cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A057031 Sequence of differences of A057030.

Original entry on oeis.org

2, 1, 2, 5, 2, 1, 8, 5, 2, 11, 2, 5, 8, 11, 2, 15, 2, 1, 24, 5, 8, 15, 2, 21, 18, 1, 2, 41, 8, 5, 34, 5, 2, 37, 18, 5, 34, 11, 2, 41, 2, 11, 34, 5, 60, 5, 8, 5, 64, 11, 18, 57, 2, 1, 54, 57, 18, 5, 8, 5, 80, 1, 2, 93, 18, 21, 34, 5, 2, 125, 2, 15, 80
Offset: 1

Views

Author

Clark Kimberling, Jul 29 2000

Keywords

Links

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

Views

Author

N. J. A. Sloane, Mira Bernstein

Keywords

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)

Examples

			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.

References

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

Links

Crossrefs

Cf. A057030 (here we have "s(1), ..., s(n)", whereas 057030 has "s(1), ..., s(n-1)").
Cf. A057032, A057033, A057063, A057064.

Programs

  • 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 *)

Formula

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

Extensions

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

Views

Author

Clark Kimberling, Jul 29 2000

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

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

Extensions

More terms from David Wasserman, Apr 22 2002

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

Views

Author

Clark Kimberling, Jul 29 2000

Keywords

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

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

Views

Author

Clark Kimberling, Aug 01 2000

Keywords

Comments

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)

Examples

			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.

Links

Crossrefs

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

Programs

  • PARI
    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

Formula

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

Views

Author

Clark Kimberling, Aug 01 2000

Keywords

Comments

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

Examples

			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.

Crossrefs

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

Programs

  • PARI
    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

Formula

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

Extensions

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

Views

Author

Clark Kimberling, Nov 14 2016

Keywords

Comments

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

Examples

			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,...

Links

Crossrefs

Cf. A000027, A277680, A007062, A057030.

Programs

  • Mathematica
    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

Views

Author

Clark Kimberling, Nov 14 2016

Keywords

Comments

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

Links

Crossrefs

Cf. A000027, A277679, A007062, A057030.

Programs

  • Mathematica
    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 *)
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