A057032 -id:A057032 - cp's OEIS frontend

A069829 a(n) = PS(n)(2n), where PS is described in A057032.

2, 5, 9, 11, 13, 19, 17, 23, 31, 26, 25, 40, 29, 47, 58, 51, 37, 69, 41, 56, 71, 67, 49, 82, 70, 73, 92, 95, 61, 123, 65, 105, 118, 94, 112, 148, 77, 107, 134, 116, 85, 143, 89, 122, 177, 127, 97, 166, 130, 133, 175, 162, 109, 211, 159, 188, 190, 154, 121, 248, 125

Offset: 1

David Wasserman, Apr 22 2002

a(n) = PS(n + c)(2n + c) for any positive integer c. Every positive integer occurs exactly once in either this sequence or in A057032.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A057032.

A069829 := proc(n) local N, k; N := n;
for k from n by -1 to 1 do
    if irem(N, k) = 0 then
        if irem(N, k)::odd then
            N := N - k;
        else
            N := N + k;
        fi;
    fi;
od;
N end:
seq(A069829(n), n = 1..61); # Peter Luschny, Sep 14 2019

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

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

A327093 Sequence obtained by swapping each (k(2n))-th element of the positive integers with the (k(2n-1))-th element, for all k > 0, in ascending order.

Original entry on oeis.org

2, 3, 7, 5, 11, 13, 15, 10, 17, 19, 23, 25, 27, 21, 40, 16, 35, 36, 39, 37, 58, 33, 47, 50, 52, 43, 45, 34, 59, 78, 63, 31, 76, 55, 82, 67, 75, 57, 99, 56, 83, 112, 87, 61, 126, 69, 95, 92, 97, 96, 133, 71, 107, 81, 142, 79, 139, 91, 119, 155, 123, 93, 122, 51, 151, 146, 135

Offset: 1

Jennifer Buckley, Sep 13 2019

nonn

Start with the sequence of positive integers [1, 2, 3, 4, 5, 6, 7, 8, ...].
Swap all pairs specified by k=1, that is, do the swaps (2,1),(4,3),(6,5),(8,7),..., resulting in [2, 1, 4, 3, 6, 5, 8, 7, ...], so the first term of the final sequence is 2 (No swaps for k>1 will affect this term).
Swap all pairs specified by k=2, that is, do the swaps (4,2),(8,6),(12,10),(16,14),..., resulting in [2, 3, 4, 1, 6, 7, 8, 5, ...], so the second term of the final sequence is 3 (No swaps for k>2 will affect this term).
Swap all pairs specified by k=3, that is, do the swaps (6,3),(12,9),(18,15),(24,21),... .
Continue for all values of k.
The complementary sequence 1, 4, 6, 8, 9, 12, 14, 18, 20, 22, 24, 26, 28, ... lists the numbers that never appear. Is there an alternative characterization of these numbers?
Equivalently, is there a characterization of the numbers (2, 3, 5, 7, 10, 11, 13, 15, 16, 17, 19, 21, 23, ...) that do appear? - N. J. A. Sloane, Sep 13 2019

Jennifer Buckley, Table of n, a(n) for n = 1..10000

For the sorted terms and the missing terms see A327445, A327446.

Cf. A327119, A064494, A064627, A057032, A069829, A327420.

func a(n int) int {
    for k := n; k > 0; k-- {
        if n%k == 0 {
            if (n/k)%2 == 0 {
                n = n - k
            } else {
                n = n + k
            }
        }
    }
    return n
}

def a(n):
    for k in srange(n, 0, -1):
        if k.divides(n):
            n += k if is_odd(n//k) else -k
    return n
print([a(n) for n in (1..67)]) # Peter Luschny, Sep 14 2019

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

A327420 Building sums recursively with the divisibility properties of their partial sums.

Original entry on oeis.org

1, 0, 2, 3, 6, 5, 9, 7, 15, 4, 14, 11, 21, 13, 16, 8, 35, 17, 26, 19, 30, 12, 28, 23, 46, 18, 38, 10, 49, 29, 45, 31, 77, 20, 50, 27, 63, 37, 52, 24, 68, 41, 54, 43, 74, 25, 64, 47, 96, 34, 62, 32, 95, 53, 70, 42, 94, 36, 86, 59, 91, 61, 88, 33, 166, 51, 85

Offset: 0

Peter Luschny, Sep 14 2019

nonn

Let R(n) = [k : n + 1 >= k >= 2] and divsign(s, k) = 0 if k does not divide s, else k if s/k is even and else -k. Compute s(k) = s(k+1) + divsign(s(k+1), k) with initial value s(n+2) = n + 1, k running down from n + 1 to 2. Then a(n) = s(2) if n > 0 and a(0) = s(n+2) = 0 + 1 = 1 as R(0) is empty in this case.
Examples: If n = 8 then R(8) = [9, 8, ..., 2] and the partial sums s are [0, 8, 8, 8, 8, 12, 15, 15] giving a(8) = 15. If p is prime, then the partial sums are [0, p, p, ..., p] since p is the only integer in R(p) diving p, i. e. the primes are the fixed points of this sequence. In the example section the computation of a(9) is traced.
Apparently the sequence is a permutation of the nonnegative integers.

			The computation of a(9) = 4:
[ k: s(k) = s(k+1) + divsign(s(k+1),k)]
[10:   0,    10,       -10]
[ 9:   9,     0,         9]
[ 8:   9,     9,         0]
[ 7:   9,     9,         0]
[ 6:   9,     9,         0]
[ 5:   9,     9,         0]
[ 4:   9,     9,         0]
[ 3:   6,     9,        -3]
[ 2:   4,     6,        -2]

Cf. A327093, A327487, A057032, A069829.

divsign(s, k) = rem(s, k) == 0 ? (-1)^div(s, k)*k : 0
function A327420(n)
    s = n + 1
    for k in n+1:-1:2 s += divsign(s, k) end
    s
end
[A327420(n) for n in 0:66] |> println

divsign := (s, k) -> `if`(irem(s, k) <> 0, 0, (-1)^iquo(s,k)*k):
A327420 := proc(n) local s, k; s := n + 1;
    for k from s by -1 to 2 do
        s := s + divsign(s, k) od;
return s end:
seq(A327420(n), n=0..66);

def A327420(n):
    s = n + 1
    r = srange(s, 1, -1)
    for k in r:
        if k.divides(s):
            s += (-1)^(s//k)*k
    return s
print([A327420(n) for n in (0..66)])

For p prime, a(p) = p. - Bernard Schott, Sep 14 2019

A327487 T(n, k) are the summands given by the generating function of A327420(n), triangle read by rows, T(n,k) for 0 <= k <= n.

Original entry on oeis.org

1, 2, -2, 3, -3, 2, 4, -4, 3, 0, 5, -5, 4, 0, 2, 6, -6, 5, 0, 0, 0, 7, -7, 6, 0, 0, 3, 0, 8, -8, 7, 0, 0, 0, 0, 0, 9, -9, 8, 0, 0, 0, 4, 3, 0, 10, -10, 9, 0, 0, 0, 0, 0, -3, -2, 11, -11, 10, 0, 0, 0, 0, 5, 0, -3, 2, 12, -12, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Peter Luschny, Sep 14 2019

			Triangle starts (at the end of the line is the row sum (A327420)):
[ 0] [ 1] 1
[ 1] [ 2,  -2] 0
[ 2] [ 3,  -3,  2] 2
[ 3] [ 4,  -4,  3, 0] 3
[ 4] [ 5,  -5,  4, 0, 2] 6
[ 5] [ 6,  -6,  5, 0, 0, 0] 5
[ 6] [ 7,  -7,  6, 0, 0, 3, 0] 9
[ 7] [ 8,  -8,  7, 0, 0, 0, 0, 0] 7
[ 8] [ 9,  -9,  8, 0, 0, 0, 4, 3,  0] 15
[ 9] [10, -10,  9, 0, 0, 0, 0, 0, -3, -2] 4
[10] [11, -11, 10, 0, 0, 0, 0, 5,  0, -3, 2] 14

Cf. A327420, A327093, A057032, A069829.

def divsign(s, k):
    if not k.divides(s): return 0
    return (-1)^(s//k)*k
def A327487row(n):
    s = n + 1
    r = srange(s, 1, -1)
    S = [-divsign(s, s)]
    for k in r:
        s += divsign(s, k)
        S.append(-divsign(s, k))
    return S
# Prints the triangle like in the example section.
for n in (0..10):
    print([n], A327487row(n), sum(A327487row(n)))

Sum_{k=0..n} T(n, k) = A327420(n).

cp's OEIS Frontend

A069829 a(n) = PS(n)(2n), where PS is described in A057032.

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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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A327093 Sequence obtained by swapping each (k*(2n))-th element of the positive integers with the (k*(2n-1))-th element, for all k > 0, in ascending order.

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A327420 Building sums recursively with the divisibility properties of their partial sums.

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A327093 Sequence obtained by swapping each (k(2n))-th element of the positive integers with the (k(2n-1))-th element, for all k > 0, in ascending order.