A069545 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.
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
Offset: 1
Examples
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.
References
- 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.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- R. S. Lehman, On Liouville's function, Math. Comp., 14 (1960), 311-320.
- Eric Weisstein's World of Mathematics, Liouville Function
Programs
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Haskell
import Data.List (group) a069545 n = a069545_list !! (n-1) a069545_list = map length $ group a008836_list -- Reinhard Zumkeller, Mar 10 2014
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Mathematica
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 *)
Formula
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).
Extensions
Corrected a(46) and a(47), and added terms after that. - Alonso del Arte, Feb 29 2012
Comments