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A172436
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Smallest m such that the Moebius function takes successively, from m, n values 1,0,1,0,... ending with 1 or 0.
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0
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1, 15, 55, 159, 411, 411, 411, 3647, 15243, 15243, 15243, 113343, 1133759, 1133759, 1133759, 29149139
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OFFSET
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1,2
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COMMENTS
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It's easy to prove that a(n) for n >= 17 does not exist, because in all sequences of 17 consecutive numbers such that the first is squarefree, there are necessarily two numbers r, s where 9 divides r and s, so Moebius(r) = Moebius(s) = 0 with r - s odd.
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 826.
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 24.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 161, #16.
Deleglise, Marc and Rivat, Joel, Computing the summation of the Mobius function. Experiment. Math. 5 (1996), no. 4, 291-295.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 262 and 287.
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LINKS
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EXAMPLE
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a(3) = 55 since Moebius(55) = 1, Moebius(56) = 0, Moebius(57) = 1, and this pattern does not occur for any smaller value of n.
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PROG
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(PARI) a(n)=local(ok, m); m=1; while(1, ok=1; for(k=1, n, if(moebius(m+k-1)!=k%2, ok=0; break)); if(ok, return(m)); m++)
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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