A220098 Manhattan distances between 2n and 1 in the double spiral with positive integers and 1 at the center.

1, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 5, 6, 5, 4, 3, 4, 5, 6, 7, 6, 5, 4, 5, 6, 7, 8, 7, 6, 5, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 6, 7, 8, 9, 10, 11, 10, 9, 8, 7, 6, 7, 8, 9, 10, 11, 12, 11, 10, 9, 8, 7, 6, 7, 8, 9, 10, 11, 12, 13

Offset: 1

Author

Alex Ratushnyak, Dec 04 2012

Keywords

• nonn
• easy

Comments

Double spiral begins:

.

82---84---86---88---90---92---94---96---98

|

80 51---53---55---57---59---61---63---65

| | |

78 49 26---28---30---32---34---36 67

| | | | |

76 47 24 11---13---15---17 38 69

| | | | | | |

74 45 22 9 2----4 19 40 71

| | | | | | | | |

72 43 20 7 1 6 21 42 73

| | | | | | | | |

70 41 18 5----3 8 23 44 75

| | | | | | |

68 39 16---14---12---10 25 46 77

| | | | |

66 37---35---33---31---29---27 48 79

| | |

64---62---60---58---56---54---52---50 81

|

99---97---95---93---91---89---87---85---83

Examples

			From _Philippe Deléham_, Mar 08 2013: (Start)
As a square array, this begins:
  1,  1,  2,  2,  3,  3,  4,  4,  5, ...
  2,  3,  3,  4,  4,  5,  5,  6,  6, ...
  2,  4,  5,  5,  6,  6,  7,  7,  8, ...
  3,  4,  6,  7,  7,  8,  8,  9,  9, ...
  3,  5,  6,  8,  9,  9, 10, 10, 11, ...
  4,  5,  7,  8, 10, 11, 11, 12, 12, ...
  4,  6,  7,  9, 10, 12, 13, 13, 14, ...
  5,  6,  8,  9, 11, 12, 14, 15, 15, ..., etc.
As a triangle, this begins:
  1
  2, 1
  2, 3, 2
  3, 4, 3, 2
  3, 4, 5, 4, 3
  4, 5, 6, 5, 4, 3, etc. (End)
		

Crossrefs

Cf. A053615, A077043, A128217, A214526.

Programs

C

#include 
#define SIZE 20
int grid[SIZE][SIZE];
int direction[] = {0, -1,  1, 0, 0, 1, -1, 0};
main() {
  int i, j, x1, y1, x2, y2, stepSize;
  int direction1pos=0, direction2pos=4, val;
  x1 = y1 = x2 = y2 = SIZE/2;
  for (val=grid[y1][x1]=1, stepSize=0; ; ++stepSize) {
    if (x1<1 || x1>=SIZE-1 || x2<1 || x2>=SIZE-1) break;
    if (y1<1 || y1>=SIZE-1 || y2<1 || y2>=SIZE-1) break;
    for (i=stepSize|1; i; ++val,--i) {
      x1 += direction[direction1pos  ];
      y1 += direction[direction1pos+1];
      x2 += direction[direction2pos  ];
      y2 += direction[direction2pos+1];
      grid[y1][x1] = val*2;
      grid[y2][x2] = val*2+1;
      printf("%d, ",abs(x1-SIZE/2)+abs(y1-SIZE/2));
    }
    direction1pos = (direction1pos+2) & 7;
    direction2pos = (direction2pos+2) & 7;
  }
  for (i=0; i
				

PARI

step(v, m) = concat(v, vector(m, k, 1+v[#v-k+1]))
a(max_n) = {my(v=[0], k=1); while(#v < max_n+1, v=step(v,k); k++); v[2..max_n+1]} \\ Thomas Scheuerle, Jan 07 2025

PARI

A053615(n) = if(n<1, 0, sqrtint(n) - A053615(n - sqrtint(n)))
a(n) = A053615(floor( floor( (sqrtint(n*8) + 1)/2 )^2/2 ) + n) \\ Thomas Scheuerle, Jan 07 2025

Formula

abs( a(n) - a(n-1) ) = 1.

From Thomas Scheuerle, Jan 07 2025: (Start)

a(n*(n+1)/2 - k) = 1 + a(n*(n-1)/2 + k) with a(0) = 0 and for 0 <= k < n.

a(n) = A053615(A128217(n+1)). (End)