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
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.
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Jose Carlos Santos
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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