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Here is an attempt at analyzing this problem. It seems better to adjust the index of the sequence, so let $\;A_n = b_k \mbox{ if } n=2^k,\; 0\;$ otherwise. Let $\;c_n\; := \det \left( {{A_{i + j}}} \right)_{i,j = 1}^n.\;$ In this determinant as a sum over all permutations, only one permutation in the sum corresponds to a nonzero product of members of the sequence $\{b_k\}.\;$ That is, $c_n = \textrm{sgn}(\sigma(n))\prod_{i=1}^k b_{\sigma(n)_i}\;$ where $\;\sigma(n)\;$ is a permutation of $\;\{1,\dots,n\}.\;$ The triangluar array of $\;\{\sigma(n)\}_1^\infty\;$ has a peculiar recursive structure of triangular arrays next to other smaller triangular arrays. In our case, $\;b_1=1,\; b_k=(-1)^k,\;$ for $k>1.\;$ If we know the signature of all $\;\sigma(n),\;$ then the problem is essentially solved.

Here is an attempt at analyzing this problem. It seems better to adjust the index of the sequence, so let $\;A_n = b_k \mbox{ if } n=2^k,\; 0\;$ otherwise. Let $\;c_n\; := \det \left( {{A_{i + j}}} \right)_{i,j = 1}^n.\;$ In this determinant as a sum over all permutations, only one permutation in the sum corresponds to a nonzero product of members of the sequence $\{b_k\}.\;$ That is, $c_n = \textrm{sgn}(\sigma(n))\prod_{i=1}^k A_{i+\sigma(n)_i}\;$ where $\;\sigma(n)\;$ is a permutation of $\;\{1,\dots,n\}.\;$ The triangluar array of $\;\{\sigma(n)\}_1^\infty\;$ has a peculiar recursive structure of triangular arrays next to other smaller triangular arrays. In our case, $\;b_1=1,\; b_k=(-1)^k,\;$ for $k>1.\;$ If we know the signature of all $\;\sigma(n),\;$ then the problem is essentially solved.

Here is an attempt at analyzing this problem. It seems better to adjust the index of the sequence, so let $\;A_n = b_k \mbox{ if } n=2^k,\; 0\;$ otherwise. Let $\;c_n\; := \det \left( {{A_{i + j}}} \right)_{i,j = 1}^n.\;$ In this determinant as a sum over all permutations, only one permutation in the sum corresponds to a nonzero product of members of the sequence $\{b_k\}.\;$ That is, $c_n = \textrm{sgn}(\sigma(n))\prod_{i=1}^k b_{\sigma(n)_i}\;$ where $\;\sigma(n)\;$ is a permutation of $\;\{1,\dots,n\}.\;$ The triangluar array $\;\sigma(n)\;$ has a peculiar recursive structure of triangular arrays next to other smaller triangular arrays. In our case, $\;b_1=1,\; b_k=(-1)^k,\;$ for $k>1.\;$ If we know the signature of all $\;\sigma(n),\;$ then the problem is essentially solved.

Here is an attempt at analyzing this problem. It seems better to adjust the index of the sequence, so let $\;A_n = b_k \mbox{ if } n=2^k,\; 0\;$ otherwise. Let $\;c_n\; := \det \left( {{A_{i + j}}} \right)_{i,j = 1}^n.\;$ In this determinant as a sum over all permutations, only one permutation in the sum corresponds to a nonzero product of members of the sequence $\{b_k\}.\;$ That is, $c_n = \textrm{sgn}(\sigma(n))\prod_{i=1}^k b_{\sigma(n)_i}\;$ where $\;\sigma(n)\;$ is a permutation of $\;\{1,\dots,n\}.\;$ The triangluar array of $\;\{\sigma(n)\}_1^\infty\;$ has a peculiar recursive structure of triangular arrays next to other smaller triangular arrays. In our case, $\;b_1=1,\; b_k=(-1)^k,\;$ for $k>1.\;$ If we know the signature of all $\;\sigma(n),\;$ then the problem is essentially solved.

Here is an attempt at analyzing this problem. It seems better to adjust the index of the sequence, so let $\;A_n = b_k \mbox{ if } n=2^k,\; 0\;$ otherwise. Let $\;c_n\; := \det \left( {{A_{i + j}}} \right)_{i,j = 1}^n.\;$ In this determinant as a sum over all permutations, only one permutation in the sum corresponds to a nonzero product of members of the sequence $\{b_k\}.\;$ That is, $c_n = \textrm{sgn}(\sigma(n))\prod_{i=1}^k b_{\sigma(n)_i}\;$ where $\;\sigma(n)\;$ is a permutation of $\;\{1,\dots,n\}\;$ which has a peculiar recursive structure of triangular arrays next to other smaller trianguular arrays. In our case, $\;b_1=1,\; b_k=(-1)^k, k>1.\;$ If we know the signature of $\;\sigma(n)\;$ then the problem is essentially solved.

Here is an attempt at analyzing this problem. It seems better to adjust the index of the sequence, so let $\;A_n = b_k \mbox{ if } n=2^k,\; 0\;$ otherwise. Let $\;c_n\; := \det \left( {{A_{i + j}}} \right)_{i,j = 1}^n.\;$ In this determinant as a sum over all permutations, only one permutation in the sum corresponds to a nonzero product of members of the sequence $\{b_k\}.\;$ That is, $c_n = \textrm{sgn}(\sigma(n))\prod_{i=1}^k b_{\sigma(n)_i}\;$ where   $\;\sigma(n)\;$ is a permutation of $\;\{1,\dots,n\}.\;$ The triangluar array $\;\sigma(n)\;$ has a peculiar recursive structure of triangular arrays next to other smaller triangular arrays. In our case, $\;b_1=1,\; b_k=(-1)^k,\;$ for $k>1.\;$ If we know the signature of all $\;\sigma(n),\;$ then the problem is essentially solved.