In my answer to this question, I have obtained that the polynomial $p(x)$ of degree $2n$ with nonnegative values on $[-1,1]$ with $p(\pm1)=1$ has $\int_{-1}^1 p(x)\,dx\geq \frac{4}{(n+1)(n+2)}$, and the optimal polynomials are squares of $$ q_n(x)=\left(\sum_{i\leq n/2}\frac{2}{2(n-2i)+1}\right)^{-1} \sum_{i\leq n/2}\frac{2P_{n-2i}(x)}{2(n-2i)+1}, $$ where $P_k(x)=\frac1{2^kk!}\frac{d^k}{dx^k}(x^2-1)^k$ are the Legendre polynomials. (One may easily see that these optimal polynomials are unique.)
On the other hand, numerical experiments by Robert Israel (summarized in his answer to the same question) make it evident that the coefficients of $q_n$ are $2^{-n}$ times the elements of the Borel triangle http://oeis.org/A234950 read diagonally with alternating signs (see http://oeis.org/A062991 for the signed version), as was mentioned by Douglas Zare. Here are several first such polynomials, copied from Robert's answer: $$ \eqalign{ q_0(x) = 1, \qquad q_1(x) = x = \frac{2x}2, \qquad q_2(x) = \dfrac{5 x^2 - 1}{4}, \\ q_3(x) = \dfrac{7 x^3 - 3 x}4 = \dfrac{14 x^3 - 6 x}{8},\qquad\qquad\quad\\ q_4(x) = \dfrac{21 x^4 - 14 x^2 + 1}{8} = \dfrac{42 x^4 - 28 x^2 + 2}{16},\qquad\\ q_{5}(x) = \dfrac{33 x^4 - 30 x^2 + 5}{8} = \dfrac{132 x^4 - 120 x^2 + 20}{32}, \quad\\ q_{6}(x) = \dfrac{429 x^6 - 495 x^4 + 135 x^2 - 5}{64}.\qquad\qquad} $$
This looks a bit strange for me, at least due to a strange numerical factor in my form of the answer.
So, my question is: How can one prove (if true) that the polynomials $q_n(x)$ have these (quite good-looking) coefficients?
