A063826 Let 1, 2, 3, 4 represent moves to the right, down, left and up; this sequence describes the movements in the clockwise square spiral (a.k.a. Ulam Spiral).

1, 2, 3, 3, 4, 4, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4

Offset: 0

Author

Wai Ha Lee (Wainson(AT)hotmail.com), Aug 20 2001

Keywords

• easy
• nice
• nonn

Comments

Sequence starts with 1, 2, 3, then can be broken into groups of 8n+4 members, so if n is incremented, starting at 1, the groups follow the following pattern: 3 occurs at the beginning of the group, 4 then occurs 2n times, 1 occurs 2n+1 times, 2 occurs 2n+1 times, 3 occurs 2n+1 times; so each group has 8n+4 terms.

Simpler description: Groups of 2*(2n-1) + 2*(2n) = 8n - 2 terms, n = 1, 2, 3, ..., consisting of 2n-1 times 1, then 2n-1 times 2; then 2n times 3, then 2n times 4. The n-th group starts at index (4n - 6)n + 2 and ends at index (4n + 2)n - 1. - M. F. Hasler, Aug 08 2020

Examples

			Breaking into the groups, we have: 1, 2, 3 n=1: 3, 4, 4, 1, 1, 1, 2, 2, 2, 3, 3, 3, n=2: 3, 4, 4, 4, 4, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3 n=3: 3, 4, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3 and so on.
The spiral of numbers which shows in which order the steps in direction right, down, left and up are made, is depicted for example in sequence A174344. - _M. F. Hasler_, Aug 08 2020
		

Links

• Harry J. Smith, Table of n, a(n) for n = 0..1000
• Dario Alpern, Ulam's Spiral
• Adrian J. F. Leatherland, The mysterious Ulam spiral phenomenon

Crossrefs

Cf. A000267, A174344.

Programs

Mathematica

a[n_] := Mod[Floor[Sqrt[4*n + 1]] + 3, 4] + 1; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Nov 30 2016 adapted from PARI *)
Join[{1,2,3},Flatten[Table[{{3},PadRight[{},2n,4],Table[PadRight[{},2n+1,k],{k,3}]},{n,5}]]] (* Harvey P. Dale, Jun 29 2019 *)

PARI

A063826(n)=(sqrtint(4*n+1)+3)%4+1 \\ To see the terms: apply(A063826, [0..99])

Formula

1 <= a(n) <= 4 with a(n) == floor(sqrt(4n + 1)) (mod 4). - M. F. Hasler, Aug 08 2020