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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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
The phrases are formed by the Ziv-Lempel encoding described in A106182. - Neal Gersh Tolunsky, Nov 30 2023
LINKS
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
Sequence in context: A047130 A125778 A047110 * A093869 A329870 A057431
KEYWORD
nonn
AUTHOR
Lewis Chen, Jun 11 2017
STATUS
approved

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Last modified June 15 02:34 EDT 2024. Contains 373402 sequences. (Running on oeis4.)