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A069545
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Liouville clusters: the number of successive occurrences of signs in Liouville function lambda(k); a(2n-1) is number of successive positive signs, while a(2n) is number of successive negative signs.
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2
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1, 2, 1, 1, 1, 2, 2, 3, 3, 4, 2, 1, 3, 6, 4, 1, 3, 5, 1, 2, 1, 1, 1, 2, 5, 1, 1, 1, 1, 1, 2, 3, 1, 4, 1, 2, 1, 3, 2, 1, 5, 1, 2, 1, 4, 3, 1, 3, 1, 1, 1, 4, 1, 3, 1, 2, 2, 1, 3, 2, 1, 2, 1, 2, 5, 3, 7, 3, 1, 1, 1, 2, 2, 1, 4, 4, 1, 2, 1, 7, 2
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OFFSET
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1,2
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COMMENTS
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Related open questions. What is the limit of ratio: a(n)/n, as n->infinity? What is frequency distribution of integer k in the sequence; a(n)=k for what set of n?
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REFERENCES
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H. Gupta, On a table of values of L(n), Proceedings of the Indian Academy of Sciences. Section A, 12 (1940), 407-409.
H. Gupta, A table of values of Liouville's function L(n), Research Bulletin of East Panjab University, No. 3 (Feb. 1950), 45-55.
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LINKS
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FORMULA
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Related to summatory Liouville function (A002819): L(m)=sum_{k=1, n} (-1)^(k-1)*a(k) where m=sum_{k=1, n} a(k).
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EXAMPLE
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a(6) = 2 because the 6th Liouville cluster consists of 2 successive negative signs: lambda(7) = lambda(8) = (-1).
a(7) = 2 because the 7th Liouville cluster consists of 2 successive positive signs: lambda(9) = lambda(10) = 1.
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MATHEMATICA
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max = 227; lambdaClLens = {}; Module[{curr = 1, cl = 1, iter = 2}, While[iter < max, If[LiouvilleLambda[iter] == curr, cl++, AppendTo[lambdaClLens, cl]; curr = (-1)curr; cl = 1]; iter++]]; lambdaClLens (* Alonso del Arte, Feb 29 2012 *)
Length/@Split[LiouvilleLambda[Range[300]]] (* Harvey P. Dale, Jul 02 2017 *)
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PROG
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(Haskell)
import Data.List (group)
a069545 n = a069545_list !! (n-1)
a069545_list = map length $ group a008836_list
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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EXTENSIONS
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Corrected a(46) and a(47), and added terms after that. - Alonso del Arte, Feb 29 2012
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STATUS
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approved
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