

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 nth 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



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

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


KEYWORD

nonn


AUTHOR



STATUS

approved



