A057033 -id:A057033 - cp's OEIS frontend

A069836 Inverse permutation to A057033: a(n) is the m such that A057033(m) = n, or 0 if no such m exists.

1, 4, 2, 13, 3, 7, 40, 22, 5, 10, 6, 12, 121, 31, 8, 67, 9, 16, 17, 364, 11, 19, 202, 37, 94, 49, 14, 24, 15, 25, 1093, 58, 112, 28, 18, 27, 52, 607, 20, 283, 21, 34, 32, 76, 23, 148, 62, 73, 157, 85, 26, 175, 38, 43, 3280, 337, 29, 46, 30, 42, 850, 103, 82, 45, 33, 1822

Offset: 1

David Wasserman, Apr 23 2002

Are there any 0's in this sequence? I have computed that the first 1000 terms are all positive, the largest of these being a(980) = 1328602.

			a(4) = 13 because A057033(13) = 4.

Cf. A057033.

function n = back_cyclic(n) i = 1; while i < n i = i + 1; if mod(n, i) == 0 n = n + i - 1; else n = n - 1; end end

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).

Original entry on oeis.org

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

N. J. A. Sloane, Mira Bernstein

nonn

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)

			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.

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

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.

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

Cf. A057032, A057033, A057063, A057064.

(* 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 *)

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

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

A057032 Let P(n) of a sequence s(1), s(2), s(3), ... be obtained by leaving s(1), ..., s(n-1) fixed and forward-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) as n -> oo is this sequence.

Original entry on oeis.org

1, 3, 4, 7, 6, 10, 8, 16, 15, 21, 12, 22, 14, 27, 28, 36, 18, 33, 20, 43, 35, 39, 24, 53, 34, 45, 46, 50, 30, 66, 32, 78, 52, 57, 55, 81, 38, 63, 59, 88, 42, 86, 44, 96, 87, 75, 48, 119, 64, 111, 76, 101, 54, 103, 79, 144, 83, 93, 60, 141, 62, 99, 113, 173, 91, 136, 68, 139

Offset: 1

Clark Kimberling, Jul 29 2000

Conjecture: a(n) - 1 is prime if and only if a(n) = n + 1. - Mikhail Kurkov, Mar 10 2022

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

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A007062, A057030, A057033, A057063, A057064, A069829.

function m = A057032(i) m = PS(i, i); function m = PS(i, n) if i == 1 m = n; elseif n < i m = PS(i - 1, n); else if mod(n, i) == 0 m = PS(i - 1, n + i - 1); else m = PS(i - 1, n - 1); end end

PS[i_, n_] := If[i == 1, n, If[n < i, PS[i-1, n], If[Mod[n, i] == 0, PS[i-1, n+i-1], PS[i-1, n-1]]]]; a[n_] := PS[n, n]; Table[a[n], {n, 1, 68}] (* Jean-François Alcover, Oct 20 2011, after MATLAB *)

a(n) = { my (p=0); forstep (d=n, 1, -1, if (p%d==0, p+=d)); p } \\ Rémy Sigrist, Aug 25 2020

Conjecture: a(n) = A057064(n+1) - 1 for n > 0. - Mikhail Kurkov, Mar 10 2022

More terms from David Wasserman, Apr 22 2002

A057030 Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...,s(n-1) 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. The limit of PS(n) is A057030.

Original entry on oeis.org

1, 3, 4, 6, 11, 13, 14, 22, 27, 29, 40, 42, 47, 55, 66, 68, 83, 85, 86, 110, 115, 123, 138, 140, 161, 179, 180, 182, 223, 231, 236, 270, 275, 277, 314, 332, 337, 371, 382, 384, 425, 427, 438, 472, 477, 537, 542, 550, 555, 619, 630

Offset: 1

Clark Kimberling, Jul 29 2000

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

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

Cf. A007062, A057032, A057033, A057063, A057064.

a057030=Range[x=3500]; Do[a057030=Flatten[Join[{Take[a057030,n-1]},Map[Reverse,Partition[Drop[a057030,n-1],n]]]],{n,2,NestWhile[#+1&,1,(x=# Floor[x/#])>0&]-1}]; a057030 (* Peter J. C. Moses, Nov 10 2016 *)

Conjecture: a(n) = A007062(n+1) - 1 for n > 0. - Mikhail Kurkov, Mar 10 2022

A057063 Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...,s(n) 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 A057063.

Original entry on oeis.org

1, 2, 4, 6, 3, 10, 12, 7, 16, 18, 11, 22, 13, 5, 28, 30, 19, 20, 36, 23, 40, 42, 9, 46, 29, 31, 52, 37, 35, 58, 60, 15, 44, 66, 43, 70, 72, 25, 54, 78, 8, 82, 61, 55, 88, 65, 59, 68, 96, 27, 100, 102, 38, 106, 108, 71, 112, 85, 33, 89, 94, 79, 48, 126, 83, 130

Offset: 1

Clark Kimberling, Aug 01 2000

nonn

It appears that this is a permutation of the integers. - Michel Marcus, Feb 19 2016
The fact that this is a permutation is proved at the MathOverflow link below. Also from that link: a(n)+1 is prime if and only if a(n) = 2*(n-1). - Ilya I. Bogdanov, Feb 15 2022
From Jianing Song, Sep 27 2023: (Start)
Let {b(n)} be the inverse permutation of this sequence, then each number n >= 3 is moved for b(n)-2 times during the process.
Proof: suppose that this number is k, which is well-defined since PS(1), PS(2), ... has a limit. Suppose that PS(i+1)(c_i) = n for each i >= 0, that is, c_i is the index of n after i steps. In the (i+1)-th step, each group of PS(i+1) contains i+2 elements, and every element is moved if and only if it has index at least i+3. We obtain that k = #{i >= 0 : c_i >= i+3}.
Note that if c_i <= i+2 for some i, then from the (i+1)-th step on, each group contains at least i+2 numbers so the first i+2 numbers in the sequence remain fixed, which means that n = PS(i+1)(c_i) = PS(i+2)(c_i) = ... = a(c_i), so c_i = c_{i+1} = ... = b(n). This shows that {i >= 0 : c_i >= i+3} = {0, 1, ..., k-1}, which implies that k is the smallest number such that c_k <= k+2. On one hand, since c_k <= k+2, we have c_k = b(n), so b(n) <= k+2. On the other hand, from the (b(n)-1)-th step on, each group contains at least b(n) numbers so the first b(n) numbers in the sequence remain fixed, which means that PS(b(n)-1)(b(n)) = PS(b(n))(b(n)) = ... = a(b(n)) = n, so c_{b(n)-2} = b(n), and k <= b(n)-2. In conclusion, we have k = b(n)-2.
By the MathOverflow link, we have a(n) <= 2*n-2 for all n, where the equality holds if and only if a(n)+1 is prime. On the other hand, it is hard to get a lower bound for {a(n)}, so it is infeasible to calculate the inverse permutation of this sequence. (End)

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

MathOverflow, Prime numbers from permutation, 2022.

Cf. A007062, A057030, A057032, A057033, A057064.

get(v, iv) = if (iv > #v, 0, v[iv]);
rcp(nbn, nbp, startv, v) = {w = vector(nbn); for (k=1, nbn, if (k % nbp, jv = startv+k, jv = startv+k-nbp); w[k] = get(v, jv);); w;}
lista(nn) = {v = vector(nn, n, n); print1(v[1], ", ", v[2], ", "); startv = 3; for (n=3, nn, w = rcp(nn-n+1, n-1, startv, v); startv = 2; if (w[1] == 0, break); print1(w[1], ", "); v = w;);} \\ Michel Marcus, Feb 19 2016

Conjecture: a(n) = A057033(n-1) + 1 for n > 1 with a(1) = 1. - Mikhail Kurkov, Mar 10 2022

A057064 Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...,s(n) fixed and forward-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 A057064.

Original entry on oeis.org

1, 2, 4, 5, 8, 7, 11, 9, 17, 16, 22, 13, 23, 15, 28, 29, 37, 19, 34, 21, 44, 36, 40, 25, 54, 35, 46, 47, 51, 31, 67, 33, 79, 53, 58, 56, 82, 39, 64, 60, 89, 43, 87, 45, 97, 88, 76, 49, 120, 65, 112, 77, 102, 55, 104, 80, 145, 84, 94, 61, 142, 63, 100, 114, 174

Offset: 1

Clark Kimberling, Aug 01 2000

nonn

It appears that this is not a permutation of the integers: 3, 6, 10, 12, 14, 18, 20, 24, ... are not terms. - Michel Marcus, Feb 19 2016

Indeed, see the first formula here and the first comment in A069829. - Mikhail Kurkov, Mar 08 2023

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

Cf. A007062, A057030, A057032, A057033, A057063, A069829.

get(v, iv) = if (iv > #v, 0, v[iv]);
fcp(nbn, nbp, startv, v) = {w = vector(nbn); for (k=1, nbn, j = k % nbp; if (j == 1, jv = startv+k+nbp-2, jv = startv+k-2); w[k] = get(v, jv);); w;}
lista(nn) = {v = vector(nn, n, n); print1(v[1], ", ", v[2], ", "); startv = 3; for (n=3, nn, w = fcp(nn-n+1, n-1, startv, v); startv = 2; if (w[1] == 0, break); print1(w[1], ", "); v = w;);} \\ Michel Marcus, Feb 19 2016

a(n) = A057032(n-1) + 1 for n > 1. - Sean A. Irvine, May 19 2022

More terms from Michel Marcus, Feb 19 2016

cp's OEIS Frontend

A069836 Inverse permutation to A057033: a(n) is the m such that A057033(m) = n, or 0 if no such m exists.

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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).

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A057030 Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...,s(n-1) 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. The limit of PS(n) is A057030.

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