A007062 -id:A007062 - cp's OEIS frontend

A309977 Position of n in row n of McGarvey's array; see A007062.

1, 1, 6, 4, 5, 11, 2, 22, 13, 15, 30, 4, 23, 11, 12, 58, 43, 45, 68, 28, 97, 37, 8, 64, 35, 95, 66, 10, 105, 21, 82, 54, 55, 155, 196, 126, 171, 93, 142, 184, 27, 235, 16, 290, 157, 65, 124, 18, 81, 131, 72, 288, 239, 243, 304, 36, 375, 209, 144, 276, 211

Offset: 1

Clark Kimberling, Aug 25 2019

			McGarvey's array (from A007062):
   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  8  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
In row 3, the entry 3 occurs in position 6, so that a(3) = 6.

Cf. A007062.

s[0] = Range[10000]; s[n_] := Flatten[Map[Reverse, Partition[s[n - 1], n]]];
Flatten[Table[Position[s[n], n], {n, 1, 500}]]

A362813 Number of numbers that occur more than once in column n of McGarvey's array (A007062).

Original entry on oeis.org

0, 2, 3, 4, 6, 6, 7, 9, 11, 12, 14, 13, 14, 16, 15, 17, 21, 20, 22, 25, 24, 25, 29, 26, 26, 31, 29, 31, 36, 33, 33, 40, 37, 36, 44, 41, 39, 48, 42, 44, 50, 44, 48, 53, 49, 49, 56, 54, 53, 62, 55, 56, 69, 61, 59, 66, 61, 66, 73, 65, 64, 75, 71, 70, 80, 72, 71

Offset: 1

Clark Kimberling, May 04 2023

nonn

			Corner of McGarvey's array (from A007062):
   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  8  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
(Column 3) = (3,4,2,1,1,2,4,5,7,12,23,...) in which the numbers 1,2,4 occur more than once.

Cf. A007062, A309977.

a = Range[x = 40000]; rows = Table[a = Flatten[Map[Reverse, Partition[a, n]]], {n,
   NestWhile[# + 1 &, 1, (x = # Floor[x/#]) > 0 &] - 1}];
Flatten[Table[{Length[Select[SortBy[Tally[Map[#[[col]] &, rows]], #[[2]] &], #[[2]] > 1 &]]}, {col, 100}]]  (* Peter J. C. Moses, May 01 2023 *)

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

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.

Original entry on oeis.org

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

Clark Kimberling, Jul 29 2000

			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.

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

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 *)

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

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

A277679 Start with 1,2,3,4,5,6,.... For n >=1, remove the first n terms and reverse the remaining terms n+1 at a time. Concatenate the terms removed. (See the example.)

Original entry on oeis.org

1, 3, 2, 7, 4, 5, 13, 6, 9, 8, 17, 10, 11, 14, 15, 23, 16, 19, 12, 25, 18, 33, 26, 27, 20, 21, 24, 31, 49, 32, 39, 22, 29, 28, 35, 34, 53, 36, 43, 30, 37, 40, 41, 50, 51, 59, 52, 55, 42, 45, 38, 61, 44, 67, 54, 85, 68, 69, 62, 63, 46, 47, 56, 57, 60, 77, 95

Offset: 1

Clark Kimberling, Nov 14 2016

This is a permutation of the natural numbers, with inverse permutation A277680.

			Remove 1 from A000027, leaving 2,3,4,5,6,7,8,...; reverse these 2 at a time, leaving 3,2,5,4,7,6,9,8,... Remove the first 2 terms and reverse the rest 3 at a time, leaving 7,4,5,8,9,6,13,10,11,14,15,12,... Remove the first 3 terms, and so on. The removed terms, taken in order, are 1,3,2,7,4,5,...

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000027, A277680, A007062, A057030.

x = Range[500];
NestWhile[# + 1 &, 1, (t = 1/2 # (1 + #);
x = Flatten[{Take[x, t],
Map[Reverse, Partition[Drop[x, t], # + 1]]}];
Length[x] > t) &]; x (* A277679 *)
Take[Ordering[#],Position[Differences[Sort[#]],Except[1]][[2]][[1]]]&[x] (* A277680 *) (* Peter J. C. Moses, Nov 13 2016 *)

A277680 Inverse of the permutation A277679 of the natural numbers.

Original entry on oeis.org

1, 3, 2, 5, 6, 8, 4, 10, 9, 12, 13, 19, 7, 14, 15, 17, 11, 21, 18, 25, 26, 32, 16, 27, 20, 23, 24, 34, 33, 40, 28, 30, 22, 36, 35, 38, 41, 51, 31, 42, 43, 49, 39, 53, 50, 61, 62, 72, 29, 44, 45, 47, 37, 55, 48, 63, 64, 70, 46, 65, 52, 59, 60, 74, 73, 84, 54

Offset: 1

Clark Kimberling, Nov 14 2016

This is a permutation of the natural numbers, with inverse permutation A277679.

Clark Kimberling, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers

Cf. A000027, A277679, A007062, A057030.

x = Range[500];
NestWhile[# + 1 &, 1, (t = 1/2 # (1 + #);
x = Flatten[{Take[x, t],
Map[Reverse, Partition[Drop[x, t], # + 1]]}];
Length[x] > t) &]; x (* A277679 *)
Take[Ordering[#],Position[Differences[Sort[#]],Except[1]][[2]][[1]]]&[x] (* A277680 *) (* Peter J. C. Moses, Nov 13 2016 *)

A366619 Group the natural numbers into blocks of size 2: [1,2], [3,4], ... and reverse the order of the numbers within each block. Then group into blocks of size 3 and reverse the order in each block.

Original entry on oeis.org

4, 1, 2, 5, 6, 3, 10, 7, 8, 11, 12, 9, 16, 13, 14, 17, 18, 15, 22, 19, 20, 23, 24, 21, 28, 25, 26, 29, 30, 27, 34, 31, 32, 35, 36, 33, 40, 37, 38, 41, 42, 39, 46, 43, 44, 47, 48, 45, 52, 49, 50, 53, 54, 51, 58, 55, 56, 59, 60, 57, 64, 61, 62, 65, 66, 63, 70, 67, 68, 71, 72, 69

Offset: 1

Wesley Ivan Hurt, Oct 14 2023

Row 3 of the array in A007062.

			Group natural numbers into blocks of size 2: [1, 2], [3, 4], [5, 6], ...
Reverse the order in each block: [2, 1], [4, 3], [6, 5], ...
Group the remaining sequence into blocks of size 3: [2, 1, 4], [3, 6, 5], ...
Reverse the order in each block to get a(n): 4, 1, 2, 5, 6, 3, ...

Index entries for linear recurrences with constant coefficients, signature (0,1,1,0,-1).

Cf. A007062.

Table[1 + Mod[n, 2] + 2 Floor[3 Floor[(n - 1)/3]/2 + Mod[-n, 3]/2], {n, 100}]

a(n) = 1 + (n mod 2) + 2*floor(3*floor((n - 1)/3)/2 + (-n mod 3)/2).
G.f.: x*(4 + x - 2*x^2 + 3*x^4)/((1 - x)^2*(1 + x)*(1 + x + x^2)). - Stefano Spezia, Oct 14 2023
a(n+6) = a(n) + 6. - Joerg Arndt, Oct 15 2023
From Wesley Ivan Hurt, Oct 15 2023: (Start)
a(n) = n - (-1)^n + 2*cos(2*(n - 1)*Pi/3) + 2*sin(2*(n - 1)*Pi/3)/sqrt(3).
a(n) = a(n-2) + a(n-3) - a(n-5) for n >= 6. (End)

cp's OEIS Frontend

A309977 Position of n in row n of McGarvey's array; see A007062.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

A362813 Number of numbers that occur more than once in column n of McGarvey's array (A007062).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

MATLAB

Mathematica

PARI

Formula

Extensions

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.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

Extensions

A277679 Start with 1,2,3,4,5,6,.... For n >=1, remove the first n terms and reverse the remaining terms n+1 at a time. Concatenate the terms removed. (See the example.)

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A277680 Inverse of the permutation A277679 of the natural numbers.

Views