A172436 Smallest m such that the Moebius function takes successively, from m, n values 1,0,1,0,... ending with 1 or 0.

1, 15, 55, 159, 411, 411, 411, 3647, 15243, 15243, 15243, 113343, 1133759, 1133759, 1133759, 29149139

Offset: 1

Author

Michel Lagneau, Feb 02 2010

Keywords

• nonn
• fini
• full

Comments

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.

Examples

			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.
		

References

• 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.

Links

• Ed Pegg Jr., The Mobius function (and squarefree numbers)
• G. Villemin's Almanac of Numbers, Nombres de Moebius et de Mertens
• Eric Weisstein's World of Mathematics, Moebius Function.

Crossrefs

Moebius (or Mobius) function mu(n): A008683, A002321, A045882.

Programs

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++)

Extensions

Edited by Franklin T. Adams-Watters, May 15 2010