A057033 Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...s(n-1) fixed and reverse-cyclically permuting every n consecutive terms thereafter; apply P(2) to 1,2,3,... to get PS(2), then apply P(3) to PS(2) to get PS(3), then apply P(4) to PS(3), etc. The limit of PS(n) is A057033.

1, 3, 5, 2, 9, 11, 6, 15, 17, 10, 21, 12, 4, 27, 29, 18, 19, 35, 22, 39, 41, 8, 45, 28, 30, 51, 36, 34, 57, 59, 14, 43, 65, 42, 69, 71, 24, 53, 77, 7, 81, 60, 54, 87, 64, 58, 67, 95, 26, 99, 101, 37, 105, 107, 70, 111, 84, 32, 88, 93, 78, 47

Offset: 1

Author

Clark Kimberling, Jul 29 2000

Keywords

• nonn
• nice
• easy

Examples

			PS(2) begins with 1,3,2,5,4,7,6; PS(3) with 1,3,5,4,2,6,9; PS(4) with 1,3,5,2,6,9,4.
		

Crossrefs

Cf. A007062, A057030, A057032, A057063, A057064, A069836.

Programs

Mathematica

terms = 62; maxTerms = terms^2; Clear[PS]; PS[n_] := PS[n] = If[n == 1, Range[maxTerms], Join[PS[n-1][[1 ;; n-1]], RotateLeft /@ Partition[ PS[n-1][[n ;; All]], n]] // Flatten]; PS[1]; PS[n = 2]; While[PS[n][[1 ;; terms]] != PS[n-1][[1 ;; terms]], n++]; A057033 = PS[n][[1 ;; terms]] (* Jean-François Alcover, Apr 24 2017 *)

Formula

Conjecture: a(n) = A057063(n+1) - 1 for n > 0. - Mikhail Kurkov, Mar 08 2023