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

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

Author

N. J. A. Sloane, May 02 2000

Keywords

• nonn

Comments

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<=k

A reordering of the natural numbers.

The sequence is self-inverse in that a(a(n))=n.

(End)

Examples

			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).
		

References

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

Links

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

Crossrefs

A self-inverse permutation of the natural numbers.

Cf. A000027, A038722.

Cf. A132666, A132664, A132665, A132674.

Programs

Mathematica

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 *)

PARI

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

Formula

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)

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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