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.

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

Author

Jennifer Buckley, Sep 13 2019

Keywords

• nonn

Comments

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

Links

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

Crossrefs

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

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

Programs

Go

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
}

SageMath

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