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I am interested in the continued fraction $$\sum\limits_k {{z^{{2^k} - 1}}} = \frac{1}{{1 - \frac{{{T_0}z}}{{1 - \frac{{{T_1}z}}{{1 - \frac{{{T_2}z}}{{1 -{ \ddots }}}}}}}}}.$$

OEIS A104977 states without proof that ${T_k} = {( - 1)^{b(k + 2) + 1}}$ (see A090678) if $b(n)$ denotes the number of non-squashing partitions of $n$ into distinct parts (see A088567).

I obtained the sequence ${T_k}$ with another method which has nothing to do with partitions, but I would be interested to learn why there is a connection with those partitions. Can anyone help me?

Edit: Let me give some more information about the sequence $T_n.$
Let $$D(n) = \det \left( {{a_{i + j}}} \right)_{i,j = 0}^{n - 1}$$ be a Hankel determinant of the sequence $(a_n), $ where $a_n=1$ if $n+2$ is a power of $2$ and $a_n=0$ else.

Then $$ T_n=D(n)D(n+2).$$ These numbers satisfy $T_{2n}=T_{2n-1}T_{n-1}$ and $T_{2n+1}=-T_{2n}$with initial values $T_0=1$ and $T_1=-1.$

The non-squashing partitions with distinct parts $b(n)$ satisfy $b(2n)=b(2n-1)+b(n)-1$ and $b(2n+1)=b(2n)+1.$

Thus a posteriori it can be observed that $${T_k} = {( - 1)^{b(k + 2) + 1}}.$$
Question: Is this a happy coincidence or is there really a connection between the Hankel determinants and the partitions.

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    $\begingroup$ oeis.org/A090678 links to two papers on "non-squashing partitions". Was there nothing useful at those links? $\endgroup$
    – Gerry Myerson
    Commented Feb 5, 2018 at 22:39
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    $\begingroup$ A088567 gives more links. $\endgroup$
    – Alexey Ustinov
    Commented Feb 6, 2018 at 6:01
  • $\begingroup$ Since $z^{2^k-1}=z^{2^0}z^{2^1}\dots z^{2^{k-1}}$, there is another immediate continued fraction given by the Euler continued fraction formula en.wikipedia.org/wiki/… $\endgroup$
    – Pietro Majer
    Commented Feb 6, 2018 at 9:02
  • $\begingroup$ @Pietro Majer: Thank you for this information. But I am only interested in this specific continued fraction which is related to the Hankel determinants of the sequence $(a_n),$ where $a_n=1$ if $n+2$ is a power of $2$ and $a_n=0$ else. $\endgroup$
    – Johann Cigler
    Commented Feb 6, 2018 at 12:05
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    $\begingroup$ @Gerry Myerson: Sorry for the typo in your name. But the website has not allowed me to edit my comment. $\endgroup$
    – Johann Cigler
    Commented Feb 6, 2018 at 12:30

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