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Let $a=\sum_{n=1}^\infty(1 \mod \phi^{-n})2^{n/3},$ where $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio and $1 \mod x=1-x\lfloor x^{-1}\rfloor$.

How can I prove that $a$ satisfies the equation $2255 a^6 - 2340 a^5 - 3174 a^4 - 672 a^3 + 180 a^2 + 36 a - 36=0?$

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    @GerryMyerson The only reason was that numerical approximation suggested so (I used$a$computer program to guess the coefficients of the polynomial)2012-01-21

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The following approach should work, although the details are messy. Let $\psi = (1 - \sqrt{5})/2 = -1/\phi$ be the Galois conjugate of $\phi$. Then the quantity $b_n = \phi^n + \psi^{n}$ is an integer, because it's both an algebraic integer and invariant under the Galois automorphism $\sqrt{5} \mapsto -\sqrt{5}$. But $|\psi|<1$, and $\psi$ is negative, so that $\lfloor{\phi^n}\rfloor$ equals $b_n$ if $n$ is odd, or $b_n - 1$ if $n$ is even. Hence we have $(1\,\mathrm{ mod }\,\phi^{-n}) = 1 - \frac{\lfloor{\phi^n}\rfloor}{\phi^n} = 1 - \frac{\phi^n + \psi^n - e_n}{\phi^n} = \frac{-\psi^n+ e_n}{\phi^n}$ where $e_n$ is 0 if $n$ is odd, 1 if $n$ is even. Now that we've gotten rid of the floor, we can substitute this expression into the sum, which produces some geometric series which can easily be evaluated. The result is clearly an algebraic number in $\mathbb{Q}(\phi, 2^{1/3})$, and therefore has degree at most 6. Finding the exact minimal polynomial will be messy though.

Here are more details: Since $\psi^n = (-\phi)^{-n}$, the sum becomes $\sum_{n=1}^{\infty} -(-2^{1/3} \phi^{-2})^n + (2^{2/3} \phi^{-2})^n$ Both of the quantities in parentheses above have absolute value less than 1, so the two separate geometric series converge individually. Letting $u = -2^{1/3} \phi^{-2}$ and $v = 2^{2/3}\phi^{-2}$, the sum is equal to $a = \frac{v}{1-v} - \frac{u}{1-u} \approx 1.86511275789174$ which, at least numerically, appears to satisfy the polynomial given: $2255 a^6 - 2340 a^5 - 3174 a^4 - 672 a^3 + 180 a^2 + 36 a - 36 \approx -4.8 \cdot 10^{-12}.$ I also numerically calculated the Galois conjugates of $a$ and then used those values to compute the characteristic polynomial, which gives the above polynomial. (All calculations done in Sage.)

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    You don't need to compute the minimal polynomial; the question is to show that a polynomial of degree $6$ is satisfied, so it suffices to take the _characteristic polynomial_ of our algebraic number acting by left multiplication on $\mathbb{Q}(\phi, \sqrt[3]{2})$, and this reduces in principle to computing a determinant.2012-01-17