A288533 Parse A004736 into distinct phrases [1], [2], [1,3], [2,1], [4], [3], [2,1,5], [4,3], [2,1,6], ...; a(n) is the length of the n-th phrase.

1, 1, 2, 2, 1, 1, 3, 2, 3, 1, 3, 2, 1, 2, 2, 2, 1, 2, 4, 1, 1, 2, 3, 3, 2, 3, 5, 1, 3, 3, 3, 1, 1, 2, 2, 4, 3, 2, 3, 4, 4, 1, 3, 4, 4, 2, 1, 2, 2, 5, 5, 1, 2, 4, 3, 5, 1, 1, 2, 3, 4, 5, 2, 2, 3, 5, 5, 3, 1, 3, 3, 3, 4, 5, 1, 2, 2, 4, 5, 6, 1, 2, 4, 4, 6, 4, 1, 2, 3, 4, 4, 6, 2, 1, 2, 3, 3, 5, 5, 4, 1, 2, 3, 5, 6, 6, 1, 1, 2, 3, 4, 5, 7, 3, 2, 3, 4, 4, 7, 6, 1, 3, 3, 4, 5, 6, 5, 1, 2, 2

Offset

1,3

Comments

The phrases are formed by the Ziv-Lempel encoding described in A106182. - Neal Gersh Tolunsky, Nov 30 2023

Links

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Example

Consider the infinite sequence [1,2,1,3,2,1,4,3,2,1,5,4,3,2,1,...], i.e., A004736. We can first take [1] since we've never used it before. Then [2]. For the third term, we've already used [1], so we must instead take [1,3].

Prog

(Python)
# you should use program from internal format
a = set()
i = 2
s = "1"
seq = ""
while i < 100:
    j = i
    while j > 0:
        if s not in a:
            seq = seq + ", " + str(len(s)-len(s.replace(", ", ""))+1)
            a.add(s)
            s = str(j)
        else:
            s = s + ", " + str(j)
        j -= 1
    i += 1
print(seq[1:])

Crossrefs

Cf. A109337, A106182, A187180, A187181, A187182, A187183, A187184, A187185, A187186, A187187, A187188, A187199, A187200.

Sequence in context: A047130 A125778 A047110 * A093869 A329870 A057431

Adjacent sequences: A288530 A288531 A288532 * A288534 A288535 A288536

Keyword

nonn

Author

Lewis Chen, Jun 11 2017

Status

approved