I believe that the circulant Hadamard conjecture (that there are no circulant Hadamard matrices of size greater than $4\times4$) is still open.

I also know that examples of $(n/2) \times n$ matrices which are partial Hadamard circulant have been found experimentally for moderate values of $n.$

To clarify, a partial circulant $m(n)\times n$ Hadamard matrix $A$ has entries from $\{-1,+1\}$ and if its first row is $a,$ then its subsequent rows are $T(a),\ldots,T^{m(n)-1}(a)$ where $T$ denotes, say, the left cyclic shift operator and $T^k$ is $T$ composed with itself $k$ times.

Is there a known infinite construction for partial circulant Hadamard matrices? Equivalently does a sequence of $m_k(n_k)\times n_k$ Hadamard matrices exist, where $n_k\rightarrow \infty$ and so does $m_k(n_k)$. Note that $m_k(n_k)=o(n_k)$ is allowed, under this definition.

I am aware of references arXiv:1201.4021, Armario et al and arxiv:1003.4003, De Launey and Levin, which address generic partial Hadamard matrices, as opposed to those that are partial circulant.

Edit: To make things explicit, the best results I have found without the circulant constraint are the following: For any $\varepsilon>0$ and for $n$ large enough, $t\equiv 0~(mod~4)$ there is a $n\times t$ partial Hadamard matrix if $n\leq \frac{t}{2}-t^{\frac{113}{132}+\varepsilon}.$ Subject to the extended Riemann hypothesis the $\frac{113}{132}$ in the exponent can be replaced by $\frac{7}{12}.$ This matches computational results, see http://codegolf.stackexchange.com/questions/55726/an-optimization-version-of-the-hadamard-problem, that have found $\frac{n}{2}\times n$ partial Hadamard matrices up to $n$ somewhere between 50 and 100.

I believe that the circulant Hadamard conjecture (that there are no circulant Hadamard matrices of size greater than $4\times4$) is still open.

I also know that examples of $(n/2) \times n$ matrices which are partial Hadamard circulant have been found experimentally for moderate values of $n.$

To clarify, a partial circulant $m(n)\times n$ Hadamard matrix $A$ has entries from $\{-1,+1\}$ and if its first row is $a,$ then its subsequent rows are $T(a),\ldots,T^{m(n)-1}(a)$ where $T$ denotes, say, the left cyclic shift operator and $T^k$ is $T$ composed with itself $k$ times.

Is there a known infinite construction for partial circulant Hadamard matrices? Equivalently does a sequence of $m_k(n_k)\times n_k$ Hadamard matrices exist, where $n_k\rightarrow \infty$ and so does $m_k(n_k)$. Note that $m_k(n_k)=o(n_k)$ is allowed, under this definition.

I am aware of references arXiv:1201.4021, Armario et al and arxiv:1003.4003, De Launey and Levin, which address generic partial Hadamard matrices, as opposed to those that are partial circulant.

Edit: To make things explicit, the best results I have found without the circulant constraint are the following: For any $\varepsilon>0$ and for $n$ large enough, $t\equiv 0~(mod~4)$ there is a $n\times t$ partial Hadamard matrix if $n\leq \frac{t}{2}-t^{\frac{113}{132}+\varepsilon}.$ Subject to the extended Riemann hypothesis the $\frac{113}{132}$ in the exponent can be replaced by $\frac{7}{12}.$ This matches computational results, see https://codegolf.stackexchange.com/questions/55726/an-optimization-version-of-the-hadamard-problem, that have found $\frac{n}{2}\times n$ partial Hadamard matrices up to $n$ somewhere between 50 and 100.

I believe that the circulant Hadamard conjecture (that there are no circulant Hadamard matrices of size greater than $4\times4$) is still open.

I also know that examples of $(n/2) \times n$ matrices which are partial Hadamard circulant have been found experimentally for moderate values of $n.$

To clarify, a partial circulant $m(n)\times n$ Hadamard matrix $A$ has entries from $\{-1,+1\}$ and if its first row is $a,$ then its subsequent rows are $T(a),\ldots,T^{m(n)-1}(a)$ where $T$ denotes, say, the left cyclic shift operator and $T^k$ is $T$ composed with itself $k$ times.

Is there a known infinite construction for partial circulant Hadamard matrices? Equivalently does a sequence of $m_k(n_k)\times n_k$ Hadamard matrices exist, where $n_k\rightarrow \infty$ and so does $m_k(n_k)$. Note that $m_k(n_k)=o(n_k)$ is allowed, under this definition.

I am aware of references here and here, which address generic partial Hadamard matrices, as opposed to those that are partial circulant.

Edit: To make things explicit, the best results I have found without the circulant constraint are the following: For any $\varepsilon>0$ and for $n$ large enough, $t\equiv 0~(mod~4)$ there is a $n\times t$ partial Hadamard matrix if $n\leq \frac{t}{2}-t^{\frac{113}{132}+\varepsilon}.$ Subject to the extended Riemann hypothesis the $\frac{113}{132}$ in the exponent can be replaced by $\frac{7}{12}.$ This matches computational results, see here, that have found $\frac{n}{2}\times n$ partial Hadamard matrices up to $n$ somewhere between 50 and 100.

I believe that the circulant Hadamard conjecture (that there are no circulant Hadamard matrices of size greater than $4\times4$) is still open.

I also know that examples of $(n/2) \times n$ matrices which are partial Hadamard circulant have been found experimentally for moderate values of $n.$

To clarify, a partial circulant $m(n)\times n$ Hadamard matrix $A$ has entries from $\{-1,+1\}$ and if its first row is $a,$ then its subsequent rows are $T(a),\ldots,T^{m(n)-1}(a)$ where $T$ denotes, say, the left cyclic shift operator and $T^k$ is $T$ composed with itself $k$ times.

Is there a known infinite construction for partial circulant Hadamard matrices? Equivalently does a sequence of $m_k(n_k)\times n_k$ Hadamard matrices exist, where $n_k\rightarrow \infty$ and so does $m_k(n_k)$. Note that $m_k(n_k)=o(n_k)$ is allowed, under this definition.

I am aware of references arXiv:1201.4021, Armario et al and arxiv:1003.4003, De Launey and Levin, which address generic partial Hadamard matrices, as opposed to those that are partial circulant.

Edit: To make things explicit, the best results I have found without the circulant constraint are the following: For any $\varepsilon>0$ and for $n$ large enough, $t\equiv 0~(mod~4)$ there is a $n\times t$ partial Hadamard matrix if $n\leq \frac{t}{2}-t^{\frac{113}{132}+\varepsilon}.$ Subject to the extended Riemann hypothesis the $\frac{113}{132}$ in the exponent can be replaced by $\frac{7}{12}.$ This matches computational results, see http://codegolf.stackexchange.com/questions/55726/an-optimization-version-of-the-hadamard-problem, that have found $\frac{n}{2}\times n$ partial Hadamard matrices up to $n$ somewhere between 50 and 100.