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%I #7 Sep 16 2017 00:30:04
%S 1,15,55,159,411,411,411,3647,15243,15243,15243,113343,1133759,
%T 1133759,1133759,29149139
%N Smallest m such that the Moebius function takes successively, from m, n values 1,0,1,0,... ending with 1 or 0.
%C 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.
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 826.
%D T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 24.
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 161, #16.
%D Deleglise, Marc and Rivat, Joel, Computing the summation of the Mobius function. Experiment. Math. 5 (1996), no. 4, 291-295.
%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 262 and 287.
%H Ed Pegg Jr., <a href="http://www.maa.org/editorial/mathgames/mathgames_11_03_03.html">The Mobius function (and squarefree numbers)</a>
%H G. Villemin's Almanac of Numbers, <a href="http://villemin.gerard.free.fr/TABLES/aaaFArit/MobiusMe.htm">Nombres de Moebius et de Mertens</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MoebiusFunction.html">Moebius Function</a>.
%e 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.
%o (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++)
%Y Moebius (or Mobius) function mu(n): A008683, A002321, A045882.
%K nonn,fini,full
%O 1,2
%A _Michel Lagneau_, Feb 02 2010
%E Edited by _Franklin T. Adams-Watters_, May 15 2010
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