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Seven consecutive numbers

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Tim

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Apr 29, 2003, 6:17:14 AM4/29/03
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Is there a block of seven consecutive numbers starting from the
natural number n such that :

mu(n) + mu(n+1) + mu(n+2) + mu(n+3) + mu(n+4) + mu(n+5) + mu(n+6) = 6 ?

(Where mu() is the Mobius function) .


Thanks

José Carlos Santos

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Apr 29, 2003, 6:33:49 AM4/29/03
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Yes: take n = 213. Then:

n = 3.71

n + 1 = 2.107

n + 2 = 5.43

n + 3 = 2^3.3^3

n + 3 = 7.31

n + 5 = 2.109

n + 6 = 3.73

and so mu takes the value 1 at all these numbers, except for
216, where mu takes the value 0.

Best regards

Jose Carlos Santos
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Ignacio Larrosa Cañestro

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Apr 29, 2003, 6:47:53 AM4/29/03
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"Tim" <tim_b...@my-deja.com> escribió en el mensaje
news:93880d6b.0304...@posting.google.com...
Yes, 213.

213 = 3*71 ==> 1
214 = 2*107 ==> 1
215 = 5*43 ==> 1
216 = 2^3*3^3 ==> 0
217 = 7*31 ==> 1
218 = 2*109 ==> 1
219 = 3*73 ==> 1

Also for
1937, 3093, 3097, 4529, 6401, 7165, 9753, 9933,
10117, 10541, 10577, 11301, 13385, 15005, 18641, 18829, 18965, ..

(up to 20000)


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Best regards,


Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUIT...@mundo-r.com

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