A007062 Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...,s(n) fixed and reversing 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. This sequence is the limit of PS(n). (Formerly M0966)

1, 2, 4, 5, 7, 12, 14, 15, 23, 28, 30, 41, 43, 48, 56, 67, 69, 84, 86, 87, 111, 116, 124, 139, 141, 162, 180, 181, 183, 224, 232, 237, 271, 276, 278, 315, 333, 338, 372, 383, 385, 426, 428, 439, 473, 478, 538, 543, 551, 556, 620

Offset

1,2

Comments

From Gerald McGarvey, Aug 05 2004: (Start)

Consider the following array:

.1..2..3..4..5..6..7..8..9.10.11.12.13.14.15.16.17.18.19.20

.2..1..4..3..6..5..8..7.10..9.12.11.14.13.16.15.18.17.20.19

.4..1..2..5..6..3.10..7..8.11.12..9.16.13.14.17.18.15.22.19

.5..2..1..4..7.10..3..6..9.12.11..8.17.14.13.16.19.22.15.18

.7..4..1..2..5.12..9..6..3.10.13.14.17..8.11.18.15.22.19.16

12..5..2..1..4..7.14.13.10..3..6..8.22.15.18.11..8.17.24.23

14..7..4..1..2..5.12.15.22..8..6..3.10.13.20.23.24.17..8.11

15.12..5..2..1..4..7.14.23.20.13.10..3..6..8.22.25.28.31.18

23.14..7..4..1..2..5.12.15.28.25.22..8..6..3.10.13.20.33.30

28.15.12..5..2..1..4..7.14.23.30.33.20.13.10..3..6..8.22.25

which is formed as follows:

. first row is the positive integers

. second row: group the first row in pairs of two and reverse the order within groups; e.g., 1 2 -> 2 1 and 3 4 -> 4 3

. n-th row: group the (n-1)st row in groups of n and reverse the order within groups

This sequence is the first column of this array, as well as the diagonal excluding the diagonal's first term. It is also various other 'partial columns' and 'partial diagonals'.

To calculate the i-th column / j-th row value, one can work backwards to find which column of the first row it came from. For each row, first reverse its position within the group, then go up. It appears that lim_{n->oo} a(n)/n^2 exists and is ~ 0.22847 ~ sqrt(0.0522). (End)

References

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Peter J. C. Moses, Table of n, a(n) for n = 1..5000

Clark Kimberling and David Callan, Problem E3163, Amer. Math. Monthly, 96 (1989), 57.

Formula

Conjecture: a(n) = A057030(n-1) + 1 for n > 1 with a(1) = 1. - Mikhail Kurkov, Feb 24 2023

Example

PS(2) begins with 1,2,4,3,6,5,8; PS(3) with 1,2,4,5,6,3,10; PS(4) with 1,2,4,5,7,10,3.

Mathematica

(* works per the name description *)
a007062=Range[x=3500]; Do[a007062=Flatten[Join[{Take[a007062, n]}, Map[Reverse, Partition[Drop[a007062, n], n]]]], {n, 2, NestWhile[#+1&, 1, (x=# Floor[x/#])>0&]-1}]; a007062
(* works by making McGarvey's array *) a=Range[x=10000]; rows=Table[a=Flatten[Map[Reverse, Partition[a, n]]], {n, NestWhile[#+1&, 1, (x=# Floor[x/#])>0&]-1}]; a007062=Map[First, rows] (* Peter J. C. Moses, Nov 10 2016 *)

Crossrefs

Cf. A057030 (here we have "s(1), ..., s(n)", whereas 057030 has "s(1), ..., s(n-1)").

Cf. A057032, A057033, A057063, A057064.

Sequence in context: A154686 A358511 A165196 * A121817 A359337 A116432

Adjacent sequences: A007059 A007060 A007061 * A007063 A007064 A007065

Keyword

nonn

Author

N. J. A. Sloane, Mira Bernstein

Extensions

More terms and better description from Clark Kimberling, Jul 28 2000

Status

approved