A308619 Negative van Eck's sequence: For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = n-m, otherwise a(n+1) = -a(n). Start with a(1)=0.

0, 0, 1, -1, 1, 2, -2, 2, 2, 1, 5, -5, 5, 2, 5, 2, 2, 1, 8, -8, 8, 2, 5, 8, 3, -3, 3, 2, 6, -6, 6, 2, 4, -4, 4, 2, 4, 2, 2, 1, 22, -22, 22, 2, 5, 22, 3, 20, -20, 20, 2, 7, -7, 7, 2, 4, 19, -19, 19, 2, 5, 16, -16, 16, 2, 5, 5, 1, 28, -28, 28, 2, 7, 19, 15, -15, 15, 2, 6, 48, -48, 48, 2, 5, 17, -17, 17, 2, 5, 5, 1, 23, -23, 23, 2, 7, 23, 3, 51, -51

Offset

1,6

Comments

Similar logic as van Eck's sequence except that if a(n) hasn't appeared before, then a(n+1) = -a(n) instead of being 0.

Links

Table of n, a(n) for n=1..100.

Formula

If a(n) < 0, then a(n+1) = -a(n) and a(n+2) = 2. - Rémy Sigrist, Jul 14 2019

Example

We start with a(1) = 0.
0 has not occurred before, so our rule says that a(2) = -a(1) = 0.
Now 0 has occurred before, at a(1), which is 1 term before, so a(3) = 1.
1 has not occurred before, so a(4) = -a(3) = -1.
-1 has not occurred before, so a(5) = -a(4) = 1.
Now 1 has occurred before, at a(3), which is 2 term before, so a(6) = 2.
2 has not occurred before, so a(6) = -a(5) = -2.
And so on...

Prog

(Python)
import numpy as np
def A308619(n):
    NegVanEck = np.array([0])
    #first value is 0
    i=NegVanEck[0]
    while i < n:
        last = NegVanEck[-1]
        pos = np.where(NegVanEck == last)
        if pos[0].size == 1:
            NegVanEck = np.append(NegVanEck, -last)
        else:
            NegVanEck = np.append(NegVanEck, pos[0][-1] - pos[0][-2])
        i += 1
    return NegVanEck
print(A308619(1000)) #prints first 1000 integers of the sequence

Crossrefs

Cf. A181391.

Sequence in context: A064741 A281659 A345202 * A114294 A371929 A242050

Adjacent sequences: A308616 A308617 A308618 * A308620 A308621 A308622

Keyword

sign

Author

João D. R. Camacho, Jun 11 2019

Status

approved