Let $(a_n)$ be a sequence. We define the sequence $(b_n)$ to be $b_n=\det(A)$ where $A$ is a square matrix of order $n$ such that $$A_{1,1}=a_1,\dots,A_{1,n}=a_n \\ \vdots\\ A_{n,1}=a_{n^2-n+1},\dots,A_{n,n}=a_{n^2} .$$
I conjecture that if $(a_n)$ converges to a real number (or maybe even bounded) then $(b_n)$ converges to zero. The reason for this conjecture is simple. It seems to be the "limit version" of the fact that if you have a square matrix with all the entries being the same then the determinant is zero.
Any thoughts?
There exists the following counterexample for the conjecture. Let $a_1=a_2=1$ and $n(1)=0$. Suppose that we already constructed the members $a_1,\dots,a_{2^{n(k)}+1}$ for some natural $k$. Let these members constitute the first row of the square matrix. Fill all entries of the first column of the matrix but the first by zeroes. Finally, fill its bottom right corner by any Hadamard_matrix of order $2^{n(k)}$, multiplied by $2^{-n(k)/2}$. Then the matrix will have the determinant $1$. Finally, pick any $n(k+1)$ such that $2^{n(k)}+1$ is not smaller than the number $N$ of already determined members of the sequence $(a_n)$ and put $a_{N+1}=\dots =a_{2^{n(k+1)}+1}=0$.