Consider the sequence $s(n)=$ $$\small \frac{\left(\prod _{b=2}^{\frac{n}{2^1}} \prod _{a=2}^{\frac{n}{2^1}} (1+i [a b\leq n])\right) \left(\prod _{d=2}^{\frac{n}{2^3}} \prod _{c=2}^{\frac{n}{2^3}} \prod _{b=2}^{\frac{n}{2^3}} \prod _{a=2}^{\frac{n}{2^3}} (1+i [a b c d\leq n])\right) \left( \prod _{f=2}^{\frac{n}{2^5}} \prod _{e=2}^{\frac{n}{2^5}} \prod _{d=2}^{\frac{n}{2^5}} \prod _{c=2}^{\frac{n}{2^5}} \prod _{b=2}^{\frac{n}{2^5}} \prod _{a=2}^{\frac{n}{2^5}} (1+i [a b c d e f \leq n])\right)...}{\left(\prod _{a=2}^{\frac{n}{2^0}} (1+i[a\leq n])\right) \left(\prod _{c=2}^{\frac{n}{2^2}} \prod _{b=2}^{\frac{n}{2^2}} \prod _{a=2}^{\frac{n}{2^2}} (1+i [a b c\leq n])\right) \left( \prod _{e=2}^{\frac{n}{2^4}} \prod _{d=2}^{\frac{n}{2^4}} \prod _{c=2}^{\frac{n}{2^4}} \prod _{b=2}^{\frac{n}{2^4}} \prod _{a=2}^{\frac{n}{2^4}} (1+i [a b c d e\leq n])\right)...}$$
Starting:
$$1,\frac{1}{2}-\frac{i}{2},-\frac{i}{2},-\frac{i}{2},-\frac{1}{4}-\frac{i}{4},-\frac{i}{2},-\frac{1}{4}-\frac{i}{4},-\frac{1}{4}-\frac{i}{4},-\frac{1}{4}-\frac{i}{4},-\frac{i}{2},-\frac{1}{4}-\frac{i}{4},-\frac{1}{4}-\frac{i}{4},-\frac{1}{4},-\frac{1}{4}-\frac{i}{4}\,...$$
Question:
Are the terms in $s(n)$ in one to one correspondence with the Mertens function?
Mathematica program suggesting that they are, at least up to $n=127$:
(*start*)
Clear[a, b, c, d, f, g, n, nn];
nn = 127;
table = Table[
1/Product[1 + I*If[a <= n, 1, 0], {a, 2, n/2^0}]*
Product[Product[1 + I*If[a*b <= n, 1, 0], {a, 2, n/2^1}], {b, 2,
n/2^1}]/Product[
Product[Product[1 + I*If[a*b*c <= n, 1, 0], {a, 2, n/2^2}], {b,
2, n/2^2}], {c, 2, n/2^2}]*
Product[Product[
Product[Product[
1 + I*If[a*b*c*d <= n, 1, 0], {a, 2, n/2^3}], {b, 2,
n/2^3}], {c, 2, n/2^3}], {d, 2, n/2^3}]/
Product[Product[
Product[Product[
Product[1 + I*If[a*b*c*d*f <= n, 1, 0], {a, 2, n/2^4}], {b, 2,
n/2^4}], {c, 2, n/2^4}], {d, 2, n/2^4}], {f, 2, n/2^4}]*
Product[Product[
Product[Product[
Product[Product[
1 + I*If[a*b*c*d*f*g <= n, 1, 0], {a, 2, n/2^5}], {b, 2,
n/2^5}], {c, 2, n/2^5}], {d, 2, n/2^5}], {f, 2, n/2^5}], {g,
2, nn/2^5}], {n, 1, nn}]
ListLinePlot[Arg[table]]
ListLinePlot[Accumulate[MoebiusMu[Range[Length[table]]]]]
(Arg[table]/Pi*4 - Accumulate[MoebiusMu[Range[Length[table]]]] - 3)/4
(*end*)
In the latex the square bracket $[ ]$ is the Iverson bracket.
