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Let $t_n$ denote the $n^{\rm th}$ term in the Thue Morse sequence. Note that $t_n=1$ if the number of $1$s in the binary expansion of $n$ is odd, $0$ otherwise. Now define a variant of the Riemann Zeta function as follows:

$$ \zeta_{TM}(s) = \sum_{n\geq0} \frac{t_{n}}{(n+1)^s} $$ for $ \rm{Re}(s)>1$. Are any values of $\zeta_{TM}(s)$ known? Is there some kind of closed form formula for this (I highly doubt there is, but one never knows).

A closely related sum for which a value is known is:

$$ \sum_{n\geq1} \frac{s_{n}}{n(n+1)} = 2\ln2, $$

where $s_n$ is the binary sum-of-digits function. Another related sum gives the Prouhet-Thue-Morse constant, which has been shown to be transcendental:

$$ \tau =\sum _{{n\geq0}}{\frac {t_{n}}{2^{{n+1}}}}=0.412454033640\ldots $$

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  • $\begingroup$ "but one never knows" you can disprove the existence of closed form formulas $\endgroup$ – mathworker21 May 14 at 9:27
  • $\begingroup$ Sure. What I meant by that was an informal "I don't know if such a formula exists in the vast ocean of published literature". $\endgroup$ – Klangen May 14 at 9:31
  • $\begingroup$ Related post on Mathematica.Stackexchange: mathematica.stackexchange.com/questions/198604/… $\endgroup$ – Klangen May 18 at 8:49
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There's probably no "closed form" for any $s>1$. Curiously, though, the sum can be evaluated for integers $s \leq 0$, in the following sense. The function $\zeta_{TM}$ extends to an analytic function on ${\bf C} \backslash \{ 1 \}$, with a simple pole at $s=1$ of residue $1/2$, and taking rational values at integers $s \leq 0$, starting $\zeta_{TM}(0) = -1/4$, $\zeta_{TM}(-1) = -1/24$, $\zeta_{TM}(-2) = 0$, $\zeta_{TM}(-3) = +1/240$, and in general $\zeta_{TM}(s) = \zeta(s) / 2$ for integers $s \leq 0$ (so in particular $\zeta_{TM}$ inherits the "trivial zeros" of $\zeta$ at $s = -2, -4, -6, \ldots$).

It is more convenient to work with the Dirichlet series whose $(n+1)^{-s}$ coefficient is $1 - 2 t_n = (-1)^{t_n}$, because the generating function for $(-1)^{t_n}$, call it $$ T(z) = \sum_{n=0}^\infty (-1)^{t_n} z^n, $$ factors as an infinite product: $$ T(z) = (1-z) (1-z^2) (1-z^4) (1-z^8) \cdots = \prod_{m=0}^\infty \bigl( 1 - z^{2^m} \bigr). $$ So define $$ Z_{TM}(s) = \zeta(s) - 2 \zeta_{TM}(s) = \sum_{n=0}^\infty \frac{(-1)^{t_n}}{(n+1)^s}. $$ The usual Mellin-transform trick gives an integral formula: $$ \Gamma(s) Z_{TM}(s) = \sum_{n=0}^\infty (-1)^{t_n} \! \int_0^\infty x^{s-1} e^{-(n+1)x} \, dx = \int_0^\infty x^{s-1} e^{-x} T(e^{-x}) \, dx. $$ This gives an analytic continuation of $\Gamma(s) Z_{TM}(s)$ to the entire complex plane, because $T(e^{-x})$ decays faster than any power of $x$ as $x \to 0$: each factor $1 - e^{-2^m x}$ of the infinite product is $O_m(x)$ and in $(0,1)$. Since $\Gamma(s)$ has no zeros, but does have simple poles at $s = 0, -1, -2, -3, \ldots$, it follows that $Z_{TM}$ is an entire function with simple zeros at the same $s$, and no other real zeros (the integral for $\Gamma(s) Z_{TM}(s)$ is plainly positive for all real $s$). Since $Z_{TM} = \zeta - 2 \zeta_{TM}$, we conclude that $\zeta_{TM}(s) = \frac12 \zeta(s)$ at those $s$, as claimed.

[The integral formula can also be used to compute $Z_{TM}(s)$, and thus also $\zeta_{TM}(s)$, to high precision; for example, using gp's "intnum" function we find $Z_{TM}(2) = 0.6931534522\ldots$ (this is not $\log 2$, though it's quite close -- the difference is $\lt 10^{-5}$), so $\zeta_{TM}(2) = (\zeta(2) - Z_{TM}(2)) / 2 = 0.4758903073\ldots$.]

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  • $\begingroup$ Fantastic answer, thank you! $\endgroup$ – Klangen May 19 at 9:12

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