return -k+(m-n), k
else:
return k, k-(m-n-t)
The squares are labeled using their taxicab geometry distance from the origin:
.
+----+----+----+----+----+----+----+
| 6 | 5 | 4 | 3 | 4 | 5 | 6 |
+----+----+----+----+----+----+----+
| 5 | 4 | 3 | 2 | 3 | 4 | 5 |
+----+----+----+----+----+----+----+
| 4 | 3 | 2 | 1 | 2 | 3 | 4 |
+----+----+----+----+----+----+----+
| 3 | 2 | 1 | 0 | 1 | 2 | 3 |
+----+----+----+----+----+----+----+
| 4 | 3 | 2 | 1 | 2 | 3 | 4 |
+----+----+----+----+----+----+----+
| 5 | 4 | 3 | 2 | 3 | 4 | 5 |
+----+----+----+----+----+----+----+
| 6 | 5 | 4 | 3 | 4 | 5 | 6 |
+----+----+----+----+----+----+----+
.
If the knight has a choice of two or more squares with the same number which also have the same distance from the origin, then the square with the minimum square spiral number, as shown in A316667, is chosen.
A328928(n)=normlp(coords(A328908(n)),1) \\ with coords() defined e.g. in A296030. - M. F. Hasler, Nov 04 2019
The squares are numbered as in the spiral given in A174344 (upside down to get a counterclockwise spiral, but this is irrelevant here). The knight starts at a(0) = 0 with coordinates (0, 0). It jumps to a(1) = 9 with coordinates (2, -1): all 8 available squares (+-2, +-1) and (+-1, +-2) are at the same sup-norm and Euclidean distance from the origin, but square number 9 has the smallest number.
{local(coords(n, m=sqrtint(n), k=m\/2)=if(m<=n-=4*k^2, [n-3*k, -k], n>=0, [-k, k-n], n>=-m, [-k-n, k], [k, 3*k+n]), U=[]/*used squares*/, K=vector(8, i, [(-1)^(i\2)<<(i>4), (-1)^i<<(i<5)])/* knight moves*/, pos(x,y)=if(y>=abs(x), 4*y^2-y-x, -x>=abs(y),4*x^2-x-y, -y>=abs(x),(4*y-3)*y+x, (4*x-3)*x+y), t(x, p=pos(x[1],x[2]))=if(p<=U[1]||setsearch(U, p), oo, [vecmax(abs(x)), norml2(x), p]), nxt(p, x=coords(p))=vecsort(apply(K->t(x+K), K))[1][3]); my(A=List(0)/*list of positions*/); for(n=1, oo, U=setunion(U, [A[n]]); while(#U>1&&U[2]==U[1]+1, U=U[^1]); iferr(listput(A, nxt(A[n])), E, break)); print("Index of last term: ", #A-1); A328909(n)=A[n+1];} \\ To compute the infinite extension of the sequence, set en upper limit to the for() loop and replace "break" by listput(A, A[n-1])
{local(coords(n, m=sqrtint(n), k=m\/2)=if(m<=n-=4*k^2, [n-3*k, -k], n>=0, [-k, k-n], n>=-m, [-k-n, k], [k, 3*k+n]), U=[]/* used squares */, K=vector(8, i, [(-1)^(i\2)<<(i>4), (-1)^i<<(i<5)])/* knight moves */, pos(x,y)=if(y>=abs(x),4*y^2-y-x, -x>=abs(y),4*x^2-x-y, -y>=abs(x),(4*y-3)*y+x, (4*x-3)*x+y), t(x, p=pos(x[1],x[2]))=if(p<=U[1]||setsearch(U, p), oo, [norml2(x),p]), nxt(p, x=coords(p))=vecsort(apply(K->t(x+K), K))[1][2]); my(A=List(0)/*list of positions*/); for(n=1, oo, U=setunion(U, [A[n]]); while(#U>1&&U[2]==U[1]+1, U=U[^1]); iferr(listput(A, nxt(A[n])), E, break)); print("Index of last term: ", #A-1); A326924(n)=A[n+1];} \\ To compute the infinite extension, set upper bound in for() loop and replace "break" by listput(A, A[n-1])
The squares are numbered using single digits of the spiral number ordering as:
.
2---2---2---1---2---0---2 :
| | :
3 1---2---1---1---1 9 3
| | | | |
2 3 4---3---2 0 1 1
| | | | | | |
4 1 5 0---1 1 8 3
| | | | | |
2 4 6---7---8---9 1 0
| | | |
5 1---5---1---6---1---7 3
| |
2---6---2---7---2---8---2---9
If the knight has a choice of two or more squares in this spiral with the same number which also have the same distance from the origin, then the square with the minimum standard spiral number, as shown in A316667, is chosen.
The squares are numbered as in the spiral given in A174344 (upside down to get a counterclockwise spiral, but this is irrelevant here). The knight starts at a(0) = 0 with coordinates (0, 0). It jumps to a(1) = 9 with co-ordinates (2, -1): all 8 available squares (+-2, +-1) and (+-1, +-2) are at the same taxicab (2 + 1 = 3) and Euclidean distance (sqrt(2^2 + 1^2) = sqrt(5)) from the origin, but square number 9 has the smallest number. a(73) = 175 with coordinates (7, 0) is the first destination which is preferred due to the 1-norm (= 7) over A326924(73) = 81 with coordinates (5, -4), having 1-norm 5 + 4 = 9 but Euclidean or 2-norm sqrt(41) smaller than 7. a(1000) = 816 with coordinates (-10, -14). a(2000) = 2568 with coordinates (-7, -25). a(5000) = 4476 with coordinates (21, -33). a(10000) = 15560 with coordinates (-2, -62). a(20000) = 19566 with coordinates (-36, 70). a(50000) = 62092 with coordinates (125, -33). a(10^5) = 135634 with coordinates (-184, -26), taxicab distance 210 from the origin. a(200'000) = 259798 with coordinates (47, 255). a(500'000) = 713534 with coordinates (-68, -422). a(1'000'000) = 995288 with coordinates (217, 499). a(1'092'366) = 1165672 with coordinates (188, 540), taxicab norm 728 from the origin, is the last square visited by the knight before there is no unvisited square within reach. By then the earliest square on the spiral not yet visited is number 629641 at (397, 396), taxicab norm 793, and the unvisited square closest to the origin is number 1794929 at (1, 670), taxicab norm 671.
{Nmax=1e5;/* Full seq. with > 10^6 terms takes long to compute. */ local( K=[[(-1)^(i\2)<<(i>4),(-1)^i<<(i<5)]|i<-[1..8]], pos(x,y)=if(y>=abs(x),4*y^2-y-x,-x>=abs(y),4*x^2-x-y,-y>=abs(x),(4*y-3)*y+x,(4*x-3)*x+y), coords(n, m=sqrtint(n), k=m\/2)=if(m<=n-=4*k^2, [n-3*k, -k], n>=0, [-k, k-n], n>=-m, [-k-n, k], [k, 3*k+n]), t(x, p=pos(x[1],x[2]))=if(pt(x+K), K))[1][3], U=0,Umin=0); my(A=List(0)); for(n=1, Nmax, U+=1<<(A[n]-Umin); while(bittest(U,0), U>>=1;Umin++); iferr(listput(A, nxt(A[n])), E, break)); print("Index of the last term: ", #A-1); A328908(n)=A[n+1];}
The digit-square spiral is
.
.
2---2---2---1---2---0---2 2
| | |
3 1---2---1---1---1 9 3
| | | | |
2 3 4---3---2 0 1 1
| | | | | | |
4 1 5 0---1 1 8 3
| | | | | |
2 4 6---7---8---9 1 0
| | | |
5 1---5---1---6---1---7 3
| |
2---6---2---7---2---8---2---9
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