Bernoulli Numbers and Tangent numbers.

Good evening. I am looking to see if there is a proof online to help guide me with the understanding that the Tangent Numbers, denoted $T_n$ and the Bernoulli numbers, denoted $B_n$ are related. It is known that

$$T_n=\frac{2^{2n}(2^{2n}-1)|B_{2n}|}{2n}$$

I know that Tangent Numbers and Euler numbers, denoted $E_n$, are related through the "zig zag" numbers, or alternating permutations of $n$ integers. For example, if $n=4$, we have a permutation $\sigma=(a_1, a_2, a_3, a_4)$ which is called "Zig Zag" or "Up Down" if

$$a_1<a_2>a_3<a_4$$

and for $n=4$, we have

$$(1,3,2,4),(1,4,2,3), (2,3,1,4),(2,4,1,3),(3,4,1,2)$$

These up down numbers are produced by the generating function $$\sec{(x)}+\tan{(x)}$$ and the even indexed coefficients are the Euler Numbers and the odd indexed coefficients are the Tangent Numbers. I know that the Bernoulli numbers are related to the Cotangent function but I'm not sure how to start or approach getting the Bernoulli numbers to relate to the Tangent function.

Solution 1:

(A bit to long for a comment.) Knuth and Buckholtz, Math. Comp. 21 (1967), 663-688 (http://www.ams.org/journals/mcom/1967-21-100/S0025-5718-1967-0221735-9/S0025-5718-1967-0221735-9.pdf) write the following $$\tan z=\frac{\sin z}{\cos z} = \frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})} =\frac{1}{z}\left(\frac{2iz}{e^{2iz}+1} -iz \right) =\frac{1}{z}\left(\frac{2iz}{e^{2iz}-1} - \frac{4iz}{e^{4iz}-1} -iz \right) $$

$$\tan z=\frac{1}{z}\left( -iz +\sum_{n\ge 0}\Big((2iz)^n-(4iz)^n\Big)\frac{B_n}{n!}\right)$$ From which they give the relation (without absolute values and a slightly different definition of $T_n$ via $\tan z = \sum_{n\ge 0} T_n z^n/n!)$ $$ B_n = -i^n \frac{nT_{n-1}}{2^n(2^n-1)}$$ If you use the generating function $$\tan z = \sum_{n=1}^{\infty} T_n \frac{z^{2n-1}}{(2n-1)!}$$ and absolute values you get your initial relation.

Solution 2:

The Bernoulli numbers are obtained thanks to the function $$\Phi(x) =\frac{x}{e^x-1} = \sum_{n=0}^{+\infty} B_n \frac{x^n}{n!}.$$ Note that $$\tan(x)=\frac{\sin(x)}{\cos(x)}=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}=\frac{e^{2ix}-1}{i(e^{2ix}+1)}.$$ I let you do some computations to get $$\tan(x) = \frac{1}{x}(\Phi(2ix)-\Phi(4ix)-ix).$$ Now use the first formula to get the Taylor expansion of $\tan$ thanks to Bernoulli numbers.