A069829 -id:A069829 - cp's OEIS frontend

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.

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

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

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).

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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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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.

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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.