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

1,2

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.

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

Table of n, a(n) for n=1..16.

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.

Example

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.

Prog

(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++)

Crossrefs

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

Sequence in context: A144312 A062025 A211797 * A119134 A119658 A072745

Adjacent sequences: A172433 A172434 A172435 * A172437 A172438 A172439

Keyword

nonn,fini,full

Author

Michel Lagneau, Feb 02 2010

Extensions

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

Status

approved