A308673 For n > 2, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = a(n)+(n-m); otherwise a(n+1) = 1. Start with a(1)=0, a(2)=0.

0, 0, 1, 1, 2, 1, 3, 1, 3, 5, 1, 4, 1, 3, 8, 1, 4, 9, 1, 4, 7, 1, 4, 7, 10, 1, 5, 22, 1, 4, 11, 1, 4, 7, 17, 1, 5, 15, 1, 4, 11, 21, 1, 5, 12, 1, 4, 11, 18, 1, 5, 12, 19, 1, 5, 9, 47, 1, 5, 9, 13, 1, 5, 9, 13, 17, 48, 1, 7, 42

Offset

1,5

Comments

It appears that if you choose "otherwise a(n+1) = x" to be a different integer, that the sequence starts to look the same, but offset in position and value. For example, if you compare the case where "otherwise a1(n+1) = 1" and "otherwise a2(n+1) = 2" then a1(n) = a2(n+3)-2.

Links

Table of n, a(n) for n=1..70.

Mathematica

a[1] = a[2] = 0; a[n_] := a[n] = Module[{k = n-2}, While[k > 0 && a[k] != a[n-1], k--]; If[k==0, 1, a[n-1] + n - k - 1]]; Array[a, 100] (* Amiram Eldar, Jul 12 2019 *)

Prog

(Python)
def Prog(length):
    L = 2
    seq = [0, 0]
    while L < length:
        x = len(seq)-1
        while x > 0:
            if seq[-1] == seq[x-1]:
                m = len(seq)-x
                a_n = seq[L-1]
                seq.append(a_n+m)
                x = -1
            else:
                x -= 1
            if x == 0:
                seq.append(1)
        L += 1
    return seq

Crossrefs

Cf. A181391.

Sequence in context: A163313 A337178 A321893 * A324287 A213594 A335105

Adjacent sequences: A308670 A308671 A308672 * A308674 A308675 A308676

Keyword

nonn

Author

Philip Kalisman, Jul 07 2019

Status

approved