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
Keywords
Examples
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].
Links
- Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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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:])
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