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4 added 113 characters in body; edited tags; deleted 6 characters in body

Please prove, give more symbolic or numeric support (counterexample!?), simplify or drop me a reference (or some vague hunch).

We have

$$B_n = n!\sum_r n!\sum_{\lambda} \frac{r^{2n}}{p'_n(r)}$$ frac{\lambda^{2n}}{p'_n(\lambda)}$$where r \lambda ranges over the roots of the polynomial p_n and p_n' is the derivative of p_n. B_n is the n-th bernoulli number. The polynomial p_n is defined as follows.$$p_n(x):=x^{n+2}t_n(1/x)$$where$$t_n(x) = \sum_{k=0}^{n+2} \frac{x^k}{k!} -1$$is the truncated taylor polynomial of \exp-1 to power n+2. Then p_n(x):=x^{n+2}t_n(1/x) - like p_n is the reciprocal polynomial of t_n just without conjugation (though it probably makes no difference if there is conjugation or not - since p is a polynomial with real coefficients and the roots come in conjugate pairs). Examples (already tested for n=0..18 with a symbolic solver and for n=0..62 numerically): n=0$$t_0(x)=\frac{x^2}{2}+xp_0(x)=x^2t(1/x)=x^2(\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{2}+x$$root of p_0 is -1/2.$$p_0'(x)=1B_0=0!\cdot \frac{(-1/2)^{2\cdot 0}}{1}=1$$n=1$$t_1(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x=\frac{x^3}{6}+\frac{x^2}{2}+xp_1(x)=x^3t_1(1/x)=x^3(\frac{1}{6x^3}+\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{6}+\frac{x}{2}+x^2$$Roots of p_1 are$$r_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt ${\lambda}_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$We have$$r_{1,2}^2=-\frac{1}{24}\frac{-}{+}\frac{\sqrt{15}i}{24}$$\lambda_{1,2}^2=-\frac{1}{24}\frac{-}{+}\frac{\sqrt{15}i}{24}$$ $$p_1'(x)=2x+\frac{1}{2}$$so $$\frac{1}{p_1'(r_{1,2})}=\frac{-}{+}\frac{2 \frac{1}{p_1'(\lambda_{1,2})}=\frac{-}{+}\frac{2 \sqrt{15}i}{5}$$ $$\frac{r_{1,2}^2}{p_{1}'(r_{1,2})}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt{15}i}{60}$$$\frac{\lambda_{1,2}^2}{p_{1}'(\lambda_{1,2})}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt{15}i}{60}B_1=1!(\frac{r_{1}^2}{p'_{1}(r_1)}+\frac{r_{2}^2}{p'_{2}(r_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$B_1=1!(\frac{\lambda_{1}^2}{p'_{1}(\lambda_1)}+\frac{\lambda_{2}^2}{p'_{2}(\lambda_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$

3 edited body

Please prove, give more symbolic or numeric support (counterexample!?), simplify or drop me a reference (or some vague hunch).

We have

$$B_n = n!\sum_r \frac{r^{2n}}{p_n'(r)}$$ frac{r^{2n}}{p'_n(r)}$$where r ranges over the roots of the polynomial p_n and p_n' is the derivative of p_n. B_n is the n-th bernoulli number. The polynomial p_n is defined as follows.$$t_n(x) = \sum_{k=0}^{n+2} \frac{x^k}{k!} -1$$is the truncated taylor polynomial of \exp-1 to power n+2. Then p_n(x):=x^{n+2}t_n(1/x) - like the reciprocal polynomial just without conjugation (though it probably makes no difference if there is conjugation or not - since p is a polynomial with real coefficients and the roots come in conjugate pairs). Examples (already tested for n=0..18 with a symbolic solver and for n=0..62 numerically): n=0$$t_0(x)=\frac{x^2}{2}+xp_0(x)=x^2t(1/x)=x^2(\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{2}+x$$root of p_0 is -1/2.$$p_0'(x)=1B_0=0!\cdot \frac{(-1/2)^{2\cdot 0}}{1}=1$$n=1$$t_1(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x=\frac{x^3}{6}+\frac{x^2}{2}+xp_1(x)=x^3t_1(1/x)=x^3(\frac{1}{6x^3}+\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{6}+\frac{x}{2}+x^2$$Roots of p_1 are$$r_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$We have$$r_{1,2}^2=-\frac{1}{24}\frac{-}{+}\frac{\sqrt{15}i}{24}p_1'(x)=2x+\frac{1}{2}$$so$$\frac{1}{p_1'(r_{1,2})}=\frac{-}{+}\frac{2 \sqrt{15}i}{5}\frac{r_{1,2}^2}{p_{1}'(r_{1,2})}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt{15}i}{60}B_1=1!(\frac{r_{1}^2}{p'_{1}(r_1)}+\frac{r_{2}^2}{p'_{2}(r_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$2 edited tags; edited body Please prove, give more symbolic or numeric support (counterexample!?), simplify or drop me a reference (or some vague hunch). We have$$B_n = n!\sum_r \frac{r^{2n}}{p_n'(r)}$$where r ranges over the roots of the polynomial p_n and p_n' is the derivative of p_n. B_n is the n-th bernoulli number. The polynomial p_n is defined as follows.$$t_n(x) = \sum_{k=0}^{n+2} \frac{x^k}{k!} -1$$is the truncated taylor polynomial of \exp-1 to power n+2. Then p_n(x):=x^{n+2}t_n(1/x) - like the reciprocal polynomial just without conjugation (though it probably makes no difference if there is conjugation or not - since p is a polynomial with real coefficients and the roots come in conjugate pairs). Examples (already tested for n=0..18 with a symbolic solver and for n=0..62 numerically): n=0$$t_0(x)=\frac{x^2}{2}+xp_0(x)=x^2t(1/x)=x^2(\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{2}+x$$root of p_0 is -1/2.$$p_0'(x)=1B_0=0!\cdot \frac{(-1/2)^{2\cdot 0}}{1}=1$$n=1$$t_1(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x=\frac{x^3}{6}+\frac{x^2}{2}+xp_1(x)=x^3t_1(1/x)=x^3(\frac{1}{6x^3}+\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{6}+\frac{x}{2}+x^2$$Roots of p_1 are$$r_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$We have$$r_{1,2}^2=-\frac{1}{24}\frac{-}{+}\frac{\sqrt{15}i}{24}p_1'(x)=2x+\frac{1}{2}$$so$$\frac{1}{p_1'(r_{1,2})}=\frac{+}{-}\frac{2 \$\frac{1}{p_1'(r_{1,2})}=\frac{-}{+}\frac{2 \sqrt{15}i}{5}\frac{r_{1,2}^2}{p_{1}'(r_{1,2})}=-\frac{1}{4}\frac{-}{+}\frac{\sqrt{15}i}{60}$$\frac{r_{1,2}^2}{p_{1}'(r_{1,2})}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt{15}i}{60}$$

$$B_1=1!(\frac{r_{1}^2}{p'_{1}(r_1)}+\frac{r_{2}^2}{p'_{2}(r_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$

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