A090915 Permutation of natural numbers arising from a square spiral.

1, 8, 7, 6, 5, 4, 3, 2, 9, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 25, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 49, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58

Offset: 1

Author

Felix Tubiana, Feb 26 2004

Keywords

• easy
• nonn

Comments

Write out the natural numbers in a square counterclockwise spiral:

.

17--16--15--14--13

| |

18 5---4---3 12

| | | |

19 6 1---2 11

| | |

20 7---8---9--10

|

21--22--23--24--25

.

Now read off the numbers in a square clockwise spiral: 1 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 9 -> etc.

Links

• Eric M. Schmidt, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A020703, A090861, A090925, A090928, A090929, A090930.

Programs

Mathematica

With[{x = Floor[(Floor[Sqrt[n-1]]+1)/2]}, Table[If[n==(2*x+1)^2, n, 8*x^2 -n+2], {n, 1, 75}]] (* G. C. Greubel, Feb 05 2019 *)

PARI

{s(n) = ((sqrtint(n-1)+1)/2)\1};
for(n=1,75, print1(if(n == (2*s(n)+1)^2, n, 8*s(n)^2-n+2), ", ")) \\ G. C. Greubel, Feb 05 2019

Sage

def a(n):
    x = (isqrt(n-1)+1)//2
    return n if n == (2*x+1)^2 else 8*x^2 + 2 - n
[a(n) for n in (1..75)] # Eric M. Schmidt, May 18 2016

Extensions

Offset corrected by Eric M. Schmidt, May 18 2016