A304586 -id:A304586 - cp's OEIS frontend

A326413 Successive squares visited by a knight on the single-digit square spiral, with ties resolved towards the left.

0, 0, 1, 0, 1, 0, 1, 1, 2, 2, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 2, 2, 1, 1, 0, 2, 3, 2, 2, 1, 3, 1, 1, 1, 1, 1, 2, 2, 3, 2, 3, 1, 4, 3, 5, 6, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0

Offset: 1

N. J. A. Sloane, Oct 17 2019

Take the standard counterclockwise square spiral starting at 0, as in A304586, but only write one digit at a time in the cells of the spiral: 0,1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,...
Place a chess knight at cell 0. Move it to the lowest-numbered cell it can attack, and if there is a tie, move it to the cell closest (in Euclidean distance) to the start, and if there is still a tie, move to the left(*).
No cell can be visited more than once.
Inspired by the Trapped Knight video and A316667.
Just as for A316667, the sequence is finite. After a while, the knight has no unvisited squares it can reach, and the sequence ends with a(1217) = 4.
(*)Moving to the left means choose the point with the lowest x-coordinate. This leads to an unambiguous choice of tied squares only for the 'move left' case.

			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

Luca Petrone, Table of n, a(n) for n = 1..1217
Eric Angelini, Kneil's Knumberphile Knight, Cinquante signes, May 04 2019.
Eric Angelini, Kneil's Knumberphile Knight, Cinquante signes, May 04 2019. [Cached copy, pdf file, with permission]
M. F. Hasler, Knight tours, OEIS wiki, Nov. 2019.
N. J. A. Sloane and Brady Haran, The Trapped Knight, Numberphile video (2019).

Cf. A304586, A316667, A328698.

Cf. A326916, A326918; A316328; A326924, A326922; A328908, A328928; A328909, A328929.

More terms from Luca Petrone

Corrected and extended by Eric Angelini, Oct 24 2019

A334742 Pascal's spiral: starting with a(1) = 1, proceed in a square spiral, computing each term as the sum of horizontally and vertically adjacent prior terms.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 10, 12, 12, 14, 17, 20, 20, 23, 27, 32, 37, 37, 42, 48, 55, 62, 62, 69, 77, 87, 99, 111, 111, 123, 137, 154, 174, 194, 194, 214, 237, 264, 296, 333, 370, 370, 407, 449, 497, 552, 614, 676, 676, 738, 807, 884, 971, 1070

Offset: 1

Alec Jones and Peter Kagey, May 09 2020

This is the square spiral analogy of Pascal's triangle thought of as a table read by antidiagonals.

			Spiral begins:
  111--99--87--77--69--62
                        |
   12--12--10---8---7  62
    |               |   |
   14   2---2---1   7  55
    |   |       |   |   |
   17   3   1---1   6  48
    |   |           |   |
   20   3---4---5---5  42
    |                   |
   20--23--27--32--37--37
a(15) = 10 = 8 + 2, the sum of the cells immediately to the right and below. The term to the left is not included in the sum because it has not yet occurred in the spiral.

Peter Kagey, Table of n, a(n) for n = 1..10000
Peter Kagey, Bitmap illustrating the parity of the first 2^20=1048576 terms. (Even and odd numbers are represented with black and white pixels respectively.)

Cf. A007318, A141481, A334688.
Cf. A180714, A214526, A274640, A278180, A304586, A307834, A334741.
x- and y-coordinates are given by A174344 and A274923, respectively.

a(A033638(n)) = a(A002620(n)) for n > 1.

A304584 A linear mapping a(n) = x + d*n of pairs of nonnegative integers (x,d), where the pairs are enumerated by antidiagonals.

Original entry on oeis.org

0, 1, 2, 2, 5, 10, 3, 9, 17, 27, 4, 14, 26, 40, 56, 5, 20, 37, 56, 77, 100, 6, 27, 50, 75, 102, 131, 162, 7, 35, 65, 97, 131, 167, 205, 245, 8, 44, 82, 122, 164, 208, 254, 302, 352, 9, 54, 101, 150, 201, 254, 309, 366, 425, 486, 10, 65, 122, 181, 242, 305, 370, 437, 506, 577, 650, 11

Offset: 0

Hugo Pfoertner, May 15 2018

nonn

The sequence solves the following riddle, which has been communicated by Klaus Nagel: A flea starts to jump on the nonnegative integers at time = 0 at an unknown location x >= 0 making jumps of unknown, but constant distance d >= 0 at every subsequent time step. By which strategy can the flea be captured with 100% certainty in a finite number of trials? The solution is to hit a(n) at time = n. This works for all enumerations of pairs (x,d) of integers, because eventually any combination of starting location x and jump width d will be addressed.

			  d:
  5 |  20
  4 |  14  19
  3 |   9  13  18
  2 |   5   8  12  17
  1 |   2   4   7  11  16
  0 |   0   1   3   6  10  15
    |________________________
  x:    0   1   2   3   4   5
.
a(13) = 1 + 13*3 = 40 because the 13th position in the enumeration corresponds to x=1 and d=3.

Rainer Rosenthal, Table of n, a(n) for n = 0..10000

Cf. A304585, A304586, A304587, A305260.

pos2pair:=proc(n) local w,k,e;w:=floor(sqrt(2*n));if w*(w+1)>2*n then k:=w-1;else k:=w;fi;e:=n-k*(k+1)/2;return [k-e,e];end:WhereFlea:=proc(n) local x,d,pair; pair:=pos2pair(n);x:=pair[1];d:=pair[2];return x+d*n;end:
seq(WhereFlea(n),n=0..66);# Rainer Rosenthal, May 23 2018

A304585 A linear mapping a(n) = x + d*n of pairs of nonnegative integers (x,d), where the pairs are enumerated by meandering antidiagonals.

Original entry on oeis.org

0, 1, 2, 6, 5, 2, 3, 9, 17, 27, 40, 34, 26, 16, 4, 5, 20, 37, 56, 77, 100, 126, 111, 94, 75, 54, 31, 6, 7, 35, 65, 97, 131, 167, 205, 245, 288, 260, 230, 198, 164, 128, 90, 50, 8, 9, 54, 101, 150, 201, 254, 309, 366, 425, 486, 550, 505, 458, 409, 358, 305, 250, 193, 134, 73, 10, 11

Offset: 0

Hugo Pfoertner, May 16 2018

nonn

The sequence is an alternative solution to the riddle described in the comments of A304584.

			  d:
  5 |  20
  4 |  10  19
  3 |   9  11  18
  2 |   3   8  12  17
  1 |   2   4   7  13  16
  0 |   0   1   5   6  14  15
    |________________________
  x:    0   1   2   3   4   5
.
a(13)= 3 + 13*1 = 16 because the 13th position in the enumeration corresponds to x=3 and d=1.

Rainer Rosenthal, Table of n, a(n) for n = 0..10000

Cf. A304584, A304586, A304587.

pos2pM:=proc(n) local w,k,e;w:=floor(sqrt(2*n));if w*(w+1)>2*n then k:=w-1;else k:=w;fi;e:=n-k*(k+1)/2;if modp(k,2)=1 then return [k-e,e];else return [e,k-e];fi end: WhereFlea:=proc(n) local x,d,pair; pair:=pos2pM(n);x:=pair[1];d:=pair[2];return x+d*n;end: seq(WhereFlea(n),n=0..66); # Rainer Rosenthal, May 23 2018

A304587 A linear mapping a(n) = x + d*n of pairs of integers (x,d), where the pairs are enumerated by a number spiral along antidiagonals.

Original entry on oeis.org

0, 1, 2, -1, -4, 2, 7, 14, 7, -2, -11, -22, -11, 3, 16, 31, 48, 33, 16, -3, -22, -43, -66, -45, -22, 4, 29, 56, 85, 116, 89, 60, 29, -4, -37, -72, -109, -148, -113, -76, -37, 5, 46, 89, 134, 181, 230, 187, 142, 95, 46, -5, -56, -109, -164, -221, -280, -227, -172, -115, -56, 6, 67

Offset: 0

Hugo Pfoertner, May 16 2018

The sequence is an alternative solution to the riddle described in the comments of A304584 without the restriction of x and d to nonnegative numbers.

			   d:
   3 |              16  28
     |             /   \   \
   2 |          17   7  15  27
     |         /   /   \   \   \
   1 |      18   8   2   6  14  26
     |     /   /   /   \   \   \   \
   0 |  19   9   3   0---1   5  13  25
     |     \   \   \    --> --> -->
  -1 |      20  10   4  12  24
     |         \   \  /   /
  -2 |          21  11  23
     |             \   /
  -3 |              22
    __________________________________
  x:    -3  -2  -1   0   1   2   3   4
.
a(10) = -1 + 10*(-1) = -11 because the 10th position in the spiral corresponds to x = -1 and d = -1,
a(15) = 1 + 15*2 = 31 because the 15th position in the spiral corresponds to x = 1 and d = 2,
a(25) = 4 + 25*0 = 4 because the 25th position in the spiral corresponds to x = 4 and d = 0.

Rainer Rosenthal, Table of n, a(n) for n = 0..10000

Cf. A001844 (where the spiral jumps to next ring), A304584, A304585, A304586.

n2left := proc(n)local w,k;return floor(sqrt((n-1)/2));end:pos2pH:=proc(n)local k,q,Q,e,E,sp;k:=n2left(n);q:=2*k^2+1;Q:=2*(k+1)^2+1;e:=n-q;E:=Q-n;if n<2 then return[n,0];fi;if e<=k then return[-k+e,-e];elif e<2*k then return[-k+e,-2*k+e];elif E<=k+1 then return[-(k+1)+E,E];else return[E-(k+1),2*(k+1)-E];fi;end:WhereFlea:=proc(n) local x,d,pair; pair:=pos2pH(n);x:=pair[1];d:=pair[2];return x+d*n;end: seq(WhereFlea(n),n=0..62);# Rainer Rosenthal, May 28 2018

def a(n):
    if n<2: return n
    k = isqrt((n-1)/2)
    e = n-k*(2*k+1)-1
    x = e if ePeter Luschny, May 29 2018

a(1) corrected by Rainer Rosenthal, May 28 2018

A334741 Fill an infinite square array by following a spiral around the origin; in the central cell enter a(0)=1; thereafter, in the n-th cell, enter the sum of the entries of those earlier cells that are in the same row or column as that cell.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 8, 11, 21, 40, 47, 93, 180, 203, 397, 796, 1576, 1675, 3305, 6636, 13192, 14004, 27607, 55029, 110192, 220024, 226740, 450123, 898661, 1798700, 3594248, 3704800, 7354303, 14681369, 29349536, 58710640, 117394896, 119196748, 237492079

Offset: 0

Alec Jones and Peter Kagey, May 09 2020

nonn

The spiral track being used here is the same as in A274640, except that the starting cell here is indexed 0 (as in A274641).

The central cell gets index 0 (and we fill it in with the value a(0)=1).

			Spiral begins:
     3----2----1
     |         |
     5    1----1   47
     |              |
     8---11---21---40
a(11) = 47 = 1 + 1 + 5 + 40, the sum of the cells in its row and column.

Peter Kagey, Table of n, a(n) for n = 0..2499

Cf. A280027.
Cf. A180714, A214526, A274640, A278180, A304586, A307834, A334742.
x- and y-coordinates are given by A174344 and A274923, respectively.

\\ here P(n) returns A174344 and A274923 as pair.
P(n)={my(m=sqrtint(n), k=ceil(m/2)); n -= 4*k^2; if(n<0, if(n<-m, [k, 3*k+n], [-k-n, k]), if(nAndrew Howroyd, May 09 2020

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A326413 Successive squares visited by a knight on the single-digit square spiral, with ties resolved towards the left.

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A334742 Pascal's spiral: starting with a(1) = 1, proceed in a square spiral, computing each term as the sum of horizontally and vertically adjacent prior terms.

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A304584 A linear mapping a(n) = x + d*n of pairs of nonnegative integers (x,d), where the pairs are enumerated by antidiagonals.

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A304585 A linear mapping a(n) = x + d*n of pairs of nonnegative integers (x,d), where the pairs are enumerated by meandering antidiagonals.

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A304587 A linear mapping a(n) = x + d*n of pairs of integers (x,d), where the pairs are enumerated by a number spiral along antidiagonals.

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A334741 Fill an infinite square array by following a spiral around the origin; in the central cell enter a(0)=1; thereafter, in the n-th cell, enter the sum of the entries of those earlier cells that are in the same row or column as that cell.

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