A068225 -id:A068225 - cp's OEIS frontend

A121496 Run lengths of successive numbers in A068225.

1, 2, 2, 1, 3, 4, 4, 3, 5, 6, 6, 5, 7, 8, 8, 7, 9, 10, 10, 9, 11, 12, 12, 11, 13, 14, 14, 13, 15, 16, 16, 15, 17, 18, 18, 17, 19, 20, 20, 19, 21, 22, 22, 21, 23, 24, 24, 23, 25, 26, 26, 25, 27, 28, 28, 27, 29, 30, 30, 29, 31, 32, 32, 31, 33, 34, 34, 33, 35, 36, 36, 35, 37, 38, 38

Offset: 1

Rick L. Shepherd, Aug 03 2006

A000027 and A103889 are bisections.

			The fifth run of successive numbers in A068225 is 8, 9, 10 with run length three so a(5) = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A000027, A068225, A103889.

Rest@ CoefficientList[Series[x (1 + x - x^3 + x^4)/((1 - x)^2*(1 + x) (1 + x^2)), {x, 0, 75}], x] (* Michael De Vlieger, Oct 02 2017 *)

a(n) = if(n%2==1,(n+1)/2,if(n%4==0,(n/2)-1,(n/2)+1))
for(n=1,80,print1(a(n),", "))

Vec(x*(1+x-x^3+x^4)/((1-x)^2*(1+x)*(1+x^2)) + O(x^100)) \\ Colin Barker, Apr 08 2016

a(2*k-1) = k, a(4*k) = 2*k-1, a(4*k-2) = 2*k, for k >= 1.
From Colin Barker, Apr 08 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.
G.f.: x*(1+x-x^3+x^4) / ((1-x)^2*(1+x)*(1+x^2)). (End)
a(n) = (2*n+1-4*cos(n*Pi/2)-cos(n*Pi))/4. - Wesley Ivan Hurt, Oct 02 2017

A054569 a(n) = 4n^2 - 6n + 3.

Original entry on oeis.org

1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703, 813, 931, 1057, 1191, 1333, 1483, 1641, 1807, 1981, 2163, 2353, 2551, 2757, 2971, 3193, 3423, 3661, 3907, 4161, 4423, 4693, 4971, 5257, 5551, 5853, 6163, 6481, 6807, 7141, 7483, 7833, 8191

Offset: 1

Enoch Haga, G. L. Honaker, Jr., Apr 10 2000

Move in 1-7 direction in a spiral organized like A068225 etc.
Third row of A082039. - Paul Barry, Apr 02 2003
Inverse binomial transform of A036826. - Paul Barry, Jun 11 2003
Equals the "middle sequence" T(2*n,n) of the Connell sequence A001614 as a triangle. - Johannes W. Meijer, May 20 2011
Ulam's spiral (SW spoke). - Robert G. Wilson v, Oct 31 2011

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Robert G. Wilson v, Cover of the March 1964 issue of Scientific American
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A033951, A054552, A054554, A054556, A054567, A054568, A068225.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

List([1..50], n-> 4*n^2-6*n+3); # G. C. Greubel, Jul 04 2019

[4*n^2-6*n+3: n in [1..50]]; // G. C. Greubel, Jul 04 2019

f[n_]:= 4*n^2-6*n+3; Array[f, 50] (* Vladimir Joseph Stephan Orlovsky, Sep 02 2008 *)
LinearRecurrence[{3,-3,1},{1,7,21},50] (* Harvey P. Dale, Nov 17 2012 *)

a(n)=4*n^2-6*n+3 \\ Charles R Greathouse IV, Sep 24 2015

[4*n^2-6*n+3 for n in (1..50)] # G. C. Greubel, Jul 04 2019

a(n+1) = 4*n^2 + 2*n + 1. - Paul Barry, Apr 02 2003
a(n) = 4*n^2 - 6*n+3 - 3*0^n (with leading zero). - Paul Barry, Jun 11 2003
Binomial transform of [1, 6, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
a(n) = 8*n + a(n-1) - 10 (with a(1)=1). - Vincenzo Librandi, Aug 07 2010
From Colin Barker, Mar 23 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(1+x)*(1+3*x)/(1-x)^3. (End)
a(n) = A000384(n) + A000384(n-1). - Bruce J. Nicholson, May 07 2017
E.g.f.: -3 + (3 - 2*x + 4*x^2)*exp(x). - G. C. Greubel, Jul 04 2019
Sum_{n>=1} 1/a(n) = A339237. - R. J. Mathar, Jan 22 2021

Edited by Frank Ellermann, Feb 24 2002

A054554 a(n) = 4n^2 - 10n + 7.

Original entry on oeis.org

1, 3, 13, 31, 57, 91, 133, 183, 241, 307, 381, 463, 553, 651, 757, 871, 993, 1123, 1261, 1407, 1561, 1723, 1893, 2071, 2257, 2451, 2653, 2863, 3081, 3307, 3541, 3783, 4033, 4291, 4557, 4831, 5113, 5403, 5701, 6007, 6321, 6643, 6973, 7311, 7657, 8011, 8373, 8743

Offset: 1

Enoch Haga, G. L. Honaker, Jr., Apr 10 2000

Move in 1-3 direction in a spiral organized like A068225 etc.
Equals binomial transform of [1, 2, 8, 0, 0, 0, ...]. - Gary W. Adamson, May 03 2008
Ulam's spiral (NE spoke). - Robert G. Wilson v, Oct 31 2011
Number of ternary strings of length 2*(n-1) that have one or no 0's, one or no 1's, and an even number of 2's. For n=2, the 3 strings of length 2 are 01, 10 and 22. For n=3, the 13 strings of length 4 are the 12 permutations of 0122 and 2222. - Enrique Navarrete, Jul 25 2025

Ivan Panchenko, Table of n, a(n) for n = 1..1000
James Grime and Brady Haran, Prime Spirals, Numberphile video (2013).
Scientific American, Cover of the March 1964 issue
Leo Tavares, Illustration: Hexagonal Dual Rays
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A054553, A068225, A054552, A054556, A054567, A054569, A033951.
Cf. A014105.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A003215, A227776.

A054554:=n->4*n^2 -10*n + 7; seq(A054554(k),k=1..100); # Wesley Ivan Hurt, Nov 05 2013

f[n_] := 4n^2 -10n + 7; Array[f, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *)
LinearRecurrence[{3,-3,1},{1,3,13},60] (* Harvey P. Dale, Jan 20 2025 *)

a(n)=4*n^2-10*n+7 \\ Charles R Greathouse IV, Nov 05 2013

a(n) = 8*n + a(n-1) - 14 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: -x*(7*x^2+1)/(x-1)^3. - Colin Barker, Sep 21 2012
For n > 2, a(n) = A014105(n) + A014105(n-1). - Bruce J. Nicholson, May 07 2017
From Leo Tavares, Feb 21 2022: (Start)
a(n) = A003215(n-2) + 2*A000217(n-1). See Hexagonal Dual Rays illustration in links.
a(n) = A227776(n-1) - 4*A000217(n-1). (End)
a(k+1) = 4k^2 - 2k + 1 in the Numberphile video. - Frank Ellermann, Mar 11 2020
E.g.f.: exp(x)*(7 - 6*x + 4*x^2) - 7. - Stefano Spezia, Apr 24 2024

Edited by Frank Ellermann, Feb 24 2002

A054556 a(n) = 4n^2 - 9n + 6.

Original entry on oeis.org

1, 4, 15, 34, 61, 96, 139, 190, 249, 316, 391, 474, 565, 664, 771, 886, 1009, 1140, 1279, 1426, 1581, 1744, 1915, 2094, 2281, 2476, 2679, 2890, 3109, 3336, 3571, 3814, 4065, 4324, 4591, 4866, 5149, 5440, 5739, 6046, 6361, 6684, 7015, 7354, 7701, 8056, 8419, 8790

Offset: 1

Enoch Haga, G. L. Honaker, Jr., Apr 10 2000

Move in 1-4 direction in a spiral organized like A068225 etc.
Equals binomial transform of [1, 3, 8, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008
Ulam's spiral (N spoke). - Robert G. Wilson v, Oct 31 2011
Also, numbers of the form m*(4*m+1)+1 for nonpositive m. - Bruno Berselli, Jan 06 2016

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Robert G. Wilson v, Cover of the March 1964 issue of Scientific American
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A054555, A068225, A054552, A054554, A054567, A054569, A033951.
Cf. A266883: m*(4*m+1)+1 for m = 0,-1,1,-2,2,-3,3,...
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A001105, A058331, A130883.

[4*n^2-9*n+6 : n in [1..50]]; // Vincenzo Librandi, Mar 10 2018

a:=n->4*n^2-9*n+6: seq(a(n),n=1..50); # Muniru A Asiru, Mar 09 2018

a[n_] := 4*n^2 - 9*n + 6; Array[a, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *)
LinearRecurrence[{3,-3,1},{1,4,15},50] (* Harvey P. Dale, Sep 06 2015 *)
CoefficientList[Series[-(6x^2 + x + 1)/(x - 1)^3, {x, 0, 49}], x] (* Robert G. Wilson v, Mar 12 2018 *)

a(n)=4*n^2-9*n+6 \\ Charles R Greathouse IV, Sep 24 2015

a(n)^2 = Sum_{i = 0..2*(4*n-5)} (4*n^2-13*n+9+i)^2*(-1)^i = ((n-1)*(4*n-5)+1)^2. - Bruno Berselli, Apr 29 2010
From Harvey P. Dale, Aug 21 2011: (Start)
a(0)=1, a(1)=4, a(2)=15; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -x*(6*x^2+x+1)/(x-1)^3. (End)
From Franck Maminirina Ramaharo, Mar 09 2018: (Start)
a(n) = binomial(2*n - 2, 2) + 2*(n - 1)^2 + 1.
a(n) = A000384(n-1) + A058331(n-1).
a(n) = A130883(n-1) + A001105(n-1). (End)
E.g.f.: exp(x)*(6 - 5*x + 4*x^2) - 6. - Stefano Spezia, Apr 24 2024

Edited by Frank Ellermann, Feb 24 2002

Incorrect formula deleted by N. J. A. Sloane, Aug 02 2009

A054567 a(n) = 4n^2 - 7n + 4.

Original entry on oeis.org

1, 6, 19, 40, 69, 106, 151, 204, 265, 334, 411, 496, 589, 690, 799, 916, 1041, 1174, 1315, 1464, 1621, 1786, 1959, 2140, 2329, 2526, 2731, 2944, 3165, 3394, 3631, 3876, 4129, 4390, 4659, 4936, 5221, 5514, 5815, 6124, 6441, 6766, 7099, 7440, 7789, 8146, 8511, 8884

Offset: 1

Enoch Haga, G. L. Honaker, Jr., Apr 10 2000

The number 1 is placed in the middle of a sheet of squared paper and the numbers 2, 3, 4, 5, 6, etc. are written in a clockwise spiral around 1, as in A068225 etc. This sequence is read off along one of the rays from 1.
Ulam's spiral (W spoke of A054552). - Robert G. Wilson v, Oct 31 2011
Also, numbers of the form m*(4*m+1)+1 for nonnegative m. - Bruno Berselli, Jan 06 2016
The sequence forms the 1x2 diagonal of the square maze arrangement in A081344. - Jarrod G. Sage, Jul 17 2024

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Robert G. Wilson v, Cover of the March 1964 issue of Scientific American
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A054566, A068225, A054552, A054554, A054556, A054569, A033951, A204674.
Cf. A266883: m*(4*m+1)+1 for m = 0,-1,1,-2,2,-3,3,...
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Table[4 n^2 - 7 n + 4, {n, 100}] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *)

Vec(-x*(4*x^2+3*x+1)/(x-1)^3 + O(x^100)) \\ Colin Barker, Oct 25 2014

a(n) = 8*n+a(n-1)-11 for n>1, a(1)=1. - Vincenzo Librandi, Aug 07 2010
a(n) = A204674(n-1) / n. - Reinhard Zumkeller, Jan 18 2012
From Colin Barker, Oct 25 2014: (Start)
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3).
G.f.: -x*(4*x^2+3*x+1) / (x-1)^3. (End)
E.g.f.: exp(x)*(4 - 3*x + 4*x^2) - 4. - Stefano Spezia, Apr 24 2024
a(n) = A016742(n-1) + n. - Jarrod G. Sage, Jul 17 2024

Edited by Frank Ellermann, Feb 24 2002

Typo fixed by Charles R Greathouse IV, Oct 28 2009

A068608 Path of a knight's tour on an infinite chessboard.

Original entry on oeis.org

1, 10, 3, 16, 19, 22, 9, 12, 15, 18, 7, 24, 11, 14, 5, 20, 23, 2, 13, 4, 17, 6, 21, 8, 25, 50, 27, 54, 31, 60, 35, 64, 67, 40, 71, 74, 45, 78, 49, 52, 29, 56, 59, 34, 63, 66, 39, 70, 43, 76, 47, 80, 51, 28, 55, 58, 33, 62, 37, 68

Offset: 0

Hans Secelle and Albrecht Heeffer (albrecht.heeffer(AT)pandora.be), Mar 09 2002

One of eight possible knight's tours. Squares are numbered in a clockwise spiral. Enumerates all positive integers.

A description of the method to construct the tour is provided in A306659. - Hugo Pfoertner, May 11 2019

Hugo Pfoertner, Table of n, a(n) for n = 0..1088
Frank Ellermann, Sequences based on a spiral of the natural numbers
Hugo Pfoertner, Illustration of tours corresponding to A068608-A068615, (2019).
Dan Thomasson, Knight's Tour Art, (2001-2014).

Cf. A068609, A068610, A068611, A068612, A068613, A068614, A068615, A306659, A306660.

\\Ellermann's clockwise square spiral, first step (0,0) -> (0,1)
y=vector(10000);L=0;d=1;n=0;
for(r=1, 100, d=-d; k=floor(r/2)*d; for(j=1, L++, y[n++]=k); forstep(j=k-d, -floor((r+1)/2)*d+d, -d, y[n++]=j));
x=vector(10100);L=1;d=-1;n=0;
for(r=1, 100, d=-d; k=floor(r/2)*d; for(j=1, L++, x[n++]=-k); forstep(j=k-d, -floor((r+1)/2)*d+d, -d, x[n++]=-j));
\\ Position in spiral
findpos(i,j)={my(size=(2*max(abs(i),abs(j))+1)^2);forstep(k=size,1,-1, if(i==x[k]&&j==y[k], return(k)))};
atan2(y,x)=if(x>0,atan(y/x),if(x==0,if(y>0,Pi/2,-Pi/2),if(y>=0,atan(y/x)+Pi,atan(y/x)-Pi)));
angle(v,w)=atan2(v[1]*w[2]-v[2]*w[1],v[1]*w[1]+v[2]*w[2]);
move=[2,1;1,2;-1,2;-2,1;-2,-1;-1,-2;1,-2;2,-1]; \\ 8 Knight moves
m=6;b=matrix(2*m+1, 2*m+1, i, j, 0);setb(pos)={b[pos[1]+m+1, pos[2]+m+1]=1};
getb(pos)=b[pos[1]+m+1, pos[2]+m+1];
inring(n, p)=!(abs(p[1])angmin, jmin=j; angmin=adiff;jlast=j)))));if(jmin>0,p+=move[jmin,];setb(p);););p+=move[jlast,];setb(p)); \\ Hugo Pfoertner, May 11 2019

A068613 Path of a knight's tour on an infinite chessboard.

Original entry on oeis.org

1, 24, 7, 18, 15, 12, 9, 22, 19, 16, 3, 10, 23, 20, 5, 14, 11, 8, 21, 6, 17, 4, 13, 2, 25, 80, 47, 76, 43, 70, 39, 66, 63, 34, 59, 56, 29, 52, 49, 78, 45, 74, 71, 40, 67, 64, 35, 60, 31, 54, 27, 50, 79, 46, 75, 72, 41, 68, 37, 62

Offset: 0

Hans Secelle and Albrecht Heeffer (albrecht.heeffer(AT)pandora.be), Mar 09 2002

One of eight possible knight's tours. Squares are numbered in a clockwise spiral. Enumerates all positive integers.

Hugo Pfoertner, Table of n, a(n) for n = 0..1088
Frank Ellermann, Sequences based on a spiral of the natural numbers

Cf. A068608, A068609, A068610, A068611, A068612, A068614, A068615.

A020703 Take the sequence of natural numbers (A000027) and reverse successive subsequences of lengths 1,3,5,7,...

Original entry on oeis.org

1, 4, 3, 2, 9, 8, 7, 6, 5, 16, 15, 14, 13, 12, 11, 10, 25, 24, 23, 22, 21, 20, 19, 18, 17, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 81, 80, 79, 78, 77

Offset: 1

N. J. A. Sloane, May 02 2000

nonn

Arrange A000027, the natural numbers, into a (square) spiral, say clockwise as shown in A068225. Read the numbers from the resulting counterclockwise spiral of the same shape that also begins with 1 and this sequence results. - Rick L. Shepherd, Aug 04 2006
Contribution from Hieronymus Fischer, Apr 30 2012: (Start)
The sequence may also be defined as follows: a(1)=1, a(n)=m^2 (where m^2 is the least square > a(k) for 1<=kA reordering of the natural numbers.
The sequence is self-inverse in that a(a(n))=n.
(End)

			a(2)=4=2^2, since 2^2 is the least square >2=a(1) and the minimal number not yet in the sequence is 2>1=a(1);
a(8)=6=a(7)-1, since the minimal number not yet in the sequence (=5) is <=7=a(7).

R. Honsberger, "Ingenuity in Mathematics", Table 10.4 on page 87.
Suggested by correspondence with Michael Somos.

Hieronymus Fischer, Table of n, a(n) for n = 1..11131
Index entries for sequences that are permutations of the natural numbers

A self-inverse permutation of the natural numbers.
Cf. A000027, A038722.
Cf. A132666, A132664, A132665, A132674.

Flatten[Table[Range[n^2,(n-1)^2+1,-1],{n,10}]] (* Harvey P. Dale, Jan 10 2016 *)
With[{nn=20},Flatten[Reverse/@TakeList[Range[nn^2],Range[1,nn,2]]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Jan 28 2019 *)

a(n)=local(t); if(n<1,0,t=sqrtint(n-1); 2*(t^2+t+1)-n)

Contribution from Hieronymus Fischer, Apr 30 2012: (Start)

a(n)=a(n-1)-1, if a(n-1)-1 > 0 is not in the set {a(k)| 1<=k

a(n)=n for n=k(k+1)+1, k>=0.

a(n+1)=(sqrt(a(n)-1)+2)^2, if a(n)-1 is a square, a(n+1)=a(n)-1, else.

a(n)=2*(floor(sqrt(n-1))+1)*floor(sqrt(n-1))-n+2. (End)

A068610 Path of a knight's tour on an infinite chessboard.

Original entry on oeis.org

1, 18, 7, 24, 11, 14, 5, 20, 23, 10, 3, 16, 19, 22, 9, 12, 15, 6, 21, 8, 25, 2, 13, 4, 17, 66, 39, 70, 43, 76, 47, 80, 51, 28, 55, 58, 33, 62, 37, 68, 41, 72, 75, 46, 79, 50, 27, 54, 31, 60, 35, 64, 67, 40, 71, 74, 45, 78, 49, 52

Offset: 0

Hans Secelle and Albrecht Heeffer (albrecht.heeffer(AT)pandora.be), Mar 09 2002

nonn

One of eight possible knight's tours. Squares are numbered in a clockwise spiral. Enumerates all positive integers.

Hugo Pfoertner, Table of n, a(n) for n = 0..1088 [Offset adapted by _Georg Fischer_, Jun 04 2019]
Frank Ellermann, Sequences based on a spiral of the natural numbers
Hugo Pfoertner, Illustration of initial terms.

Cf. A068608, A068609, A068611, A068612, A068613, A068614, A068615.

A068609 Path of a knight's tour on an infinite chessboard.

Original entry on oeis.org

1, 14, 5, 20, 23, 10, 3, 16, 19, 22, 9, 12, 15, 18, 7, 24, 11, 4, 17, 6, 21, 8, 25, 2, 13, 58, 33, 62, 37, 68, 41, 72, 75, 46, 79, 50, 27, 54, 31, 60, 35, 64, 67, 40, 71, 74, 45, 78, 49, 52, 29, 56, 59, 34, 63, 66, 39, 70, 43, 76

Offset: 0

Hans Secelle and Albrecht Heeffer (albrecht.heeffer(AT)pandora.be), Mar 09 2002

nonn

One of eight possible knight's tours. Squares are numbered in a clockwise spiral. Enumerates all positive integers.

Hugo Pfoertner, Table of n, a(n) for n = 0..1088 [Offset adapted by _Georg Fischer_, Jun 04 2019]
Frank Ellermann, Sequences based on a spiral of the natural numbers

Cf. A068608, A068610, A068611, A068612, A068613, A068614, A068615.

a(56), which had been shown as 37, corrected to 39 by Kevin Ryde, Jan 17 2012

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A121496 Run lengths of successive numbers in A068225.

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A054569 a(n) = 4*n^2 - 6*n + 3.

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A054554 a(n) = 4*n^2 - 10*n + 7.

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A054556 a(n) = 4*n^2 - 9*n + 6.

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A054567 a(n) = 4*n^2 - 7*n + 4.

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Mathematica

PARI

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A068608 Path of a knight's tour on an infinite chessboard.

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PARI

A068613 Path of a knight's tour on an infinite chessboard.

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A054569 a(n) = 4n^2 - 6n + 3.

A054554 a(n) = 4n^2 - 10n + 7.

A054556 a(n) = 4n^2 - 9n + 6.

A054567 a(n) = 4n^2 - 7n + 4.