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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)
12
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 (list; graph; refs; listen; history; text; internal format)
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 lim_{n->infinity} 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
Clark Kimberling and David Callan, Problem E3163, Amer. Math. Monthly, 96 (1989), 57.
FORMULA
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 *)
PROG
(PARI) a(n) = if(n==1, 1, my(n=n-1, A=n, B); for(i=1, n-1, B=n-i; A=A+B-2*(A%B)); A+1) \\ Mikhail Kurkov, Feb 24 2023
CROSSREFS
Cf. A057030 (here we have "s(1), ..., s(n)", whereas 057030 has "s(1), ..., s(n-1)").
Sequence in context: A154686 A358511 A165196 * A121817 A359337 A116432
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms and better description from Clark Kimberling, Jul 28 2000
STATUS
approved

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Last modified March 19 03:33 EDT 2024. Contains 370952 sequences. (Running on oeis4.)