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:<math>
f\left(\frac n{2^m}\right)=\frac{1}{2}-\frac1{2^{m^2+1}\,\left(\frac{1}{2};\frac{1}{2}\right)_m}\,\sum_{k=0}^m\,\frac{\left[\begin{smallmatrix}m\\[3pt] k\end{smallmatrix}\right]_{1/2}}{2^{k\,(k-1)}\,(m+k)!}\sum^{2^{m+k}-1}_{j=0}(-1)^{s_2(j)}\,\left(j-2^k\,n+\tfrac{1}{2}\right)^{m+k}\sgn\left(j-2^k\,n+\tfrac{1}{2}\right),</math>
where <math>n,\,m,\,k,\,j</math> are non-negative integers (<math>n</math>(0\le is odd, <math>0<n<2\le2^m)</math>), <math>\sgn(\cdot)</math> is the [[sign function]], <math>\left(a;q\right)_m</math> is the [[q-Pochhammer symbol]], <math>\left[\begin{smallmatrix}m\\[3pt] k\end{smallmatrix}\right]_q</math> is the [[Gaussian binomial coefficient]], and <math>s_2(j)</math> is the [[Digit sum|sum of digits]] of ''j'' in [[base-2]]. Note that <math>(-1)^{s_2(j)}=f(2j+1)=t_j</math> is just the signed [[Thue–Morse sequence]], satisfying the recurrence <math>t_0=1,\,\,t_j=(-1)^j\,t_{\lfloor j/2\rfloor}.</math>
==References==
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