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Problem #321

Consecutive Zeros Mobius
 Public ★(x11) 04/08/16 by Philippe_57721 8xp Math 14.9%

It can easily be seen that there cannot be more than 3 consecutive integers such as $\mu(n_i) = \pm 1,\quad\mu(n)$ being the Möbius function

However, we can find arbitrary long sequences of consecutive integers with a Möbius value of 0.

Find 9 consecutive integers $n+1,n+2, ..., n+9 \textrm{ such as }\mu(n+i) = 0$
(Each of them is divisible by a distinct prime square, the primes are the smallest possible).

[My timing: 10 sec]

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