Search Results

To be unambiguous about the order of multiplication, let $B(n) = A_1 A_2 \cdots A_n$. We have the D-finite recurrences $B(n)_{r,1} = (\frac{n}{n-1})^k B(n-1)_{r,1} + n^k B(n-2)_{r,1}$ $B(n)_{r,2} = B …

It's not entirely clear to me how much data your guesses are based on, so I present a table with calculated data and guessed polynomials based on that data and the assumption that $f(1) = f(-1) = 1$. …

The solutions look like a mess, so it's not too surprising that you always end up with one. If we follow Iosif Pinelis in dividing all by the last constraint by $\lambda$ and substituting $r = \frac{1 …

$\det \begin{pmatrix} 1 \end{pmatrix} = 1$ works for any $p$. $\det \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} = -1$ similarly. For $n=3$ we require $p \ge 5$. By exhaustion there's no solution for …

You're trying to solve $$(3a_xt^2 + 2b_xt + c_x)(6a_yt + 2b_y) - (6a_xt + 2b_x)(3a_yt^2 + 2b_yt + c_y) = 0$$ Expanded out, $$6(a_y b_x - a_x b_y) t^2 + 6(a_y c_x - a_x c_y)t + 2(b_y c_x - b_x c_y) = 0 …

Disclaimer: this is only a partial answer. If $S = \{x, y\}$ (considered as variables), the determinant must be a polynomial with integer coefficients and constant coefficient 1. Therefore by Gauss's …

This is only empirical observation, but I was requested to post it as an answer rather than merely a comment. Define $b(n) = \frac{\det(A_n)}{n! \, \sigma(\operatorname{lcm}(1,\ldots,n))}$ for $n \ge …

OP asked me to fill in the details of my comment, and in attempting to do so I realised that I claimed too much. However, a very similar argument proves a weaker result which is strong enough to suppo …