Let M denote the sequence of values of $\mu$(n) , the Mobius function from elementary number theory. A large table is available at OEIS [A008683] . The sequence begins M = (1,-1,-1,0,-1,1,-1,0,0,1,...) and is known to be quite irregular. This suggests that it will be difficult to find many patterns in M which hold up over significant length scales.
Questions: (1) Is there any initial interval [1,n] with n$\ge$2, for which the Mobius values form a palindrome [i.e. $\mu$(1)=$\mu$(n), $\mu$(2)=$\mu$(n-1), etc. ] ? This seems highly unlikely and even rash but is there a disproof? $\quad$In any case, it would be interesting to know how close the sum$\qquad$ $\qquad$ $\quad$ $\sum_{k=1}^n$ $\mu$(k)$\mu$(n+1-k)$\;$ can be to $\sum_{k=1}^n$ $\mu$(k)$^2$ $\,$.
(2) By contrast, it is easy to find palindromes if we are allowed to vary the starting point. For example, the interval [102,110] , with values (-1,-1,0,-1,1,-1,0,-1,-1), $\,$ already provides a length 9 palindrome. In fact, is there any absolute bound at all on the length of such a palindromic interval? [Note that the Chinese Remainder Theorem can be used to find n with 2$^2$|n, 3$^2$|(n+1), 5$^2$|(n+2), etc. leading to a long stretch of 0's in M. To avoid this trivial case, we require that the palindrome not be identically zero.]
(3) Despite its irregular appearance, M cannot fully mimic a random sequence drawn from {0,1,-1}. In fact, it is constrained in many ways. For example, each string of four (all strings and substrings are consecutive) contains a multiple of 4 and leads to a corresponding Mobius value of zero. Thus, (1,1,1,1) forms a forbidden subsequence in M. [Let f.s. = forbidden (sub)sequence = any finite string which does not occur in M .]$\,$Of course, any extension of an f.s., such as (0,1,1,1,1) or (1,1,1,1,-1), is likewise forbidden so let's define a minimal f.s. to be one that contains no other f.s. within it. For instance, each length 4 string taken from {1,-1} is a minimal f.s. because (i) it's forbidden and (ii) a check (n$\le$200 suffices) shows that all such length 3 substrings do in fact arise. $\qquad$ What can one say about the collection of all minimal forbidden sequences? In particular, is it possible that there are only finitely many of them?
Thanks
(2) As you say, there are arbitrarily long sequences of the form $0, \ldots, 0$. If you include the entries before and after these, I would think that a good fraction of them will be palindromes $(1, 0, \ldots, 0, 1)$ or $(-1, 0, \ldots, 0, -1)$.
The first palindrome of length $n$ (from $1$ to $20$) starts at the following positions: $$ \matrix{ n & | & 1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12 &13 &14 &15 &16 &17 &18 &19 &20\cr position & | & 1 &2 &3 &62 &4 &61 &15 &115 &14 &116 &13 &831 &12 &37173 &597 &457472 &596 &2955661 &595 &6495574\cr}$$
EDIT: See OEIS sequence A293041.